MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 9
Liberty University
Question 1
Question
Prove the following trigonometric identity:
cos2(x)−sin2(x) = cos(2x)
Solution
To prove the trigonometric identity cos2(x)−sin2(x) = cos(2x), we will use the
basic trigonometric identities.
Step 1: Recall the double-angle identity for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 2: Substitute the right-hand side of the identity into the left-hand
side:
cos2(x)−sin2(x) = cos(2x)
Thus, we have successfully proven the trigonometric identity cos2(x)−sin2(x) =
cos(2x).
Question 2
Question
Solve the equation sin(x) + 2 cos(x) = √3for xin the interval [0,2π].
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin(x) + 2 cos(x) = √3
⇒sin(x) + 2 cos(x) = 2√3(1
2)
⇒sin(x) + 2 cos(x) = 2√3 cos (π
6)
Step 2: Express the left-hand side as a single trigonometric function. Let’s
rewrite the left-hand side using the angle addition formula for sine:
sin(x) + 2 cos(x) = 2√3 cos (π
6)
= 2√3(cos(x) cos (π
6)−sin(x) sin (π
6))
= 2√3(cos(x)√3
2−sin(x)1
2)
=√3 cos(x)−sin(x)
Step 3: Replace sin(x) + 2 cos(x)with √3 cos(x)−sin(x)in the equation.
√3 cos(x)−sin(x) = 2√3 cos (π
6)
Step 4: Simplify the equation and solve for x.
√3 cos(x)−sin(x) = √3·2·√3
2= 3
⇒√3 cos(x)−sin(x) = 3
Step 5: Divide by √3to simplify the equation further.
cos(x)−1
√3sin(x) = 1
Step 6: Use the identity tan (π
6)=1
√3to rewrite the equation as a single
trigonometric function.
cos(x)−tan (π
6)sin(x) = 1
cos (x−π
6)= 1
Step 7: Solve for x. Since cosine is equal to 1 only at x= 0 in the interval
[0,2π], the solution to the equation is x= 0.
2
Question 3
Question
Solve the trigonometric equation sin(x) = cos(x)for xin the interval [0,2π).
Solution
Step 1: To solve the equation sin(x) = cos(x), we can rewrite it in terms of sines
and cosines:
sin(x) = cos(x) =⇒sin(x)−cos(x) = 0
Step 2: Using the trigonometric identity sin(x−π
4)= sin(π
2−x)= sin(π
4)cos(x)−
cos(π
4)sin(x), we rewrite sin(x)−cos(x)in the form of sin(x−π
4):
sin(x)−cos(x) = sin(x−π
4)= 0
Step 3: Since sin(x−π
4)= 0 if and only if x−π
4=nπ for some integer n,
we have:
x−π
4=nπ
Step 4: Solve for xby adding π
4to both sides:
x=nπ +π
4
Step 5: Since we are looking for solutions in the interval [0,2π), we need to
find the values of nsuch that xlies in this interval. Thus, we have:
0≤nπ +π
4<2π
Step 6: Set the upper bound:
nπ +π
4<2π
nπ < 7π
4
n < 7
4
Step 7: Since nmust be an integer, the possible values for nare 0and 1.
Step 8: Plug n= 0 and n= 1 back into the equation x=nπ +π
4to find
the solutions in the interval [0,2π): For n= 0:x= 0 + π
4=π
4For n= 1:
x=π+π
4=5π
4
Therefore, the solutions to the equation sin(x) = cos(x)in the interval [0,2π)
are x=π
4and x=5π
4.
3
Question 4
Question
Solve the equation cos2(x)−3 cos(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s first rewrite the equation by substituting cos(x)with u:
We have u2−3u+ 2 = 0.
Step 2: Next, we need to solve this quadratic equation. Factoring gives:
u2−3u+ 2 = (u−1)(u−2) = 0.
Step 3: Setting each factor to zero, we find:
u−1 = 0 or u−2 = 0
Step 4: Solving these equations, we get:
u= 1 or u= 2
Step 5: Now, we will substitute back cos(x)for uin our solutions:
cos(x) = 1 or cos(x) = 2
Step 6: Since cos(x)can only take values between -1 and 1, the equation
cos(x) = 2 has no solutions.
Step 7: Therefore, the only valid solution is cos(x) = 1.
Step 8: Recall that cos(x) = 1 when x= 0.
Step 9: Thus, the solution to the equation cos2(x)−3 cos(x) + 2 = 0 for
0≤x≤2πis x= 0.
Question 5
Question
Solve the trigonometric equation sin(x) cos(2x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Use the double angle identity cos(2x) = 2 cos2(x)−1to rewrite the
equation:
sin(x)(2 cos2(x)−1) = 1
2
4
Step 2: Expand the left side of the equation using the identity sin(x) cos2(x)−
sin(x) = 1
2:
2 sin(x) cos2(x)−sin(x) = 1
2
Step 3: Substitute cos2(x) = 1 −sin2(x)into the equation:
2 sin(x)−sin(x) = 1
2
Step 4: Simplify the equation:
sin(x) = 1
2
Step 5: The solutions to sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the original equation are x=π
6and x=5π
6.
Question 6
Question
Prove the trigonometric identity:
cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)=2
1−sin(θ)
Solution
Step 1: Start with the left-hand side of the given equation and work to simplify
it. cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)
Step 2: Find a common denominator for the two fractions.
cos2(θ) + (1 −sin2(θ))
cos(θ)(1 −sin(θ))
Step 3: Simplify the numerator using trigonometric identities.
cos2(θ) + cos2(θ)
cos(θ)(1 −sin(θ))
Step 4: Combine like terms in the numerator.
2 cos2(θ)
cos(θ)(1 −sin(θ))
5
Step 5: Simplify the expression by canceling out a factor of cos(θ).
2 cos(θ)
1−sin(θ)
Step 6: Finally, simplify the expression to match the right-hand side of the
given equation. 2
1−sin(θ)
Therefore, we have proven the given trigonometric identity:
cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)=2
1−sin(θ)
Question 7
Question
Solve the trigonometric equation cos2(x)−2 cos(x) + 1 = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the equation in terms of cos(x)as a quadratic equation by
using the substitution cos(x) = t.
cos2(x)−2 cos(x) + 1 = 0
t2−2t+ 1 = 0
Step 2: Solve the quadratic equation t2−2t+ 1 = 0 by factoring or using
the quadratic formula. The equation factors to (t−1)2= 0. So, t−1 = 0 which
implies t= 1.
Step 3: Substitute back cos(x) = tto find the possible solutions in the
interval [0,2π). Since cos(x) = 1,x= 0 and x= 2πare solutions in the interval
[0,2π).
Step 4: Thus, the solutions to the trigonometric equation cos2(x)−2 cos(x)+
1 = 0 in the interval [0,2π)are x= 0 and x= 2π.
Question 8
Question
Simplify the following expression:
tan(x) + sin(x)
1 + tan(x) sin(x)
6
Solution
Step 1: Recall the trigonometric identity tan(x) = sin(x)
cos(x).
Step 2: Substitute tan(x) = sin(x)
cos(x)and simplify the expression:
sin(x)
cos(x)+ sin(x)
1 + sin(x)
cos(x)sin(x)
Step 3: Combine the numerators:
sin(x) + sin(x) cos(x)
cos(x) + sin(x) cos(x)
Step 4: Factor out a sin(x)from the numerator and factor out a cos(x)from
the denominator: sin(x)(1 + cos(x))
cos(x)(1 + sin(x))
Step 5: Apply the trigonometric identity sin2(x) + cos2(x) = 1 to simplify
the expression further:
sin(x)(1 + cos(x))
cos(x)(1 + sin(x)) =sin(x)(1 + cos(x))
cos(x)(1 + sin(x)) ×1 + sin(x)
1 + sin(x)
Step 6: Expand the numerator and denominator:
sin(x) + sin(x) cos(x)
cos(x) + sin(x)
Therefore, the simplified expression is sin(x)+sin(x) cos(x)
cos(x)+sin(x).
Question 9
Question
Solve the trigonometric equation sin(3x) = cos(2x)for 0≤x≤2π.
Solution
Step 1: Recall the trigonometric identities: sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = cos2(x)−sin2(x).
Step 2: Substitute these identities into the equation sin(3x) = cos(2x)to get
3 sin(x)−4 sin3(x) = cos2(x)−sin2(x)
Step 3: Using the Pythagorean identity cos2(x) + sin2(x)=1, rewrite the
equation as 3 sin(x)−4 sin3(x) = 1 −2 sin2(x).
Step 4: Rearrange the equation to get 4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0.
Step 5: Factor the cubic equation to get (2 sin(x)−1)(2 sin2(x)+sin(x)−1) =
0.
Step 6: Solve for sin(x)in each factor:
7
• For 2 sin(x)−1 = 0, we get sin(x) = 1
2.
• For 2 sin2(x) + sin(x)−1=0, we can solve by setting u= sin(x)to get
2u2+u−1 = 0. The solutions are u=−1±√1+8
4, so either u=1
2or
u=−1.
Step 7: Solve for xin each case:
• For sin(x) = 1
2, the solutions in [0,2π]are x=π
6,5π
6.
• For sin(x) = 1
2, the solutions in [0,2π]are x=π
6,5π
6.
• For sin(x) = −1, the only solution in [0,2π]is x=3π
2.
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π]are x=π
6,5π
6,3π
2.
Question 10
Question
Solve the trigonometric equation √3 sin(x) cos(x) = 1
2for xin the interval
[0,2π).
Solution
Step 1: Use the double angle identity for sine to simplify the left side of the
equation:
√3 sin(x) cos(x) = √3
2sin(2x)
Step 2: Now the equation becomes √3
2sin(2x) = 1
2.
Step 3: Divide both sides by √3
2to solve for sin(2x):
sin(2x) = 1
√3
Step 4: To find the solutions for 2x, we need to consider all possible values of
sin(2x)in the interval [−1,1]. Since sin(π
6)=1
2and sin(5π
6)=1
2, the possible
solutions for 2xare π
6and 5π
6.
Step 5: Finally, solve for xby dividing the solutions of 2xby 2:
x=π
12,5π
12
Therefore, the solutions to the given trigonometric equation in the interval
[0,2π)are x=π
12 ,5π
12 .
8
Question 11
Question
Simplify the expression sin4(x)−cos4(x)
sin2(x)−cos2(x)using trigonometric identities.
Solution
Step 1: Recall the Pythagorean identity sin2(x) + cos2(x) = 1.
