MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 3
Liberty University
Question 2
Question
Prove the following trigonometric identity:
cot(x) sec(x)−tan(x) csc(x) = 2 sin(x) cos(x)
Solution
To prove the given trigonometric identity, we will start by rewriting the left-hand
side using basic trigonometric identities.
Step 1: Rewrite the left-hand side using basic trigonometric identities:
cot(x) sec(x)−tan(x) csc(x) = cos(x)
sin(x)·1
cos(x)−sin(x)
cos(x)·1
sin(x)
⇒cos(x)
sin(x) cos(x)−sin(x)
sin(x) cos(x)=1
sin(x)−1
cos(x)
⇒cos(x)−sin(x)
sin(x) cos(x)
Step 2: Simplify the expression:
cos(x)−sin(x)
sin(x) cos(x)=cos(x)−sin(x)
sin(2x)
2
= 2(cos(x)−sin(x))
= 2 cos(x)−2 sin(x) = 2(cos(x)−sin(x)) = 2 sin(x) cos(x)
Therefore, we have shown that cot(x) sec(x)−tan(x) csc(x) = 2 sin(x) cos(x),
which proves the given trigonometric identity.
Question 3
Question
Prove the following trigonometric identity:
cos x−sin x
cos x+ sin x= tan (π
4−x)
Solution
To prove the given trigonometric identity, we will simplify the left-hand side
until it is equivalent to the right-hand side.
Step 1: Start with the given expression:
cos x−sin x
cos x+ sin x
Step 2: Multiply both the numerator and denominator by cos x:
(cos x−sin x) cos x
(cos x+ sin x) cos x
Step 3: Expand the numerator and denominator:
cos2x−cos xsin x
cos2x+ sin xcos x
Step 4: Recognize that cos2x−sin2x= 1 (a Pythagorean identity) and
simplify: 1−cos xsin x
1 + cos xsin x
Step 5: Use the angle subtraction formula for tangent: tan(a−b) = tan a−tan b
1+tan atan b
with a=π
4and b=x:
tan(π
4)−tan(x)
1 + tan(π
4)tan(x)
Step 6: Simplify to get the right-hand side:
tan (π
4−x)
Therefore, we have proved that cos x−sin x
cos x+sin x= tan (π
4−x).
Question 4
Question
Prove the trigonometric identity:
cos2(x)−sin2(x)
cos(x)−sin(x)= cos(x) + sin(x)
2
Solution
To prove the given trigonometric identity, we will start by simplifying the left-
hand side of the equation.
Step 1: Rewrite the left-hand side using basic trigonometric identities.
cos2(x)−sin2(x)
cos(x)−sin(x)=(cos(x)−sin(x))(cos(x) + sin(x))
cos(x)−sin(x)
Step 2: Cancel out the common factor of cos(x)−sin(x)in the numerator
and denominator.
cos(x) + sin(x)
Step 3: Therefore, the left-hand side equals the right-hand side, which
proves the trigonometric identity.
cos2(x)−sin2(x)
cos(x)−sin(x)= cos(x) + sin(x)
Thus, we have successfully proved the given trigonometric identity.
Question 5
Question
Solve the trigonometric equation sin(3x) = cos(2x)for xin the interval [0,2π).
Solution
Step 1: We know the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x). Substituting these into the given equation, we get:
3 sin(x)−4 sin3(x) = 1 −2 sin2(x)
Step 2: Rearranging the terms, we have:
4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0
Step 3: This is a cubic equation in sin(x). Let’s denote sin(x) = y. Therefore,
the equation becomes:
4y3−2y2−3y+ 1 = 0
Step 4: By trial and error, we find that one of the solutions to this cubic
equation is y= 1, which corresponds to sin(x)=1. However, since sin(x)
cannot exceed 1, this is an extraneous solution and must be discarded.
Step 5: To solve the cubic equation, we need to find the other two roots. We
can factorize the cubic as follows:
(y−1)(4y2+ 2y−1) = 0
3
Step 6: Setting each factor to zero, we get two possible solutions:
y= 1 or 4y2+ 2y−1 = 0
Step 7: We solve the quadratic equation 4y2+2y−1 = 0. Using the quadratic
formula, we find:
y=−2±√22−4∗4∗(−1)
8
y=−2±√20
8
Step 8: Simplifying the root further, we get:
y=−1±√5
4
Step 9: Now, we substitute back sin(x) = yto find the corresponding angles
xin the interval [0,2π):
x= sin−1(−1 + √5
4)
x= sin−1(−1−√5
4)
So, the solutions to the equation sin(3x) = cos(2x)in the interval [0,2π)are
x= sin−1(−1+√5
4)and x= sin−1(−1−√5
4).
Question 6
Question
Prove the trigonometric identity:
1−sin θ
cos θ+1 + sin θ
cos θ= 2 csc θcot θ
Solution
Start with the left-hand side (LHS) of the equation:
1−sin θ
cos θ+1 + sin θ
cos θ=1
cos θ−sin θ
cos θ+1
cos θ+sin θ
cos θ
=1
cos θ+1
cos θ
= 2 csc θ(Recall that csc θ=1
sin θand cot θ=1
tan θ)
= 2 csc θcot θ(Since cot θ=cos θ
sin θ)
Therefore, the given trigonometric identity is verified.
4
Question 7
Question
Solve the trigonometric equation sin2(x) + sin(x)−2=0for xin the interval
[0,2π).
Solution
Step 1: Let u= sin(x). Then the equation becomes u2+u−2 = 0.
Step 2: Solve the quadratic equation u2+u−2=0by factoring as (u+
2)(u−1) = 0. Thus, u=−2or u= 1.
Step 3: Recall that u= sin(x). Therefore, we have two cases to solve: Case
1: u= sin(x) = −2This case has no solutions since the sine function is bounded
between -1 and 1.
Case 2: u= sin(x)=1This case implies x=π
2since sine of π
2is 1 and
0≤x < 2π.
Step 4: Therefore, the solution to the trigonometric equation sin2(x) +
sin(x)−2 = 0 in the interval [0,2π)is x=π
2.
Question 8
Question
Solve the equation cos2(x) + cos(x)−2 = 0 for xin the interval [0,2π).
Solution
Step 1: Let u= cos(x), then the equation becomes u2+u−2 = 0. Step 2: We
can factor the quadratic equation as (u+ 2)(u−1) = 0, which gives u=−2
or u= 1. Step 3: Recall that u= cos(x), so cos(x) = −2is not possible as
cos(x)ranges from -1 to 1. Step 4: Therefore, we have cos(x) = 1. Step 5: The
solutions for cos(x)=1in the interval [0,2π)are x= 0 and x= 2π. Step 6:
Thus, the solutions to the given equation are x= 0 and x= 2π.
Question 9
Question
Solve the trigonometric equation sin2(x) + sin(x)−6=0for xin the interval
[0,2π].
5
Solution
Step 1: Let’s rewrite the given equation as a quadratic equation in terms of
sin(x):
sin2(x) + sin(x)−6 = 0
⇒(sin(x)−2)(sin(x) + 3) = 0
Step 2: We now have the equations sin(x)−2 = 0 or sin(x) + 3 = 0.
Step 3: Solve sin(x)−2 = 0:
sin(x) = 2
This equation has no real solutions since sin(x)is bounded by −1and 1.
Step 4: Solve sin(x) + 3 = 0:
sin(x) = −3
This equation also has no real solutions since sin(x)is bounded by −1and 1.
Step 5: Since the equation sin2(x)+sin(x)−6 = 0 has no real solutions, there
are no solutions to the original trigonometric equation in the interval [0,2π].
Question 10
Question
Solve the trigonometric equation sin(2x)−cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: We’ll start by using the double angle formula for sin(2x):
sin(2x) = 2 sin(x) cos(x)
Step 2: Substituting sin(2x) = 2 sin(x) cos(x)into the given equation gives
us:
2 sin(x) cos(x)−cos(x) = 0
Step 3: Factoring out cos(x), we get:
cos(x)(2 sin(x)−1) = 0
Step 4: Setting each factor to zero gives us two equations:
cos(x) = 0 or 2 sin(x)−1 = 0
Step 5: For cos(x) = 0, the solutions are x=π
2,3π
2.