Step 2: Rewrite the expression using the Pythagorean identity:
sin4(x)−cos4(x)
sin2(x)−cos2(x)=(sin2(x) + cos2(x))(sin2(x)−cos2(x))
sin2(x)−cos2(x).
Step 3: Factor out the common term:
(sin2(x) + cos2(x))(sin2(x)−cos2(x))
sin2(x)−cos2(x)= sin2(x) + cos2(x).
Step 4: Apply the Pythagorean identity again to simplify:
sin2(x) + cos2(x) = 1.
Therefore, the simplified expression is 1.
Question 12
Question
Prove the trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)
Solution
To prove the given trigonometric identity, we will start from the left-hand side
and manipulate the expression until it simplifies to the right-hand side.
Step 1: Rewrite the left-hand side of the equation.
sin3(x)
cos(x)+cos3(x)
sin(x)
Step 2: Multiply the first term by sin(x)
sin(x)to get a common denominator.
sin4(x)
sin(x) cos(x)+cos3(x)
sin(x)
9
Step 3: Combine the terms by finding a common denominator.
sin4(x) + cos3(x) cos(x)
sin(x) cos(x)
Step 4: Apply trigonometric identity sin2(x) + cos2(x) = 1.
(1 −cos2(x))2+ cos3(x) cos(x)
sin(x) cos(x)
Step 5: Expand and simplify the numerator.
1−2 cos2(x) + cos4(x) + cos4(x)
sin(x) cos(x)
Step 6: Combine like terms in the numerator.
1−2 cos2(x) + 2 cos4(x)
sin(x) cos(x)
Step 7: Factor out a common factor in the numerator.
1−2 cos2(x) + 2 cos2(x) cos2(x)
sin(x) cos(x)
Step 8: Simplify the expression further.
1 + cos2(x)−2 cos2(x)
sin(x) cos(x)
=1−cos2(x)
sin(x) cos(x)
=sin2(x)
sin(x) cos(x)
=1
sin(x) cos(x)
Therefore, we have shown that the given trigonometric identity is true.
Question 13
Question
Prove the trigonometric identity: tan(θ) sec(θ) = sec(θ) + tan(θ).
10
Solution
Step 1: Start with the left-hand side of the identity.
tan(θ) sec(θ)
Step 2: Rewrite tan(θ)and sec(θ)in terms of sine and cosine.
tan(θ) = sin(θ)
cos(θ)and sec(θ) = 1
cos(θ)
Step 3: Substitute these values back into the left-hand side of the identity.
sin(θ)
cos(θ)·1
cos(θ)
Step 4: Multiply the fractions.
sin(θ)
cos(θ)·cos(θ)
Step 5: Simplify the expression.
sin(θ)
cos2(θ)
Step 6: Recall that tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=sin(θ)
cos(θ)·1
1= sec(θ)·tan(θ)
Step 7: Hence, the left-hand side is equal to sec(θ)·tan(θ).
Step 8: Now, simplify the right-hand side of the identity.
sec(θ) + tan(θ)
Step 9: Substitute the values for sec(θ)and tan(θ).
1
cos(θ)+sin(θ)
cos(θ)
Step 10: Combine the fractions.
1 + sin(θ)
cos(θ)
Step 11: Recall that sec(θ) = 1
cos(θ)and tan(θ) = sin(θ)
cos(θ).
Step 12: Hence, the right-hand side simplifies to sec(θ)·tan(θ).
Step 13: Therefore, the left-hand side of the identity is equal to the right-
hand side, and the trigonometric identity tan(θ) sec(θ) = sec(θ) + tan(θ)is
proven.
11
Question 14
Question
Solve the trigonometric equation sin2x−√3 sin x−1 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let u= sin x, then the equation becomes a quadratic equation in u:
u2−√3u−1 = 0
Step 2: Solve the quadratic equation u2−√3u−1 = 0 using the quadratic
formula:
u=−(−√3) ±√(−√3)2−4(1)(−1)
2(1)
u=√3±√7
2
Step 3: Since u= sin x, the solutions for xare obtained by finding the angles
whose sine is the solutions u=√3±√7
2.
x= sin−1(√3 + √7
2
,x= sin−1(√3−√7
2)
Step 4: Use a calculator to find the approximate values of the angles:
x≈70.53◦,229.47◦,19.11◦,160.89◦
Therefore, the solutions to the trigonometric equation sin2x−√3 sin x−1 = 0
for 0◦≤x≤360◦are x≈70.53◦,229.47◦,19.11◦,160.89◦.
Question 15
Question
Prove the identity:
sin(α−β) sin(α+β) = sin2α−sin2β
12
Solution
Step 1: Expand the left-hand side of the given identity using the angle differ-
ence identities:
sin(α−β) = sin αcos β−cos αsin β
sin(α+β) = sin αcos β+ cos αsin β
Step 2: Multiply the two expressions together:
sin(α−β) sin(α+β) = (sin αcos β−cos αsin β)(sin αcos β+ cos αsin β)
= sin2αcos2β−sin αcos βcos αsin β−cos αsin βsin αcos β−cos2αsin2β
Step 3: Simplify the expression:
sin2αcos2β−sin αcos βcos αsin β−cos αsin βsin αcos β−cos2αsin2β
= sin2α(cos2β−sin2β)−2 sin αcos βcos αsin β−cos2αsin2β
= sin2αcos(2β)−2 sin αcos αsin βcos β−cos2αsin2β
Step 4: Recall the double angle formula for cosine: cos(2θ) = 2 cos2θ−1
and substitute it into the expression:
= sin2α(2 cos2β−1) −2 sin αcos αsin βcos β−cos2αsin2β
= 2 sin2αcos2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
Step 5: Use the Pythagorean identity sin2θ+ cos2θ= 1 and rearrange
terms:
2 sin2αcos2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= 2(sin2α(1 −sin2β)) −sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= 2 sin2α−2 sin2αsin2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= sin2α−sin2αsin2β−2 sin αcos αsin βcos β−cos2αsin2β
Step 6: Recognize that the obtained expression is equal to sin2α−sin2β,
which completes the proof of the given identity. Thus, we have shown that:
sin(α−β) sin(α+β) = sin2α−sin2β
Question 16
Question
Prove the trigonometric identity: sin4(x)−cos4(x) = 1 −2 cos2(2x).
13
Solution
Step 1: Recall the double angle formula for cosine:
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Express cos(2x)in terms of sin(x)and cos(x):
cos(2x) = cos2(x)−sin2(x)
Step 3: Rearrange the above equation to express sin2(x)in terms of cos(2x)
and cos2(x):
sin2(x) = cos2(x)−cos(2x)
Step 4: Substitute the expression for sin2(x)into the given identity:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
(cos2(x)−cos(2x))2−cos4(x) = 1 −2 cos2(2x)
Step 5: Expand the left side of the equation:
cos4(x)−2 cos2(x) cos(2x) + cos2(2x)−cos4(x) = 1 −2 cos2(2x)
Step 6: Simplify the equation by canceling out cos4(x)terms:
−2 cos2(x) cos(2x) + cos2(2x) = 1 −2 cos2(2x)
Step 7: Further simplify the equation by combining like terms:
−2 cos2(x) cos(2x) = 1 −3 cos2(2x)
Step 8: Recall the double angle formula for cosine and use it to express
cos(2x)in terms of cos(x)and sin(x):
cos(2x) = 2 cos2(x)−1
Step 9: Substitute cos(2x) = 2 cos2(x)−1into the equation:
−2 cos2(x)(2 cos2(x)−1) = 1 −3(2 cos2(x)−1)
Step 10: Simplify the equation by expanding and combining like terms:
−4 cos4(x) + 2 cos2(x) = 1 −6 cos2(x)+3
−4 cos4(x) + 2 cos2(x) = −6 cos2(x)+4
Step 11: Finally, simplify the equation further to prove the identity:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
14
Question 17
Question
Solve the trigonometric equation tan(2x) + √3 = 0 for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation using the double angle identity for tangent:
tan(2x) = −√3
Step 2: Use the identity tan(2x) = 2 tan(x)
1−tan2(x)to rewrite the equation:
2 tan(x)
1−tan2(x)=−√3
Step 3: Let u= tan(x), so the equation becomes:
2u
1−u2=−√3
Step 4: Multiply both sides by 1−u2to get rid of the denominator:
2u=−√3(1 −u2)
Step 5: Expand and rearrange the equation:
2u=−√3 + √3u2
0 = √3u2−2u−√3
Step 6: Solve the quadratic equation √3u2−2u−√3 = 0 for u:
u=−(−2) ±√(−2)2−4(√3)(−√3)
2(√3)
u=2±√4 + 12
2√3
u=2±√16
2√3
u=2±4
2√3
Step 7: Solve for uto find the possible values of x: 1. When u=2+4
2√3=
6
2√3=3
√3=√3: Since u= tan(x), we have tan(x) = √3. This gives x=π
3.
2. When u=2−4
2√3=−2
2√3=−1
√3=−√3
3: Since u= tan(x), we have
tan(x) = −√3
3. This gives x=5π
6.
Therefore, the solutions to the equation are x=π
3and x=5π
6.
15
Question 18
Question
Solve the trigonometric equation tan2(x)−tan(x)−6=0for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the quadratic equation in terms of tan(x):
The equation is tan2(x)−tan(x)−6=0. We can treat this as a quadratic
equation in terms of tan(x). Let tan(x) = t. Then the equation becomes
t2−t−6 = 0.
Step 2: Factor the quadratic equation:
We need to factor the quadratic equation t2−t−6 = 0 to solve for t. The
factors of -6 that add up to -1 are -3 and 2. Therefore, we can rewrite the
equation as (t−3)(t+ 2) = 0.
Step 3: Find the roots of the quadratic equation:
Setting each factor to zero gives us t−3=0or t+ 2 = 0. Solving these
equations gives t= 3 or t=−2.
Step 4: Translate back to the original variable x:
Since tan(x) = t, we have tan(x)=3or tan(x) = −2. Recall that tan(x) =
sin(x)
cos(x). Therefore, sin(x)
cos(x)= 3 or sin(x)
cos(x)=−2.
Step 5: Solve for x:
For sin(x)
cos(x)= 3, we have sin(x) = 3 cos(x). Dividing both sides by cos(x), we
get tan(x) = 3. This gives x= arctan(3).
For sin(x)
cos(x)=−2, we have sin(x) = −2 cos(x). Dividing both sides by cos(x),
we get tan(x) = −2. This gives x= arctan(−2).
Thus, the solutions to the trigonometric equation tan2(x)−tan(x)−6=0
in the interval [0,2π)are x= arctan(3) and x= arctan(−2).
Question 19
Question
Prove the following trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x)
Solution
Step 1: Start with the identity cos(3x) = cos(2x+x).
Step 2: Use the angle addition formula for cosine: cos(a+b) = cos(a) cos(b)−
sin(a) sin(b). Applying this formula, we have:
cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x)
16
Step 3: Recall the double angle formulas for cosine and sine:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
Substitute these formulas into the expression from Step 2:
cos(2x) cos(x)−sin(2x) sin(x) = (cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x)
Step 4: Simplify the expression by expanding and combining like terms:
(cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x) = cos3(x)−sin2(x) cos(x)−2 sin(x) cos(x) sin(x)
Step 5: Recall the Pythagorean identity sin2(x) + cos2(x)=1. Rearrange
this to solve for sin2(x):sin2(x) = 1−cos2(x). Substitute this into the expression
in Step 4:
cos3(x)−(1 −cos2(x)) cos(x)−2 sin(x) cos(x) sin(x)
Step 6: Further simplify the expression:
cos3(x)−cos(x) + cos3(x)−2 sin(x) cos2(x)
Step 7: Use the Pythagorean identity to replace sin(x)in terms of cos(x):
sin(x) = √1−cos2(x). Substitute this into the expression in Step 6 and sim-
plify:
2 cos3(x)−cos(x)−2√1−cos2(x) cos2(x)
Step 8: Simplify further to get the desired result:
cos(3x) = 4 cos3(x)−3 cos(x)
Therefore, the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x)is proven.
Question 20
Question
Prove the following trigonometric identity:
sin(α+β) sin(α−β) = sin2α−sin2β
Solution
To prove this trigonometric identity, we will use the sum-to-product identities
and the Pythagorean identity.
Step 1: Apply the sum-to-product identities By using the sum-to-
product identities for sine functions, we have:
sin(α+β) = sin αcos β+ cos αsin β
17
sin(α−β) = sin αcos β−cos αsin β
Step 2: Calculate the left-hand side (LHS) Let’s calculate the left-hand
side of the identity:
sin(α+β) sin(α−β) = (sin αcos β+ cos αsin β)(sin αcos β−cos αsin β)
= (sin αcos β)2−(cos αsin β)2
= sin2αcos2β−cos2αsin2β
Step 3: Apply the Pythagorean identity We know that cos2θ+sin2θ=
1. Applying this Pythagorean identity, we get:
cos2β= 1 −sin2β
sin2β= 1 −cos2β
Step 4: Substitute into LHS Substitute the expressions for cos2βand
sin2βinto the LHS we calculated in Step 2:
sin2αcos2β−cos2αsin2β= sin2α(1 −sin2β)−cos2α(1 −cos2β)
= sin2α−sin2αsin2β−cos2α+ cos2αcos2β
Step 5: Simplify and use the Pythagorean identity Continuing from
Step 4, simplify the expression and use the Pythagorean identities:
sin2α−sin2αsin2β−cos2α+cos2αcos2β= sin2α−sin2α(1−cos2β)−(1−sin2β)+cos2αcos2β
= sin2α−sin2α+ sin2αcos2β−1 + sin2β+ cos2αcos2β
= sin2αcos2β+ sin2β−1 + cos2αcos2β
Step 6: Final simplification To complete the proof, we need to show
that:
sin2αcos2β+ sin2β−1 + cos2αcos2β= sin2α−sin2β
This can be further simplified as:
sin2αcos2β+ sin2β−1 + cos2αcos2β= sin2α−sin2β
This proves the given trigonometric identity.
Question 21
Question
Solve the trigonometric equation for 0≤x≤2π:
2 cos2(x)−3 cos(x)−2 = 0
18
Solution
Step 1: Let u= cos(x), then the equation becomes a quadratic equation in u:
2u2−3u−2 = 0
Step 2: Solve the quadratic equation 2u2−3u−2 = 0 by factoring:
(2u+ 1)(u−2) = 0
Step 3: Set each factor equal to zero to solve for u:
2u+ 1 = 0 ⇒2u=−1⇒u=−1
2
u−2 = 0 ⇒u= 2
Step 4: Since u= cos(x), we have two possible solutions:
cos(x) = −1
2or cos(x) = 2
Step 5: For cos(x) = −1
2, we know that xcould be in the second or third
quadrant. In these quadrants, cos(x)is negative, so we need to find the angles
where cos(x) = −1
2. Thus, we have x=2π
3and x=4π
3.
Step 6: For cos(x)=2, there are no solutions since the cosine function is
bounded by -1 and 1.
Therefore, the solutions to the trigonometric equation 2 cos2(x)−3 cos(x)−
2 = 0 for 0≤x≤2πare x=2π
3and x=4π
3.
Question 22
Question
Prove the following trigonometric identity:
cos2(θ)
1−sin(θ)+sin2(θ)
1−cos(θ)= csc(θ) + cot(θ)
19
Solution
Step 1: Start with the left side of the equation and simplify.
cos2(θ)
1−sin(θ)+sin2(θ)
1−cos(θ)
=cos2(θ)(1 + sin(θ))
(1 −sin(θ))(1 + sin(θ)) +sin2(θ)(1 + cos(θ))
(1 −cos(θ))(1 + cos(θ)) (Multiplying by conjugate)
=cos2(θ) + cos2(θ) sin(θ) + sin2(θ) + sin2(θ) cos(θ)
1−sin2(θ)+1−cos2(θ)
=(cos2(θ) + sin2(θ)) + cos(θ) sin2(θ) + cos2(θ) sin(θ)
2 cos(θ) sin(θ)(Using trigonometric identities)
=1 + cos(θ) sin(θ)(1 + cos(θ))
2 cos(θ) sin(θ)
=1 + 1
2sin(2θ) + 1
2cos(2θ)
2 cos(θ) sin(θ)(Using double angle formulas)
=1
2 cos(θ) sin(θ)+sin(2θ)
4 cos(θ) sin(θ)+cos(2θ)
4 cos(θ) sin(θ)
= csc(θ) + cot(θ) + cos(2θ)
4 cos(θ) sin(θ)(Using trigonometric identities)
Step 2: Using the double angle formulas, simplify the last term to match the
right side of the equation.
cos(2θ)
4 cos(θ) sin(θ)=2 cos2(θ)−1
4 cos(θ) sin(θ)
=2 cos(θ)−1
cos(θ)
4 cos(θ) sin(θ)(Using reciprocal identity)
=
1
sin(θ)−1
cos(θ)
4
=cot(θ)−csc(θ)
4
Therefore, the left side of the equation simplifies to:
csc(θ) + cot(θ) + (cot(θ)−csc(θ)
4)= csc(θ) + cot(θ)
Thus, the given trigonometric identity is proven.
20
Question 23
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the multiple angle formula for sine:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 1: Find sin(2x)using the double angle formula:
sin(2x) = 2 sin(x) cos(x)
Step 2: Find cos(2x)using the double angle formula:
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 3: Substitute sin(2x)and cos(2x)into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
= 2 sin(x) cos2(x) + sin(x)−2 sin3(x)
= 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, sin(3x) = 3 sin(x)−4 sin3(x), which proves the trigonometric
identity.
Question 24
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to arrive at the right-hand side.
21
Step 1: Start with the left-hand side:
sin4(x)−cos4(x)
Step 2: Rewrite sin4(x)and cos4(x)in terms of sin2(x)and cos2(x)using
the Pythagorean identity:
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
Step 3: Apply the difference of squares factorization formula: a2−b2=
(a+b)(a−b)
(sin2(x))2−(cos2(x))2= (sin2(x) + cos2(x))(sin2(x)−cos2(x))
Step 4: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify:
(sin2(x) + cos2(x))(sin2(x)−cos2(x)) = 1(sin2(x)−cos2(x))
Step 5: Simplify the expression to get the right-hand side:
1(sin2(x)−cos2(x)) = sin2(x)−cos2(x)
Step 6: Use the difference of squares formula sin2(x)−cos2(x) = 2 sin(x) cos(x)
to arrive at the final result:
sin2(x)−cos2(x) = 2 sin(x) cos(x) = 2 sin2(x) cos2(x)
Therefore, we have shown that sin4(x)−cos4(x) = 2 sin2(x) cos2(x), which
proves the given trigonometric identity.
Question 25
Question
Prove the following trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
Solution
We will start by using the Pythagorean identity cos2(x) = 1 −sin2(x)to rewrite
the left side of the equation.
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x))
= (1 −sin2(x))(cos2(x)−sin2(x))
= (1 −sin2(x))(1 −2 sin2(x))
= 1 −sin2(x)−2 sin2(x) + 2 sin4(x)
= 1 −3 sin2(x) + 2 sin4(x).
22
Next, we will express cos(2x)in terms of sin(x)using the double angle for-
mula cos(2x) = cos2(x)−sin2(x).
cos(2x) = cos2(x)−sin2(x)
= (1 −sin2(x)) −sin2(x)
= 1 −sin2(x)−sin2(x)
= 1 −2 sin2(x)
= 1 −3 sin2(x) + 2 sin4(x).
Therefore, we have shown that cos4(x)−sin4(x) = cos(2x), as required.
Question 26
Question
Solve the equation tan2(x)−√3 tan(x)−1 = 0 for xin the interval [0,2π].
Solution
Step 1: Let y= tan(x), then the equation can be rewritten as y2−√3y−1 = 0.
Step 2: Solve the quadratic equation for yusing the quadratic formula:
y=−(−√3) ±√(−√3)2−4(1)(−1)
2(1)
y=√3±√3+4
2
y=√3±√7
2
Step 3: Recall that y= tan(x), so we have two possible solutions for y:
tan(x) = √3 + √7
2
tan(x) = √3−√7
2
Step 4: To find the values of xbetween 0and 2πthat satisfy these equations,
we need to consider the signs of √3±√7in each quadrant.
Step 5: In the first quadrant, both tan(x)and √3±√7are positive.
Step 6: In the second quadrant, tan(x)is positive and √3−√7is negative,
so this case is not possible.
Step 7: In the third quadrant, both tan(x)and √3−√7are negative.
Step 8: In the fourth quadrant, tan(x)is negative and √3 + √7is positive,
so this case is not possible.
23
Step 9: Therefore, the solution to the equation tan(x) = √3+√7
2in the
interval [0,2π]is given by x= arctan (√3+√7
2).
Step 10: Similarly, the solution to the equation tan(x) = √3−√7
2in the
interval [0,2π]is given by x= arctan (√3−√7
2).