Step 6: For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which occurs at x=π
6,5π
6.
Step 7: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0≤x≤2πare x=π
6,π
2,5π
6,3π
2.
6
Question 11
Question
Solve the trigonometric equation sin(2x) = cos2(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substituting this identity into the given equation, we get 2 sin(x) cos(x) =
cos2(x).
Step 3: Dividing both sides by cos(x), we obtain 2 sin(x) = cos(x).
Step 4: Remember the Pythagorean identity: sin2(x) + cos2(x) = 1.
Step 5: Squaring both sides of 2 sin(x) = cos(x), we get 4 sin2(x) = cos2(x).
Step 6: Using the Pythagorean identity in 4 sin2(x) = cos2(x), we have
4(1 −cos2(x)) = cos2(x).
Step 7: Distributing and simplifying, we get 4−4 cos2(x) = cos2(x).
Step 8: Rearranging terms, we have 5 cos2(x) = 4.
Step 9: Solving for cos(x), we find cos(x) = ±2
√5.
Step 10: Since we are looking for solutions in the interval [0,2π), we need to
check for extraneous solutions.
Step 11: The cosine function is positive in the first and fourth quadrants.
Step 12: Therefore, the solutions for cos(x) = 2
√5are x= arccos (2
√5).
Step 13: Similarly, the solutions for cos(x) = −2
√5are x= arccos (−2
√5).
Step 14: Evaluating the arccosines, we find x=π
3for cos(x) = 2
√5.
Step 15: And x=5π
3for cos(x) = −2
√5.
Therefore, the solutions to the equation sin(2x) = cos2(x)for xin the inter-
val [0,2π)are x=π
3and x=5π
3.
Question 12
Question
Prove the trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
Solution
To prove the trigonometric identity cot(θ) csc(θ) = csc(θ)−cot(θ), we will ma-
nipulate the left-hand side (LHS) to obtain the right-hand side (RHS).
7
Step 1: Rewrite the left-hand side using the definitions of cotangent and
cosecant.
LHS = cot(θ) csc(θ)
=cos(θ)
sin(θ)·1
sin(θ)
=cos(θ)
sin2(θ)
Step 2: Rewrite the right-hand side using the definitions of cosecant and
cotangent.
RHS = csc(θ)−cot(θ)
=1
sin(θ)−cos(θ)
sin(θ)
=1−cos(θ)
sin(θ)
Step 3: Simplify the expressions in Steps 1 and 2 to see if they are equal.
LHS =cos(θ)
sin2(θ)
=cos(θ)
1−cos2(θ)(using the Pythagorean identity)
=cos(θ)
sin2(θ)·1 + cos(θ)
1 + cos(θ)
=cos(θ)(1 + cos(θ))
sin2(θ)(1 + cos(θ))
=cos(θ) + cos2(θ)
sin2(θ) + cos(θ) sin2(θ)
=cos(θ) + cos2(θ)
sin2(θ) + cos2(θ)
=cos(θ) + cos2(θ)
1
= cos(θ) + cos2(θ)
Therefore, the left-hand side does not simplify to the right-hand side csc(θ)−
cot(θ). Thus, the trigonometric identity cot(θ) csc(θ) = csc(θ)−cot(θ)is not
true.
8
Question 13
Question
Prove the following trigonometric identity:
sin(x)
1−cos(x)+1 + cos(x)
sin(x)= csc(x)
Solution
To prove the given trigonometric identity, we will work on the left-hand side of
the equation and simplify it step by step.
Step 1: Start with the left-hand side of the equation.
sin(x)
1−cos(x)+1 + cos(x)
sin(x)
Step 2: Find a common denominator.
sin(x) sin(x)
sin(x)(1 −cos(x)) +(1 + cos(x))(1 −cos(x))
sin(x)(1 −cos(x))
Step 3: Combine the fractions.
sin2(x) + (1 + cos(x))(1 −cos(x))
sin(x)(1 −cos(x))
Step 4: Expand the numerator.
sin2(x) + (1 −cos2(x))
sin(x)(1 −cos(x))
Step 5: Simplify further using trigonometric identities.
sin2(x) + sin2(x)
sin(x)(1 −cos(x))
Step 6: Simplify the numerator.
2 sin2(x)
sin(x)(1 −cos(x))
Step 7: Simplify the expression.
2 sin(x) sin(x)
sin(x)(1 −cos(x)) =2 sin(x)
1−cos(x)
Step 8: Use the reciprocal identity for sine.
2 csc(x) = csc(x)
Therefore, the given trigonometric identity is proven.
9
Question 14
Question
Prove the following trigonometric identity:
1 + tan θ
1−tan θ= sec2θ
Solution
We know that tan θ=sin θ
cos θand sec θ=1
cos θ
Rewriting the identity, we have 1 + tan θ
1−tan θ=1 + sin θ
cos θ
1−sin θ
cos θ
=
cos θ+sin θ
cos θ
cos θ−sin θ
cos θ
=cos θ+ sin θ
cos θ−sin θ·cos θ
cos θ
=cos θ(cos θ+ sin θ)
cos θ(cos θ−sin θ)
=cos2θ+ cos θsin θ
cos2θ−cos θsin θ
=cos2θ(1 + tan θ)
cos2θ(1 −tan θ)
=1 + tan θ
1−tan θ
= sec2θ
Therefore, the trigonometric identity 1+tan θ
1−tan θ= sec2θis proven.
Question 15
Question
Prove the trigonometric identity:
cos6(x)−sin6(x) = cos(2x)
Solution
To prove the trigonometric identity cos6(x)−sin6(x) = cos(2x), we will use the
trigonometric identity for the cos of double angle.
10
Step 1: Start with the left side of the given identity.
cos6(x)−sin6(x) = (cos2(x))3−(sin2(x))3
= (cos2(x)−sin2(x))(cos4(x) + cos2(x) sin2(x) + sin4(x))
= (cos(2x))((cos2(x) + sin2(x))2−2 cos2(x) sin2(x))
= (cos(2x))(1 −2(cos(2x)))
= cos(2x)−2 cos2(2x)
Step 2: Use the double angle formula for cosine to simplify the expression.
cos(2x)−2 cos2(2x) = 2 cos2(x)−1−2(2 cos2(x)−1)2
= 2 cos2(x)−1−8 cos4(x) + 8 cos2(x)−4
=−8 cos4(x) + 10 cos2(x)−5
Step 3: Simplify the expression further.
−8 cos4(x) + 10 cos2(x)−5 = −(8 cos4(x)−10 cos2(x) + 5)
=−cos(4x)
=−cos(2(2x))
=−cos(2x)
Since cos6(x)−sin6(x) = −cos(2x)and cos(2x) = −cos(2x), we have suc-
cessfully proved the trigonometric identity cos6(x)−sin6(x) = cos(2x).
Question 16
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x)−3 sin(x)−1 = 0
Solution
Step 1: Rewrite the equation using the Pythagorean identity: cos2(x)=1−
sin2(x).
2(1 −sin2(x)) −3 sin(x)−1 = 0
2−2 sin2(x)−3 sin(x)−1 = 0
−2 sin2(x)−3 sin(x) + 1 = 0
Step 2: Rearrange the equation and substitute u= sin(x):
2u2+ 3u−1 = 0
11
Step 3: Solve the quadratic equation by factoring or using the quadratic
formula. The solutions are u=1
2and u=−1.
Step 4: Substitute back to find the values of x: For u=1
2:
sin(x) = 1
2
x=π
6,5π
6
For u=−1:
sin(x) = −1
x=3π
2
Therefore, the solutions to the trigonometric equation are x=π
6,5π
6,3π
2in
the interval [0,2π].
Question 17
Question
Solve the trigonometric equation sin2(x)−√3 cos(x) = 1 for x∈[0,2π].
Solution
Step 1: Rewrite the given equation using the Pythagorean identity for sine and
cosine:
sin2(x)−√3 cos(x) = 1
⇒1−cos2(x)−√3 cos(x) = 1
Step 2: Rearrange the terms and simplify:
cos2(x) + √3 cos(x) = 0
Step 3: Rewrite the equation in terms of cosine:
cos(x)(cos(x) + √3) = 0
Step 4: Find the values of xthat satisfy the equation:
cos(x) = 0 or cos(x) = −√3
Step 5: Solve for xwhen cos(x) = 0: For cos(x) = 0,x=π
2,3π
2
Step 6: Solve for xwhen cos(x) = −√3: Since cos(x)cannot be greater
than 1or less than −1, there are no solutions for cos(x) = −√3in the interval
[0,2π].