Question 27
Question
Prove the trigonometric identity: sin(3x) = 3 sin(x)−4 sin3(x)for all real num-
bers x.
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the angle addition formula for sine: sin(A+B) = sin(A) cos(B)+cos(A) sin(B).
Step 1: Express sin(3x)using the angle addition formula.
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express sin(2x)using the double angle formula.
sin(2x) = 2 sin(x) cos(x)
Step 3: Express cos(2x)using the double angle formula.
cos(2x) = cos2(x)−sin2(x)
= 1 −2 sin2(x)
= 2 −2(1 −cos2(x))
= 2 −2 + 2 cos2(x)
= 2 cos2(x)
Step 4: Substitute the expressions for sin(2x)and cos(2x)into the
expression for sin(3x).
sin(3x) = (2 sin(x) cos(x)) cos(x) + (2 cos2(x)) sin(x)
= 2 sin(x) cos2(x) + 2 cos2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + 2(1 −sin2(x)) sin(x)
= 2 sin(x)−2 sin3(x) + 2 sin(x)−2 sin3(x)
= 4 sin(x)−4 sin3(x)
24
Step 5: Simplify the expression obtained to prove the desired iden-
tity.
sin(3x) = 4 sin(x)−4 sin3(x)
= 3 sin(x)−4 sin3(x)(proved)
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)is proven
for all real numbers x.
Question 28
Question
Prove the following trigonometric identity:
1−sin2a
1 + sin a= cos a
Solution
To prove the trigonometric identity 1−sin2a
1+sin a= cos a, we will start with the left-
hand side and manipulate it to become the right-hand side using trigonometric
identities.
Step 1: Rewrite the left-hand side of the equation.
1−sin2a
1 + sin a
Step 2: Use the Pythagorean identity sin2a+ cos2a= 1 to simplify the
numerator.
1−sin2a= cos2a
Step 3: Substitute the simplified numerator back into the expression.
cos2a
1 + sin a
Step 4: Factor out a cos afrom the numerator.
cos a·cos a
1 + sin a
Step 5: Use the Pythagorean identity again to simplify the remaining frac-
tion.
cos a·cos a
1 + sin a= cos a·cos a
1 + sin a= cos a·cos a
1 + sin a·1−sin a
1−sin a
Step 6: Further simplify by expanding the denominator.
cos a·cos a−sin acos a
1−sin2a
25
Step 7: Use the double angle identity cos 2a= cos2a−sin2ato simplify
the numerator.
cos a·cos a−sin acos a
cos 2a
Step 8: Substitute cos 2aback into the expression.
cos a·cos a−sin acos a
cos 2a= cos a
Therefore, we have shown that 1−sin2a
1+sin a= cos ais a valid trigonometric iden-
tity.
Question 29
Question
Prove the trigonometric identity:
2 cos2(x)−1 = sin2(x)
Solution
To prove the trigonometric identity 2 cos2(x)−1 = sin2(x), we will start with
the fundamental Pythagorean trigonometric identity:
sin2(x) + cos2(x) = 1
Step 1: Rearrange the Pythagorean identity to solve for sin2(x):
sin2(x) = 1 −cos2(x)
Step 2: Substitute sin2(x)with 1−cos2(x)in the given identity:
2 cos2(x)−1 = sin2(x)
2 cos2(x)−1 = 1 −cos2(x)
2 cos2(x)−1 = cos2(x)
2 cos2(x) = 1 + cos2(x)
Step 3: Subtract cos2(x)from both sides to isolate cos2(x):
2 cos2(x)−cos2(x) = 1
cos2(x) = 1
Thus, we have proven the trigonometric identity 2 cos2(x)−1 = sin2(x).
26
Question 30
Question
Prove the following trigonometric identity:
cos2θ(1 −tan2θ)
1 + sin θtan θ= cos θ−sin θ
Solution
Step 1: Start with the left-hand side (LHS) of the given identity.
LHS =cos2θ(1 −tan2θ)
1 + sin θtan θ
Step 2: Expand the terms in the numerator using the identity tan2θ=
sec2θ−1.
LHS =cos2θ(1 −sec2θ+ 1)
1 + sin θtan θ
Step 3: Simplify the numerator further.
LHS =cos2θ(cos2θ)
1 + sin θtan θ=cos4θ
1 + sin θtan θ
Step 4: Use the Pythagorean identity cos2θ= 1 −sin2θin the numerator.
LHS =(1 −sin2θ)2
1 + sin θtan θ
Step 5: Expand the numerator.
LHS =1−2 sin2θ+ sin4θ
1 + sin θtan θ
Step 6: Recognize that 1−sin2θ= cos2θ.
LHS =cos2θ−2 sin2θ+ sin4θ
1 + sin θtan θ
Step 7: Simplify the numerator further.
LHS =cos2θ(1 −2 tan2θ)
1 + sin θtan θ
Step 8: Use the identity tan2θ= sec2θ−1again in the numerator.
LHS =cos2θ(1 −2 sec2θ+ 2)
1 + sin θtan θ
27
Step 9: Simplify the numerator one more time.
LHS =2 cos2θ−2 cos2θsec2θ
1 + sin θtan θ
Step 10: Recognize that 2−2 cos2θ= 2 sin2θ.
LHS =2 sin2θ−2 cos2θsec2θ
1 + sin θtan θ
Step 11: Use the trigonometric identity sec θ=1
cos θin the numerator.
LHS =2 sin2θ−2 cos2θ(1
cos2θ)
1 + sin θtan θ
Step 12: Simplify the numerator further.
LHS =2 sin2θ−2
1 + sin θtan θ
Step 13: Factor out a 2 from the numerator.
LHS =2(sin2θ−1)
1 + sin θtan θ
Step 14: Recognize that sin2θ−1 = −cos2θ.
LHS =2(−cos2θ)
1 + sin θtan θ
Step 15: Simplify the LHS even further.
LHS =−2 cos2θ
1 + sin θtan θ
Step 16: Finally, notice that −2 cos2θ=
Question 31
Question
Prove the following trigonometric identity:
4 sin2θ−3 = cos(2θ)
28
Solution
We will prove the given trigonometric identity using the Pythagorean identity
sin2θ+ cos2θ= 1 and double angle identities.
Step 1: Rewrite cos(2θ)using double angle identities
cos(2θ) = cos2θ−sin2θ
= (1 −sin2θ)−sin2θ(Using the Pythagorean identity)
= 1 −2 sin2θ
Step 2: Substitute cos(2θ)with 1−2 sin2θin the original expression
4 sin2θ−3 = 1 −2 sin2θ
4 sin2θ+ 2 sin2θ= 4
6 sin2θ= 4
sin2θ=2
3
sin θ=±√2
3
Therefore, the given trigonometric identity is true when sin θ=±√2
3.
Question 32
Question
Prove the trigonometric identity:
1 + sin θ
1−sin θ= csc θ+ cot θ
Solution
Start with the LHS: 1 + sin θ
1−sin θ=1
1−sin θ+sin θ
1−sin θ
=1
1−sin θ+sin θ
1−sin θ×1 + sin θ
1 + sin θ
=1
cos2θ+sin θ(1 + sin θ)
cos2θ
=1 + sin θ
cos2θ
=1
cos θ×1
1−sin θ
= csc θ×csc θ
= csc θ+ cot θ
Thus, the LHS equals the RHS. Hence, the trigonometric identity is proven.
29
Question 33
Question
Prove the following trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
Solution
Step 1: Recall the definitions of the trigonometric functions.
cot(θ) = cos(θ)
sin(θ)
csc(θ) = 1
sin(θ)
Step 2: Start with the left side of the identity.
Left side = cot(θ) csc(θ)
=cos(θ)
sin(θ)·1
sin(θ)
=cos(θ)
sin2(θ)
Step 3: Now simplify the right side of the identity.
Right side = csc(θ)−cot(θ)
=1
sin(θ)−cos(θ)
sin(θ)
=1−cos(θ)
sin(θ)
Step 4: Notice that the right side can be rewritten using sin2(θ).
Right side =1−cos(θ)
sin(θ)
=sin2(θ)
sin(θ)
=sin2(θ)
sin2(θ)
Step 5: Compare the two sides of the identity. Since the left side and the
right side are equal, we have proven the trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
30
Question 34
Question
Prove the trigonometric identity:
1−cos(x)
sin(x)= cot (x
2)
Solution
1. We begin by using the double angle formula for cotangent:
cot(2θ) = cos(2θ)
sin(2θ)=cos2(θ)−sin2(θ)
2 sin(θ) cos(θ)
2. Let x= 2θto obtain :
cot(x) = cos2(x
2)−sin2(x
2)
2 sin (x
2)cos (x
2)
3. Next, we apply the half angle formulas for sine and cosine:
cos2(x
2)=1 + cos(x)
2and sin2(x
2)=1−cos(x)
2
4. Substituting these into the expression for cot(x), we get:
cot(x) =
1+cos(x)
2−1−cos(x)
2
2√1−cos(x)
2
1+cos(x)
2
5. Simplifying the right-hand side, we obtain:
cot(x) =
1
2(cos(x)+1−1 + cos(x))
2√1−cos2(x)
4
6. Further simplification gives:
cot(x) = 2 cos(x)
2 sin(x)= cot (x
2)
7. Therefore, we have proven the trigonometric identity:
1−cos(x)
sin(x)= cot (x
2)
31
Question 35
Question
Prove the trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ) cos(4θ)
Solution
Step 1: Recall the double angle formula for cosine, cos(2θ) = cos2(θ)−sin2(θ).
Step 2: Expand the left side of the equation using the difference of squares
formula:
cos4(θ)−sin4(θ) = (cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ))
Step 3: Substitute the double angle formula for cosine into the expanded equa-
tion:
(cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ)) = (cos(2θ) + sin(2θ))(cos(2θ)−sin(2θ))
Step 4: Apply the double angle formula for sine, sin(2θ) = 2 sin(θ) cos(θ), to
the equation:
(cos(2θ)+sin(2θ))(cos(2θ)−sin(2θ)) = (cos(2θ)+2 sin(θ) cos(θ))(cos(2θ)−2 sin(θ) cos(θ))
Step 5: Expand the terms in the equation:
(cos(2θ)+2 sin(θ) cos(θ))(cos(2θ)−2 sin(θ) cos(θ)) = cos2(2θ)−2 sin(2θ) sin(θ) cos(θ)
Step 6: Apply the double angle formula for cosine, cos(2θ) = cos2(θ)−sin2(θ),
and the double angle formula for sine, sin(2θ) = 2 sin(θ) cos(θ), to simplify the
equation further:
cos2(2θ)−2 sin(2θ) sin(θ) cos(θ) = cos2(θ)−sin2(θ) = cos(2θ) cos(4θ)
Therefore, we have shown that cos4(θ)−sin4(θ) = cos(2θ) cos(4θ).
32
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin(x) + 2 cos(x) = √3
⇒sin(x) + 2 cos(x) = 2√3(1
2)
⇒sin(x) + 2 cos(x) = 2√3 cos (π
6)
Step 2: Express the left-hand side as a single trigonometric function. Let’s
rewrite the left-hand side using the angle addition formula for sine:
sin(x) + 2 cos(x) = 2√3 cos (π
6)
= 2√3(cos(x) cos (π
6)−sin(x) sin (π
6))
= 2√3(cos(x)√3
2−sin(x)1
2)
=√3 cos(x)−sin(x)
Step 3: Replace sin(x) + 2 cos(x)with √3 cos(x)−sin(x)in the equation.
√3 cos(x)−sin(x) = 2√3 cos (π
6)
Step 4: Simplify the equation and solve for x.
√3 cos(x)−sin(x) = √3·2·√3
2= 3
⇒√3 cos(x)−sin(x) = 3
Step 5: Divide by √3to simplify the equation further.
cos(x)−1
√3sin(x) = 1
Step 6: Use the identity tan (π
6)=1
√3to rewrite the equation as a single
trigonometric function.
cos(x)−tan (π
6)sin(x) = 1
cos (x−π
6)= 1
Step 7: Solve for x. Since cosine is equal to 1 only at x= 0 in the interval
[0,2π], the solution to the equation is x= 0.
2
Question 3
Question
Solve the trigonometric equation sin(x) = cos(x)for xin the interval [0,2π).
Solution
Step 1: To solve the equation sin(x) = cos(x), we can rewrite it in terms of sines
and cosines:
sin(x) = cos(x) =⇒sin(x)−cos(x) = 0
Step 2: Using the trigonometric identity sin(x−π
4)= sin(π
2−x)= sin(π
4)cos(x)−
cos(π
4)sin(x), we rewrite sin(x)−cos(x)in the form of sin(x−π
4):
sin(x)−cos(x) = sin(x−π
4)= 0
Step 3: Since sin(x−π
4)= 0 if and only if x−π
4=nπ for some integer n,
we have:
x−π
4=nπ
Step 4: Solve for xby adding π
4to both sides:
x=nπ +π
4
Step 5: Since we are looking for solutions in the interval [0,2π), we need to
find the values of nsuch that xlies in this interval. Thus, we have:
0≤nπ +π
4<2π
Step 6: Set the upper bound:
nπ +π
4<2π
nπ < 7π
4
n < 7
4
Step 7: Since nmust be an integer, the possible values for nare 0and 1.
Step 8: Plug n= 0 and n= 1 back into the equation x=nπ +π
4to find
the solutions in the interval [0,2π): For n= 0:x= 0 + π
4=π
4For n= 1:
x=π+π
4=5π
4
Therefore, the solutions to the equation sin(x) = cos(x)in the interval [0,2π)
are x=π
4and x=5π
4.
3
Question 4
Question
Solve the equation cos2(x)−3 cos(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s first rewrite the equation by substituting cos(x)with u:
We have u2−3u+ 2 = 0.
Step 2: Next, we need to solve this quadratic equation. Factoring gives:
u2−3u+ 2 = (u−1)(u−2) = 0.
Step 3: Setting each factor to zero, we find:
u−1 = 0 or u−2 = 0
Step 4: Solving these equations, we get:
u= 1 or u= 2
Step 5: Now, we will substitute back cos(x)for uin our solutions:
cos(x) = 1 or cos(x) = 2
Step 6: Since cos(x)can only take values between -1 and 1, the equation
cos(x) = 2 has no solutions.
Step 7: Therefore, the only valid solution is cos(x) = 1.
Step 8: Recall that cos(x) = 1 when x= 0.
Step 9: Thus, the solution to the equation cos2(x)−3 cos(x) + 2 = 0 for
0≤x≤2πis x= 0.
Question 5
Question
Solve the trigonometric equation sin(x) cos(2x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Use the double angle identity cos(2x) = 2 cos2(x)−1to rewrite the
equation:
sin(x)(2 cos2(x)−1) = 1
2
4
Step 2: Expand the left side of the equation using the identity sin(x) cos2(x)−
sin(x) = 1
2:
2 sin(x) cos2(x)−sin(x) = 1
2
Step 3: Substitute cos2(x) = 1 −sin2(x)into the equation:
2 sin(x)−sin(x) = 1
2
Step 4: Simplify the equation:
sin(x) = 1
2
Step 5: The solutions to sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the original equation are x=π
6and x=5π
6.
Question 6
Question
Prove the trigonometric identity:
cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)=2
1−sin(θ)
Solution
Step 1: Start with the left-hand side of the given equation and work to simplify
it. cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)
Step 2: Find a common denominator for the two fractions.
cos2(θ) + (1 −sin2(θ))
cos(θ)(1 −sin(θ))
Step 3: Simplify the numerator using trigonometric identities.
cos2(θ) + cos2(θ)
cos(θ)(1 −sin(θ))
Step 4: Combine like terms in the numerator.
2 cos2(θ)
cos(θ)(1 −sin(θ))
5
Step 5: Simplify the expression by canceling out a factor of cos(θ).
2 cos(θ)
1−sin(θ)
Step 6: Finally, simplify the expression to match the right-hand side of the
given equation. 2
1−sin(θ)
Therefore, we have proven the given trigonometric identity:
cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)=2
1−sin(θ)
Question 7
Question
Solve the trigonometric equation cos2(x)−2 cos(x) + 1 = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the equation in terms of cos(x)as a quadratic equation by
using the substitution cos(x) = t.
cos2(x)−2 cos(x) + 1 = 0
t2−2t+ 1 = 0
Step 2: Solve the quadratic equation t2−2t+ 1 = 0 by factoring or using
the quadratic formula. The equation factors to (t−1)2= 0. So, t−1 = 0 which
implies t= 1.
Step 3: Substitute back cos(x) = tto find the possible solutions in the
interval [0,2π). Since cos(x) = 1,x= 0 and x= 2πare solutions in the interval
[0,2π).
Step 4: Thus, the solutions to the trigonometric equation cos2(x)−2 cos(x)+
1 = 0 in the interval [0,2π)are x= 0 and x= 2π.
Question 8
Question
Simplify the following expression:
tan(x) + sin(x)
1 + tan(x) sin(x)
6
Solution
Step 1: Recall the trigonometric identity tan(x) = sin(x)
cos(x).
Step 2: Substitute tan(x) = sin(x)
cos(x)and simplify the expression:
sin(x)
cos(x)+ sin(x)
1 + sin(x)
cos(x)sin(x)
Step 3: Combine the numerators:
sin(x) + sin(x) cos(x)
cos(x) + sin(x) cos(x)
Step 4: Factor out a sin(x)from the numerator and factor out a cos(x)from
the denominator: sin(x)(1 + cos(x))
cos(x)(1 + sin(x))
Step 5: Apply the trigonometric identity sin2(x) + cos2(x) = 1 to simplify
the expression further:
sin(x)(1 + cos(x))
cos(x)(1 + sin(x)) =sin(x)(1 + cos(x))
cos(x)(1 + sin(x)) ×1 + sin(x)
1 + sin(x)
Step 6: Expand the numerator and denominator:
sin(x) + sin(x) cos(x)
cos(x) + sin(x)
Therefore, the simplified expression is sin(x)+sin(x) cos(x)
cos(x)+sin(x).
Question 9
Question
Solve the trigonometric equation sin(3x) = cos(2x)for 0≤x≤2π.
Solution
Step 1: Recall the trigonometric identities: sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = cos2(x)−sin2(x).
Step 2: Substitute these identities into the equation sin(3x) = cos(2x)to get
3 sin(x)−4 sin3(x) = cos2(x)−sin2(x)
Step 3: Using the Pythagorean identity cos2(x) + sin2(x)=1, rewrite the
equation as 3 sin(x)−4 sin3(x) = 1 −2 sin2(x).
Step 4: Rearrange the equation to get 4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0.
Step 5: Factor the cubic equation to get (2 sin(x)−1)(2 sin2(x)+sin(x)−1) =
0.
Step 6: Solve for sin(x)in each factor:
7
• For 2 sin(x)−1 = 0, we get sin(x) = 1
2.
• For 2 sin2(x) + sin(x)−1=0, we can solve by setting u= sin(x)to get
2u2+u−1 = 0. The solutions are u=−1±√1+8
4, so either u=1
2or
u=−1.
Step 7: Solve for xin each case:
• For sin(x) = 1
2, the solutions in [0,2π]are x=π
6,5π
6.
• For sin(x) = 1
2, the solutions in [0,2π]are x=π
6,5π
6.
• For sin(x) = −1, the only solution in [0,2π]is x=3π
2.
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π]are x=π
6,5π
6,3π
2.
Question 10
Question
Solve the trigonometric equation √3 sin(x) cos(x) = 1
2for xin the interval
[0,2π).
Solution
Step 1: Use the double angle identity for sine to simplify the left side of the
equation:
√3 sin(x) cos(x) = √3
2sin(2x)
Step 2: Now the equation becomes √3
2sin(2x) = 1
2.
Step 3: Divide both sides by √3
2to solve for sin(2x):
sin(2x) = 1
√3
Step 4: To find the solutions for 2x, we need to consider all possible values of
sin(2x)in the interval [−1,1]. Since sin(π
6)=1
2and sin(5π
6)=1
2, the possible
solutions for 2xare π
6and 5π
6.
Step 5: Finally, solve for xby dividing the solutions of 2xby 2:
x=π
12,5π
12
Therefore, the solutions to the given trigonometric equation in the interval
[0,2π)are x=π
12 ,5π
12 .
8
Question 11
Question
Simplify the expression sin4(x)−cos4(x)
sin2(x)−cos2(x)using trigonometric identities.
Solution
Step 1: Recall the Pythagorean identity sin2(x) + cos2(x) = 1.
Step 2: Rewrite the expression using the Pythagorean identity:
sin4(x)−cos4(x)
sin2(x)−cos2(x)=(sin2(x) + cos2(x))(sin2(x)−cos2(x))
sin2(x)−cos2(x).
Step 3: Factor out the common term:
(sin2(x) + cos2(x))(sin2(x)−cos2(x))
sin2(x)−cos2(x)= sin2(x) + cos2(x).