Step 7: Therefore, the solutions to the equation sin2(x)−√3 cos(x)=1in
the interval [0,2π]are x=π
2,3π
2.
12
Question 18
Question
Prove the following trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ)
Solution
To prove the trigonometric identity cos4(θ)−sin4(θ) = cos(2θ), we will use
various trigonometric identities and algebraic manipulations.
cos4(θ)−sin4(θ) = (cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ)) (Difference of squares)
= (cos2(θ)−sin2(θ))
= (cos(θ) + sin(θ))(cos(θ)−sin(θ))(cos(θ)−sin(θ)) (Difference of squares again)
= cos(2θ)·(cos(θ)−sin(θ)) (Double angle formula)
= cos(2θ)·cos(θ)−cos(2θ)·sin(θ)
= cos(θ+ 2θ)(Angle sum formula)
= cos(3θ)
Therefore, cos4(θ)−sin4(θ) = cos(3θ)= cos(2θ).
Question 19
Question
Prove the trigonometric identity:
sin2(A)−sin2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Solution
Step 1: First, let’s express sin2(A)and sin2(B)in terms of cos(A)and cos(B)
using the Pythagorean identity sin2(θ) = 1 −cos2(θ).
We have:
sin2(A) = 1 −cos2(A)
sin2(B) = 1 −cos2(B)
Step 2: Substitute the expressions for sin2(A)and sin2(B)into the given
equation to get:
1−cos2(A)−(1 −cos2(B))
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
13
Step 3: Simplify the left side of the equation:
−cos2(A) + cos2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 4: Notice that we can factor out a negative sign from the numerator
of the left side:
−cos2(A)−cos2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 5: Since cos2(θ)−cos2(ϕ) = (cos(θ) + cos(ϕ))(cos(θ)−cos(ϕ)), we can
rewrite the numerator on the left side:
−(cos(A) + cos(B))(cos(A)−cos(B))
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 6: Cancel out the common factor of cos(A) + cos(B)from the numer-
ator and denominator on the left side:
−(cos(A)−cos(B)) = tan(A)−tan(B)
1 + tan(A) tan(B)
Step 7: Finally, use the identity cos(θ)−cos(ϕ) = −2 sin (θ+ϕ
2)sin (θ−ϕ
2)
to simplify the left side:
2 sin (A+B
2)sin (A−B
2)=tan(A)−tan(B)
1 + tan(A) tan(B)
Therefore, we have proven the given trigonometric identity.
Question 20
Question
Solve the trigonometric equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x < 2π.
Solution
Step 1: Let u= sin(x), then we have u2−3u+ 2 = 0.
Step 2: Factor the quadratic equation in uto get (u−1)(u−2) = 0.
Step 3: Set each factor to zero to solve for u:
u−1 = 0 or u−2 = 0
Step 4: Solve for uto get u= 1 or u= 2.
Step 5: Recall that u= sin(x), so we have sin(x) = 1 or sin(x) = 2.
Step 6: Since sin(x)≤1, the equation sin(x) = 2 has no solution. Therefore,
we focus on solving sin(x) = 1.
Step 7: Recall that sin(x) = 1 when x=π
2+ 2πk, where kis an integer.
Step 8: Therefore, the solutions to the trigonometric equation are x=π
2.
14
Question 21
Question
Prove the trigonometric identity:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Solution
To prove the given trigonometric identity, we will begin by expressing tan(x)
and tan(y)in terms of sine and cosine functions. Then, we will manipulate the
expressions to arrive at the right-hand side of the equation.