Step 4: Apply the Pythagorean identity again to simplify:
sin2(x) + cos2(x) = 1.
Therefore, the simplified expression is 1.
Question 12
Question
Prove the trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)
Solution
To prove the given trigonometric identity, we will start from the left-hand side
and manipulate the expression until it simplifies to the right-hand side.
Step 1: Rewrite the left-hand side of the equation.
sin3(x)
cos(x)+cos3(x)
sin(x)
Step 2: Multiply the first term by sin(x)
sin(x)to get a common denominator.
sin4(x)
sin(x) cos(x)+cos3(x)
sin(x)
9
Step 3: Combine the terms by finding a common denominator.
sin4(x) + cos3(x) cos(x)
sin(x) cos(x)
Step 4: Apply trigonometric identity sin2(x) + cos2(x) = 1.
(1 −cos2(x))2+ cos3(x) cos(x)
sin(x) cos(x)
Step 5: Expand and simplify the numerator.
1−2 cos2(x) + cos4(x) + cos4(x)
sin(x) cos(x)
Step 6: Combine like terms in the numerator.
1−2 cos2(x) + 2 cos4(x)
sin(x) cos(x)
Step 7: Factor out a common factor in the numerator.
1−2 cos2(x) + 2 cos2(x) cos2(x)
sin(x) cos(x)
Step 8: Simplify the expression further.
1 + cos2(x)−2 cos2(x)
sin(x) cos(x)
=1−cos2(x)
sin(x) cos(x)
=sin2(x)
sin(x) cos(x)
=1
sin(x) cos(x)
Therefore, we have shown that the given trigonometric identity is true.
Question 13
Question
Prove the trigonometric identity: tan(θ) sec(θ) = sec(θ) + tan(θ).
10
Solution
Step 1: Start with the left-hand side of the identity.
tan(θ) sec(θ)
Step 2: Rewrite tan(θ)and sec(θ)in terms of sine and cosine.
tan(θ) = sin(θ)
cos(θ)and sec(θ) = 1
cos(θ)
Step 3: Substitute these values back into the left-hand side of the identity.
sin(θ)
cos(θ)·1
cos(θ)
Step 4: Multiply the fractions.
sin(θ)
cos(θ)·cos(θ)
Step 5: Simplify the expression.
sin(θ)
cos2(θ)
Step 6: Recall that tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=sin(θ)
cos(θ)·1
1= sec(θ)·tan(θ)
Step 7: Hence, the left-hand side is equal to sec(θ)·tan(θ).
Step 8: Now, simplify the right-hand side of the identity.
sec(θ) + tan(θ)
Step 9: Substitute the values for sec(θ)and tan(θ).
1
cos(θ)+sin(θ)
cos(θ)
Step 10: Combine the fractions.
1 + sin(θ)
cos(θ)
Step 11: Recall that sec(θ) = 1
cos(θ)and tan(θ) = sin(θ)
cos(θ).
Step 12: Hence, the right-hand side simplifies to sec(θ)·tan(θ).
Step 13: Therefore, the left-hand side of the identity is equal to the right-
hand side, and the trigonometric identity tan(θ) sec(θ) = sec(θ) + tan(θ)is
proven.
11
Question 14
Question
Solve the trigonometric equation sin2x−√3 sin x−1 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let u= sin x, then the equation becomes a quadratic equation in u:
u2−√3u−1 = 0
Step 2: Solve the quadratic equation u2−√3u−1 = 0 using the quadratic
formula:
u=−(−√3) ±√(−√3)2−4(1)(−1)
2(1)
u=√3±√7
2
Step 3: Since u= sin x, the solutions for xare obtained by finding the angles
whose sine is the solutions u=√3±√7
2.
x= sin−1(√3 + √7
2
,x= sin−1(√3−√7
2)
Step 4: Use a calculator to find the approximate values of the angles:
x≈70.53◦,229.47◦,19.11◦,160.89◦
Therefore, the solutions to the trigonometric equation sin2x−√3 sin x−1 = 0
for 0◦≤x≤360◦are x≈70.53◦,229.47◦,19.11◦,160.89◦.
Question 15
Question
Prove the identity:
sin(α−β) sin(α+β) = sin2α−sin2β
12
Solution
Step 1: Expand the left-hand side of the given identity using the angle differ-
ence identities:
sin(α−β) = sin αcos β−cos αsin β
sin(α+β) = sin αcos β+ cos αsin β
Step 2: Multiply the two expressions together:
sin(α−β) sin(α+β) = (sin αcos β−cos αsin β)(sin αcos β+ cos αsin β)
= sin2αcos2β−sin αcos βcos αsin β−cos αsin βsin αcos β−cos2αsin2β
Step 3: Simplify the expression:
sin2αcos2β−sin αcos βcos αsin β−cos αsin βsin αcos β−cos2αsin2β
= sin2α(cos2β−sin2β)−2 sin αcos βcos αsin β−cos2αsin2β
= sin2αcos(2β)−2 sin αcos αsin βcos β−cos2αsin2β
Step 4: Recall the double angle formula for cosine: cos(2θ) = 2 cos2θ−1
and substitute it into the expression:
= sin2α(2 cos2β−1) −2 sin αcos αsin βcos β−cos2αsin2β
= 2 sin2αcos2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
Step 5: Use the Pythagorean identity sin2θ+ cos2θ= 1 and rearrange
terms:
2 sin2αcos2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= 2(sin2α(1 −sin2β)) −sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= 2 sin2α−2 sin2αsin2β−sin2α−2 sin αcos αsin βcos β−cos2αsin2β
= sin2α−sin2αsin2β−2 sin αcos αsin βcos β−cos2αsin2β
Step 6: Recognize that the obtained expression is equal to sin2α−sin2β,
which completes the proof of the given identity. Thus, we have shown that:
sin(α−β) sin(α+β) = sin2α−sin2β
Question 16
Question
Prove the trigonometric identity: sin4(x)−cos4(x) = 1 −2 cos2(2x).
13
Solution
Step 1: Recall the double angle formula for cosine:
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Express cos(2x)in terms of sin(x)and cos(x):
cos(2x) = cos2(x)−sin2(x)
Step 3: Rearrange the above equation to express sin2(x)in terms of cos(2x)
and cos2(x):
sin2(x) = cos2(x)−cos(2x)
Step 4: Substitute the expression for sin2(x)into the given identity:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
(cos2(x)−cos(2x))2−cos4(x) = 1 −2 cos2(2x)
Step 5: Expand the left side of the equation:
cos4(x)−2 cos2(x) cos(2x) + cos2(2x)−cos4(x) = 1 −2 cos2(2x)
Step 6: Simplify the equation by canceling out cos4(x)terms:
−2 cos2(x) cos(2x) + cos2(2x) = 1 −2 cos2(2x)
Step 7: Further simplify the equation by combining like terms:
−2 cos2(x) cos(2x) = 1 −3 cos2(2x)
Step 8: Recall the double angle formula for cosine and use it to express
cos(2x)in terms of cos(x)and sin(x):
cos(2x) = 2 cos2(x)−1
Step 9: Substitute cos(2x) = 2 cos2(x)−1into the equation:
−2 cos2(x)(2 cos2(x)−1) = 1 −3(2 cos2(x)−1)
Step 10: Simplify the equation by expanding and combining like terms:
−4 cos4(x) + 2 cos2(x) = 1 −6 cos2(x)+3
−4 cos4(x) + 2 cos2(x) = −6 cos2(x)+4
Step 11: Finally, simplify the equation further to prove the identity:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
14
Question 17
Question
Solve the trigonometric equation tan(2x) + √3 = 0 for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation using the double angle identity for tangent:
tan(2x) = −√3
Step 2: Use the identity tan(2x) = 2 tan(x)
1−tan2(x)to rewrite the equation:
2 tan(x)
1−tan2(x)=−√3
Step 3: Let u= tan(x), so the equation becomes:
2u
1−u2=−√3
Step 4: Multiply both sides by 1−u2to get rid of the denominator:
2u=−√3(1 −u2)
Step 5: Expand and rearrange the equation:
2u=−√3 + √3u2
0 = √3u2−2u−√3
Step 6: Solve the quadratic equation √3u2−2u−√3 = 0 for u:
u=−(−2) ±√(−2)2−4(√3)(−√3)
2(√3)
u=2±√4 + 12
2√3
u=2±√16
2√3
u=2±4
2√3
Step 7: Solve for uto find the possible values of x: 1. When u=2+4
2√3=
6
2√3=3
√3=√3: Since u= tan(x), we have tan(x) = √3. This gives x=π
3.
2. When u=2−4
2√3=−2
2√3=−1
√3=−√3
3: Since u= tan(x), we have
tan(x) = −√3
3. This gives x=5π
6.
Therefore, the solutions to the equation are x=π
3and x=5π
6.
15
Question 18
Question
Solve the trigonometric equation tan2(x)−tan(x)−6=0for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the quadratic equation in terms of tan(x):
The equation is tan2(x)−tan(x)−6=0. We can treat this as a quadratic
equation in terms of tan(x). Let tan(x) = t. Then the equation becomes
t2−t−6 = 0.
Step 2: Factor the quadratic equation:
We need to factor the quadratic equation t2−t−6 = 0 to solve for t. The
factors of -6 that add up to -1 are -3 and 2. Therefore, we can rewrite the
equation as (t−3)(t+ 2) = 0.
Step 3: Find the roots of the quadratic equation:
Setting each factor to zero gives us t−3=0or t+ 2 = 0. Solving these
equations gives t= 3 or t=−2.
Step 4: Translate back to the original variable x:
Since tan(x) = t, we have tan(x)=3or tan(x) = −2. Recall that tan(x) =
sin(x)
cos(x). Therefore, sin(x)
cos(x)= 3 or sin(x)
cos(x)=−2.
Step 5: Solve for x:
For sin(x)
cos(x)= 3, we have sin(x) = 3 cos(x). Dividing both sides by cos(x), we
get tan(x) = 3. This gives x= arctan(3).
For sin(x)
cos(x)=−2, we have sin(x) = −2 cos(x). Dividing both sides by cos(x),
we get tan(x) = −2. This gives x= arctan(−2).
Thus, the solutions to the trigonometric equation tan2(x)−tan(x)−6=0
in the interval [0,2π)are x= arctan(3) and x= arctan(−2).
Question 19
Question
Prove the following trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x)
Solution
Step 1: Start with the identity cos(3x) = cos(2x+x).
Step 2: Use the angle addition formula for cosine: cos(a+b) = cos(a) cos(b)−
sin(a) sin(b). Applying this formula, we have:
cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x)
16
Step 3: Recall the double angle formulas for cosine and sine:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
Substitute these formulas into the expression from Step 2:
cos(2x) cos(x)−sin(2x) sin(x) = (cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x)
Step 4: Simplify the expression by expanding and combining like terms:
(cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x) = cos3(x)−sin2(x) cos(x)−2 sin(x) cos(x) sin(x)
Step 5: Recall the Pythagorean identity sin2(x) + cos2(x)=1. Rearrange
this to solve for sin2(x):sin2(x) = 1−cos2(x). Substitute this into the expression
in Step 4:
cos3(x)−(1 −cos2(x)) cos(x)−2 sin(x) cos(x) sin(x)
Step 6: Further simplify the expression:
cos3(x)−cos(x) + cos3(x)−2 sin(x) cos2(x)
Step 7: Use the Pythagorean identity to replace sin(x)in terms of cos(x):
sin(x) = √1−cos2(x). Substitute this into the expression in Step 6 and sim-
plify:
2 cos3(x)−cos(x)−2√1−cos2(x) cos2(x)
Step 8: Simplify further to get the desired result:
cos(3x) = 4 cos3(x)−3 cos(x)
Therefore, the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x)is proven.
Question 20
Question
Prove the following trigonometric identity:
sin(α+β) sin(α−β) = sin2α−sin2β
Solution
To prove this trigonometric identity, we will use the sum-to-product identities
and the Pythagorean identity.
Step 1: Apply the sum-to-product identities By using the sum-to-
product identities for sine functions, we have:
sin(α+β) = sin αcos β+ cos αsin β
17
sin(α−β) = sin αcos β−cos αsin β
Step 2: Calculate the left-hand side (LHS) Let’s calculate the left-hand
side of the identity:
sin(α+β) sin(α−β) = (sin αcos β+ cos αsin β)(sin αcos β−cos αsin β)
= (sin αcos β)2−(cos αsin β)2
= sin2αcos2β−cos2αsin2β
Step 3: Apply the Pythagorean identity We know that cos2θ+sin2θ=
1. Applying this Pythagorean identity, we get:
cos2β= 1 −sin2β
sin2β= 1 −cos2β
Step 4: Substitute into LHS Substitute the expressions for cos2βand
sin2βinto the LHS we calculated in Step 2:
sin2αcos2β−cos2αsin2β= sin2α(1 −sin2β)−cos2α(1 −cos2β)
= sin2α−sin2αsin2β−cos2α+ cos2αcos2β
Step 5: Simplify and use the Pythagorean identity Continuing from
Step 4, simplify the expression and use the Pythagorean identities:
sin2α−sin2αsin2β−cos2α+cos2αcos2β= sin2α−sin2α(1−cos2β)−(1−sin2β)+cos2αcos2β
= sin2α−sin2α+ sin2αcos2β−1 + sin2β+ cos2αcos2β
= sin2αcos2β+ sin2β−1 + cos2αcos2β
Step 6: Final simplification To complete the proof, we need to show
that:
sin2αcos2β+ sin2β−1 + cos2αcos2β= sin2α−sin2β
This can be further simplified as:
sin2αcos2β+ sin2β−1 + cos2αcos2β= sin2α−sin2β
This proves the given trigonometric identity.
Question 21
Question
Solve the trigonometric equation for 0≤x≤2π:
2 cos2(x)−3 cos(x)−2 = 0
18
Solution
Step 1: Let u= cos(x), then the equation becomes a quadratic equation in u:
2u2−3u−2 = 0
Step 2: Solve the quadratic equation 2u2−3u−2 = 0 by factoring:
(2u+ 1)(u−2) = 0
Step 3: Set each factor equal to zero to solve for u:
2u+ 1 = 0 ⇒2u=−1⇒u=−1
2
u−2 = 0 ⇒u= 2
Step 4: Since u= cos(x), we have two possible solutions:
cos(x) = −1
2or cos(x) = 2
Step 5: For cos(x) = −1
2, we know that xcould be in the second or third
quadrant. In these quadrants, cos(x)is negative, so we need to find the angles
where cos(x) = −1
2. Thus, we have x=2π
3and x=4π
3.
Step 6: For cos(x)=2, there are no solutions since the cosine function is
bounded by -1 and 1.
Therefore, the solutions to the trigonometric equation 2 cos2(x)−3 cos(x)−
2 = 0 for 0≤x≤2πare x=2π
3and x=4π
3.
Question 22
Question
Prove the following trigonometric identity:
cos2(θ)
1−sin(θ)+sin2(θ)
1−cos(θ)= csc(θ) + cot(θ)
19
Solution
Step 1: Start with the left side of the equation and simplify.
cos2(θ)
1−sin(θ)+sin2(θ)
1−cos(θ)
=cos2(θ)(1 + sin(θ))
(1 −sin(θ))(1 + sin(θ)) +sin2(θ)(1 + cos(θ))
(1 −cos(θ))(1 + cos(θ)) (Multiplying by conjugate)
=cos2(θ) + cos2(θ) sin(θ) + sin2(θ) + sin2(θ) cos(θ)
1−sin2(θ)+1−cos2(θ)
=(cos2(θ) + sin2(θ)) + cos(θ) sin2(θ) + cos2(θ) sin(θ)
2 cos(θ) sin(θ)(Using trigonometric identities)
=1 + cos(θ) sin(θ)(1 + cos(θ))
2 cos(θ) sin(θ)
=1 + 1
2sin(2θ) + 1
2cos(2θ)
2 cos(θ) sin(θ)(Using double angle formulas)
=1
2 cos(θ) sin(θ)+sin(2θ)
4 cos(θ) sin(θ)+cos(2θ)
4 cos(θ) sin(θ)
= csc(θ) + cot(θ) + cos(2θ)
4 cos(θ) sin(θ)(Using trigonometric identities)
Step 2: Using the double angle formulas, simplify the last term to match the
right side of the equation.
cos(2θ)
4 cos(θ) sin(θ)=2 cos2(θ)−1
4 cos(θ) sin(θ)
=2 cos(θ)−1
cos(θ)
4 cos(θ) sin(θ)(Using reciprocal identity)
=
1
sin(θ)−1
cos(θ)
4
=cot(θ)−csc(θ)
4
Therefore, the left side of the equation simplifies to:
csc(θ) + cot(θ) + (cot(θ)−csc(θ)
4)= csc(θ) + cot(θ)
Thus, the given trigonometric identity is proven.
20
Question 23
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the multiple angle formula for sine:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 1: Find sin(2x)using the double angle formula:
sin(2x) = 2 sin(x) cos(x)
Step 2: Find cos(2x)using the double angle formula:
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 3: Substitute sin(2x)and cos(2x)into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
= 2 sin(x) cos2(x) + sin(x)−2 sin3(x)
= 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, sin(3x) = 3 sin(x)−4 sin3(x), which proves the trigonometric
identity.
Question 24
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to arrive at the right-hand side.
21
Step 1: Start with the left-hand side:
sin4(x)−cos4(x)
Step 2: Rewrite sin4(x)and cos4(x)in terms of sin2(x)and cos2(x)using
the Pythagorean identity:
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
Step 3: Apply the difference of squares factorization formula: a2−b2=
(a+b)(a−b)
(sin2(x))2−(cos2(x))2= (sin2(x) + cos2(x))(sin2(x)−cos2(x))
Step 4: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify:
(sin2(x) + cos2(x))(sin2(x)−cos2(x)) = 1(sin2(x)−cos2(x))
Step 5: Simplify the expression to get the right-hand side:
1(sin2(x)−cos2(x)) = sin2(x)−cos2(x)
Step 6: Use the difference of squares formula sin2(x)−cos2(x) = 2 sin(x) cos(x)
to arrive at the final result:
sin2(x)−cos2(x) = 2 sin(x) cos(x) = 2 sin2(x) cos2(x)
Therefore, we have shown that sin4(x)−cos4(x) = 2 sin2(x) cos2(x), which
proves the given trigonometric identity.
Question 25
Question
Prove the following trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
Solution
We will start by using the Pythagorean identity cos2(x) = 1 −sin2(x)to rewrite
the left side of the equation.
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x))
= (1 −sin2(x))(cos2(x)−sin2(x))
= (1 −sin2(x))(1 −2 sin2(x))
= 1 −sin2(x)−2 sin2(x) + 2 sin4(x)
= 1 −3 sin2(x) + 2 sin4(x).
22
Next, we will express cos(2x)in terms of sin(x)using the double angle for-
mula cos(2x) = cos2(x)−sin2(x).
cos(2x) = cos2(x)−sin2(x)
= (1 −sin2(x)) −sin2(x)
= 1 −sin2(x)−sin2(x)
= 1 −2 sin2(x)
= 1 −3 sin2(x) + 2 sin4(x).
Therefore, we have shown that cos4(x)−sin4(x) = cos(2x), as required.
Question 26
Question
Solve the equation tan2(x)−√3 tan(x)−1 = 0 for xin the interval [0,2π].
Solution
Step 1: Let y= tan(x), then the equation can be rewritten as y2−√3y−1 = 0.
Step 2: Solve the quadratic equation for yusing the quadratic formula:
y=−(−√3) ±√(−√3)2−4(1)(−1)
2(1)
y=√3±√3+4
2
y=√3±√7
2
Step 3: Recall that y= tan(x), so we have two possible solutions for y:
tan(x) = √3 + √7
2
tan(x) = √3−√7
2
Step 4: To find the values of xbetween 0and 2πthat satisfy these equations,
we need to consider the signs of √3±√7in each quadrant.
Step 5: In the first quadrant, both tan(x)and √3±√7are positive.
Step 6: In the second quadrant, tan(x)is positive and √3−√7is negative,
so this case is not possible.
Step 7: In the third quadrant, both tan(x)and √3−√7are negative.
Step 8: In the fourth quadrant, tan(x)is negative and √3 + √7is positive,
so this case is not possible.
23
Step 9: Therefore, the solution to the equation tan(x) = √3+√7
2in the
interval [0,2π]is given by x= arctan (√3+√7
2).
Step 10: Similarly, the solution to the equation tan(x) = √3−√7
2in the
interval [0,2π]is given by x= arctan (√3−√7
2).
Question 27
Question
Prove the trigonometric identity: sin(3x) = 3 sin(x)−4 sin3(x)for all real num-
bers x.
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the angle addition formula for sine: sin(A+B) = sin(A) cos(B)+cos(A) sin(B).
Step 1: Express sin(3x)using the angle addition formula.