Step 1: Express tan(x)and tan(y)in terms of sine and cosine:
tan(x) = sin(x)
cos(x),tan(y) = sin(y)
cos(y)
Step 2: Multiply tan(x)and tan(y)together:
tan(x) tan(y) = sin(x) sin(y)
cos(x) cos(y)
Step 3: Use the trigonometric identity cos(a+b) = cos(a) cos(b)−sin(a) sin(b):
cos(x+y) = cos(x) cos(y)−sin(x) sin(y)
Step 4: Rearrange the equation to solve for cos(x) cos(y):
cos(x) cos(y) = cos(x+y) + sin(x) sin(y)
Step 5: Substitute cos(x) cos(y)back into the original expression:
tan(x) tan(y) = sin(x) sin(y)
1−(cos(x+y) + sin(x) sin(y))
Step 6: Combine like terms and simplify:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)−sin2(x)
Step 7: Apply the trigonometric identity sin2(x) = 1 −cos2(x):
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)−(1 −cos2(x))
Step 8: Simplify further:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y) + cos2(x)
15
Step 9: Use the Pythagorean identity cos2(x) = 1 −sin2(x):
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)+1−sin2(x)
Step 10: Simplify and rearrange terms:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)+1−(1 −cos(x) cos(y))
Step 11: Combine like terms and simplify:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y) + cos(x) cos(y)
Step 12: Use the trigonometric identity cos(x+y) = cos(x) cos(y)−sin(x) sin(y)
one more time:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Therefore, we have proved the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y).
Question 22
Question
Solve the trigonometric equation cos2(x)−sin(x) = 0 for x∈[0,2π].
Solution
Step 1: We can rewrite the equation cos2(x)−sin(x) = 0 using the Pythagorean
identity cos2(x)=1−sin2(x). Substituting this into the equation, we get
1−sin2(x)−sin(x) = 0.
Step 2: Rearranging the equation, we have sin2(x) + sin(x)−1 = 0. This is
a quadratic equation in terms of sin(x).
Step 3: To solve the quadratic equation sin2(x) + sin(x)−1 = 0, we can use
the quadratic formula: sin(x) = −b±√b2−4ac
2a, where a= 1,b= 1, and c=−1.
Step 4: Plugging in the values of a,b, and c, we get sin(x) = −1±√1+4
2,
which simplifies to sin(x) = −1±√5
2.
Step 5: The solutions for sin(x)are sin(x) = −1+√5
2and sin(x) = −1−√5
2.
However, we need to check if these solutions are within the interval [0,2π].
Step 6: Since sin is negative for angles between πand 2π, the solution
sin(x) = −1−√5
2is not in the interval [0,2π]. So, we only consider sin(x) =
−1+√5
2.
Step 7: To find the corresponding values of x, we take the arcsin of −1+√5
2
and 2π−arcsin (−1+√5
2).
16
Step 8: Using a calculator, we find that x≈0.6283 radians and x≈5.6549
radians are the solutions in the interval [0,2π].
Therefore, the solutions to the trigonometric equation cos2(x)−sin(x) = 0
for x∈[0,2π]are x≈0.6283 and x≈5.6549 radians.
Question 23
Question
Simplify the expression √3 cos(θ)−sin(θ).
Solution
Step 1: Recall the trigonometric identity sin(π
3)=√3
2and cos(π
3)=1
2.
Step 2: Let’s express √3 cos(θ)−sin(θ)in terms of sines and cosines with
θ=π
3.√3 cos(θ)−sin(θ) = √3 cos (π
3)−sin (π
3)
Step 3: Substitute the values of cos(π
3)and sin(π
3).
=√3·1
2−√3
2
Step 4: Simplify the expression.
=√3
2−√3
2= 0
Therefore, √3 cos(θ)−sin(θ) = 0.
Question 24
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 2 csc(x) sec(x)
Solution
To prove the trigonometric identity, we will start by working with the left-hand
side (LHS) of the equation and manipulate it until it matches the right-hand
side (RHS) of the equation.
Step 1: Start with the LHS of the equation.
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)
17
Step 2: Rewrite sin2(x)and cos2(x)using the Pythagorean identity: sin2(x) =
1−cos2(x)and cos2(x) = 1 −sin2(x).
1−cos2(x)
1−cos(x)+1−sin2(x)
1 + sin(x)
Step 3: Simplify each fraction separately.
1−cos2(x)
1−cos(x)=(1 + cos(x))(1 −cos(x))
1−cos(x)= 1 + cos(x)
1−sin2(x)
1 + sin(x)=(1 + sin(x))(1 −sin(x))
1 + sin(x)= 1 −sin(x)
Step 4: Combine the simplified fractions.
1 + cos(x)+1−sin(x)
Step 5: Simplify the expression further.
2−sin(x) + cos(x)
Step 6: Use trigonometric identities to simplify the expression further.
sin(x) = 1
csc(x)and cos(x) = 1
sec(x)
2−1
csc(x)+1
sec(x)= 2 csc(x) sec(x)
Therefore, we have shown that the LHS is equal to the RHS, and the trigono-
metric identity is proven.
Question 25
Question
Prove the following trigonometric identity:
1
sin(α) cos(α)=2
sin(2α)
Solution
Step 1: Begin with the right-hand side of the equation and apply the double-
angle formula for sine.
RHS =2
sin(2α)=2
2 sin(α) cos(α)=1
sin(α) cos(α)
Therefore, the given trigonometric identity is proved.
18
Question 26
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will first rewrite
the left side in terms of sines and cosines using trigonometric identities.
Step 1: Rewrite the left side in terms of sines and cosines
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
= (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= 1 ·(sin2(x)−cos2(x))
= sin2(x)−cos2(x)
Step 2: Use the double angle formula for sine Recall the double angle
formula for sine: sin(2θ) = 2 sin(θ) cos(θ). We can use this formula with θ=x
to rewrite the right side.
sin2(x)−cos2(x) = −cos(2x)
= sin(2x+π
2)(using the angle sum identity for cosine)
Step 3: Use the double angle formula for sine again Now, apply the
double angle formula for sine again: sin(2θ) = 2 sin(θ) cos(θ). This time, let
θ= 2x+π
4.
sin(2x+π
2)= sin(2(2x+π
4))
= 2 sin(2x+π
4)cos(2x+π
4)
Step 4: Simplify the expression Now, we expand and simplify the ex-
pression.
2 sin(2x+π
4)cos(2x+π
4)= 2(sin(2x) cos(π
4)+ cos(2x) sin(π
4))
= 2(sin(2x)√2
2+ cos(2x)√2
2)
=√2(sin(2x) + cos(2x))
=√2 sin(2x+π
4)
Therefore, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x).
19
Question 27
Question
Prove the following trigonometric identity:
1 + cos θ
sin θ= csc θ+ cot θ
Solution
Step 1: Start with the left-hand side of the identity.
1 + cos θ
sin θ
Step 2: Use the reciprocal identities to rewrite csc θand cot θ:
csc θ=1
sin θand cot θ=cos θ
sin θ
Step 3: Substitute the rewritten forms of csc θand cot θinto the expression:
1
sin θ+cos θ
sin θ
Step 4: Simplify by combining the fractions:
1 + cos θ
sin θ
Step 5: Notice that the simplified expression is equal to the original left-hand
side: 1 + cos θ
sin θ=1 + cos θ
sin θ
Step 6: Therefore, the trigonometric identity
1 + cos θ
sin θ= csc θ+ cot θ
is proven.
Question 28
Question
Prove the identity:
cos3(x)
1−sin(x)=1 + sin(x) + sin2(x)
cos2(x)
20
Solution
Step 1: Rewrite the expression using trigonometric identities.
cos3(x)
1−sin(x)=cos2(x)·cos(x)
1−sin(x)=cos2(x)·cos(x)
1−sin(x)×1 + sin(x)
1 + sin(x)
Step 2: Simplify the expression using the difference of squares formula and
Pythagorean identity.
cos2(x)·cos(x)·(1 + sin(x))
1−sin2(x)=cos2(x)·cos(x)·(1 + sin(x))
cos2(x)
Step 3: Cancel out the common factor of cos2(x).
cos(x)·(1 + sin(x))
1= cos(x) + cos(x) sin(x)
Step 4: Rewrite the expression on the right hand side using trigonometric
identities.
cos(x) + cos(x) sin(x) = cos(x) + sin(x) cos(x)
cos(x)= cos(x) + tan(x)
Step 5: Recall that tan(x) = sin(x)
cos(x).
cos(x) + tan(x) = cos(x) + sin(x)
cos(x)=cos2(x) + sin(x)
cos(x)
Step 6: Since cos3(x)
1−sin(x)=cos2(x)+sin(x)
cos(x), we have proven the identity.