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express sin(2x)using the double angle formula.
sin(2x) = 2 sin(x) cos(x)
Step 3: Express cos(2x)using the double angle formula.
cos(2x) = cos2(x)−sin2(x)
= 1 −2 sin2(x)
= 2 −2(1 −cos2(x))
= 2 −2 + 2 cos2(x)
= 2 cos2(x)
Step 4: Substitute the expressions for sin(2x)and cos(2x)into the
expression for sin(3x).
sin(3x) = (2 sin(x) cos(x)) cos(x) + (2 cos2(x)) sin(x)
= 2 sin(x) cos2(x) + 2 cos2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + 2(1 −sin2(x)) sin(x)
= 2 sin(x)−2 sin3(x) + 2 sin(x)−2 sin3(x)
= 4 sin(x)−4 sin3(x)
24
Step 5: Simplify the expression obtained to prove the desired iden-
tity.
sin(3x) = 4 sin(x)−4 sin3(x)
= 3 sin(x)−4 sin3(x)(proved)
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)is proven
for all real numbers x.
Question 28
Question
Prove the following trigonometric identity:
1−sin2a
1 + sin a= cos a
Solution
To prove the trigonometric identity 1−sin2a
1+sin a= cos a, we will start with the left-
hand side and manipulate it to become the right-hand side using trigonometric
identities.
Step 1: Rewrite the left-hand side of the equation.
1−sin2a
1 + sin a
Step 2: Use the Pythagorean identity sin2a+ cos2a= 1 to simplify the
numerator.
1−sin2a= cos2a
Step 3: Substitute the simplified numerator back into the expression.
cos2a
1 + sin a
Step 4: Factor out a cos afrom the numerator.
cos a·cos a
1 + sin a
Step 5: Use the Pythagorean identity again to simplify the remaining frac-
tion.
cos a·cos a
1 + sin a= cos a·cos a
1 + sin a= cos a·cos a
1 + sin a·1−sin a
1−sin a
Step 6: Further simplify by expanding the denominator.
cos a·cos a−sin acos a
1−sin2a
25
Step 7: Use the double angle identity cos 2a= cos2a−sin2ato simplify
the numerator.
cos a·cos a−sin acos a
cos 2a
Step 8: Substitute cos 2aback into the expression.
cos a·cos a−sin acos a
cos 2a= cos a
Therefore, we have shown that 1−sin2a
1+sin a= cos ais a valid trigonometric iden-
tity.
Question 29
Question
Prove the trigonometric identity:
2 cos2(x)−1 = sin2(x)
Solution
To prove the trigonometric identity 2 cos2(x)−1 = sin2(x), we will start with
the fundamental Pythagorean trigonometric identity:
sin2(x) + cos2(x) = 1
Step 1: Rearrange the Pythagorean identity to solve for sin2(x):
sin2(x) = 1 −cos2(x)
Step 2: Substitute sin2(x)with 1−cos2(x)in the given identity:
2 cos2(x)−1 = sin2(x)
2 cos2(x)−1 = 1 −cos2(x)
2 cos2(x)−1 = cos2(x)
2 cos2(x) = 1 + cos2(x)
Step 3: Subtract cos2(x)from both sides to isolate cos2(x):
2 cos2(x)−cos2(x) = 1
cos2(x) = 1
Thus, we have proven the trigonometric identity 2 cos2(x)−1 = sin2(x).
26
Question 30
Question
Prove the following trigonometric identity:
cos2θ(1 −tan2θ)
1 + sin θtan θ= cos θ−sin θ
Solution
Step 1: Start with the left-hand side (LHS) of the given identity.
LHS =cos2θ(1 −tan2θ)
1 + sin θtan θ
Step 2: Expand the terms in the numerator using the identity tan2θ=
sec2θ−1.
LHS =cos2θ(1 −sec2θ+ 1)
1 + sin θtan θ
Step 3: Simplify the numerator further.
LHS =cos2θ(cos2θ)
1 + sin θtan θ=cos4θ
1 + sin θtan θ
Step 4: Use the Pythagorean identity cos2θ= 1 −sin2θin the numerator.
LHS =(1 −sin2θ)2
1 + sin θtan θ
Step 5: Expand the numerator.
LHS =1−2 sin2θ+ sin4θ
1 + sin θtan θ
Step 6: Recognize that 1−sin2θ= cos2θ.
LHS =cos2θ−2 sin2θ+ sin4θ
1 + sin θtan θ
Step 7: Simplify the numerator further.
LHS =cos2θ(1 −2 tan2θ)
1 + sin θtan θ
Step 8: Use the identity tan2θ= sec2θ−1again in the numerator.
LHS =cos2θ(1 −2 sec2θ+ 2)
1 + sin θtan θ
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Step 9: Simplify the numerator one more time.
LHS =2 cos2θ−2 cos2θsec2θ
1 + sin θtan θ
Step 10: Recognize that 2−2 cos2θ= 2 sin2θ.
LHS =2 sin2θ−2 cos2θsec2θ
1 + sin θtan θ
Step 11: Use the trigonometric identity sec θ=1
cos θin the numerator.
LHS =2 sin2θ−2 cos2θ(1
cos2θ)
1 + sin θtan θ
Step 12: Simplify the numerator further.
LHS =2 sin2θ−2
1 + sin θtan θ
Step 13: Factor out a 2 from the numerator.
LHS =2(sin2θ−1)
1 + sin θtan θ
Step 14: Recognize that sin2θ−1 = −cos2θ.
LHS =2(−cos2θ)
1 + sin θtan θ
Step 15: Simplify the LHS even further.
LHS =−2 cos2θ
1 + sin θtan θ
Step 16: Finally, notice that −2 cos2θ=
Question 31
Question
Prove the following trigonometric identity:
4 sin2θ−3 = cos(2θ)
28
Solution
We will prove the given trigonometric identity using the Pythagorean identity
sin2θ+ cos2θ= 1 and double angle identities.
Step 1: Rewrite cos(2θ)using double angle identities
cos(2θ) = cos2θ−sin2θ
= (1 −sin2θ)−sin2θ(Using the Pythagorean identity)
= 1 −2 sin2θ
Step 2: Substitute cos(2θ)with 1−2 sin2θin the original expression
4 sin2θ−3 = 1 −2 sin2θ
4 sin2θ+ 2 sin2θ= 4
6 sin2θ= 4
sin2θ=2
3
sin θ=±√2
3
Therefore, the given trigonometric identity is true when sin θ=±√2
3.
Question 32
Question
Prove the trigonometric identity:
1 + sin θ
1−sin θ= csc θ+ cot θ
Solution
Start with the LHS: 1 + sin θ
1−sin θ=1
1−sin θ+sin θ
1−sin θ
=1
1−sin θ+sin θ
1−sin θ×1 + sin θ
1 + sin θ
=1
cos2θ+sin θ(1 + sin θ)
cos2θ
=1 + sin θ
cos2θ
=1
cos θ×1
1−sin θ
= csc θ×csc θ
= csc θ+ cot θ
Thus, the LHS equals the RHS. Hence, the trigonometric identity is proven.
29
Question 33
Question
Prove the following trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
Solution
Step 1: Recall the definitions of the trigonometric functions.
cot(θ) = cos(θ)
sin(θ)
csc(θ) = 1
sin(θ)
Step 2: Start with the left side of the identity.
Left side = cot(θ) csc(θ)
=cos(θ)
sin(θ)·1
sin(θ)
=cos(θ)
sin2(θ)
Step 3: Now simplify the right side of the identity.
Right side = csc(θ)−cot(θ)
=1
sin(θ)−cos(θ)
sin(θ)
=1−cos(θ)
sin(θ)
Step 4: Notice that the right side can be rewritten using sin2(θ).
Right side =1−cos(θ)
sin(θ)
=sin2(θ)
sin(θ)
=sin2(θ)
sin2(θ)
Step 5: Compare the two sides of the identity. Since the left side and the
right side are equal, we have proven the trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
30
Question 34
Question
Prove the trigonometric identity:
1−cos(x)
sin(x)= cot (x
2)
Solution
1. We begin by using the double angle formula for cotangent:
cot(2θ) = cos(2θ)
sin(2θ)=cos2(θ)−sin2(θ)
2 sin(θ) cos(θ)
2. Let x= 2θto obtain :
cot(x) = cos2(x
2)−sin2(x
2)
2 sin (x
2)cos (x
2)
3. Next, we apply the half angle formulas for sine and cosine:
cos2(x
2)=1 + cos(x)
2and sin2(x
2)=1−cos(x)
2
4. Substituting these into the expression for cot(x), we get:
cot(x) =
1+cos(x)
2−1−cos(x)
2
2√1−cos(x)
2
1+cos(x)
2
5. Simplifying the right-hand side, we obtain:
cot(x) =
1
2(cos(x)+1−1 + cos(x))
2√1−cos2(x)
4
6. Further simplification gives:
cot(x) = 2 cos(x)
2 sin(x)= cot (x
2)
7. Therefore, we have proven the trigonometric identity:
1−cos(x)
sin(x)= cot (x
2)
31
Question 35
Question
Prove the trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ) cos(4θ)
Solution
Step 1: Recall the double angle formula for cosine, cos(2θ) = cos2(θ)−sin2(θ).
Step 2: Expand the left side of the equation using the difference of squares
formula:
cos4(θ)−sin4(θ) = (cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ))
Step 3: Substitute the double angle formula for cosine into the expanded equa-
tion:
(cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ)) = (cos(2θ) + sin(2θ))(cos(2θ)−sin(2θ))
Step 4: Apply the double angle formula for sine, sin(2θ) = 2 sin(θ) cos(θ), to
the equation:
(cos(2θ)+sin(2θ))(cos(2θ)−sin(2θ)) = (cos(2θ)+2 sin(θ) cos(θ))(cos(2θ)−2 sin(θ) cos(θ))
Step 5: Expand the terms in the equation:
(cos(2θ)+2 sin(θ) cos(θ))(cos(2θ)−2 sin(θ) cos(θ)) = cos2(2θ)−2 sin(2θ) sin(θ) cos(θ)
Step 6: Apply the double angle formula for cosine, cos(2θ) = cos2(θ)−sin2(θ),
and the double angle formula for sine, sin(2θ) = 2 sin(θ) cos(θ), to simplify the
equation further:
cos2(2θ)−2 sin(2θ) sin(θ) cos(θ) = cos2(θ)−sin2(θ) = cos(2θ) cos(4θ)
Therefore, we have shown that cos4(θ)−sin4(θ) = cos(2θ) cos(4θ).
32