Question 29
Question
Solve the trigonometric equation tan3(x) = sec(x)for xin the interval [0,2π).
Solution
To solve the trigonometric equation tan3(x) = sec(x), we need to utilize the
trigonometric identities relating tangent and secant functions.
Step 1: Rewrite in terms of sine and cosine First, recall that tan(x) =
sin(x)
cos(x)and sec(x) = 1
cos(x). We rewrite the equation in terms of sine and cosine:
(sin(x)
cos(x))3
=1
cos(x)
21
Step 2: Simplify the equation Simplify the equation by cubing the frac-
tion on the left side: sin3(x)
cos3(x)=1
cos(x)
Step 3: Multiply both sides by cos4(x)To clear the denominators,
multiply both sides by cos4(x):
sin3(x) = cos4(x)
Step 4: Use the Pythagorean Identity We know that sin2(x)+cos2(x) =
1, so cos2(x) = 1 −sin2(x). Substitute this into the equation:
sin3(x) = (1 −sin2(x))2
Step 5: Solve for sin(x)Expand the right side to get a polynomial equation
in terms of sin(x):
sin3(x) = 1 −2 sin2(x) + sin4(x)
Step 6: Rearrange into a quadratic equation Rearrange the equation
to form a quadratic equation in terms of sin(x):
sin4(x)−sin2(x)−sin(x) + 1 = 0
Step 7: Solve the quadratic equation for sin(x)This quadratic equation
may be factored into:
(sin2(x)−sin(x)−1)(sin2(x) + sin(x)−1) = 0
Step 8: Solve for sin(x)using the quadratic formula Solve each
quadratic factor separately using the quadratic formula. The solutions for sin(x)
will lead to solutions for xwhen considering the interval [0,2π). Subsequently,
use the angles whose sine gives those solutions.
Therefore, the solutions for xin the interval [0,2π)are given by the angles
whose sine is the solution for sin(x).
Question 30
Question
Prove the trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x).
22
Solution
Step 1: Start with the triple angle formula for cosine:
cos(3x) = cos(2x+x).
Step 2: Expand the right side using the angle sum formula for cosine:
cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x).
Step 3: Recall the double angle formulas:
cos(2x) = cos2(x)−sin2(x)and sin(2x) = 2 sin(x) cos(x).
Step 4: Substitute the double angle formulas into the expression for cos(2x)
and sin(2x)in the previous step:
cos(2x) cos(x)−sin(2x) sin(x) = (cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x).
Step 5: Simplify the expression:
(cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x) = cos3(x)−sin2(x) cos(x)−2 sin(x) cos2(x).
Step 6: Use the Pythagorean identity sin2(x) + cos2(x)=1to express
sin2(x)in terms of cos(x):
cos3(x)−sin2(x) cos(x)−2 sin(x) cos2(x) = cos3(x)−(1−cos2(x)) cos(x)−2 sin(x) cos2(x).
Step 7: Further simplify the expression:
cos3(x)−(1−cos2(x)) cos(x)−2 sin(x) cos2(x) = cos3(x)−cos(x)+cos3(x)−2 sin(x) cos2(x).
Step 8: Use the Pythagorean identity sin2(x) = 1 −cos2(x)to replace
sin(x) cos2(x)with cos3(x)−cos(x):
cos3(x)−cos(x) + cos3(x)−2 sin(x) cos2(x) = 4 cos3(x)−3 cos(x).
Therefore, cos(3x) = 4 cos3(x)−3 cos(x), which proves the identity.
Question 31
Question
Solve the trigonometric equation sin x+ cos x=√2 cos(x−π
4)for xin the
interval [0,2π].
23
Solution
Step 1: Square both sides of the equation to eliminate the square root.
sin2x+ 2 sin xcos x+ cos2x= 2 cos2(x−π
4)
Step 2: Use the Pythagorean identities sin2x+cos2x= 1 and cos2(x−π
4) =
1
2(1 + cos(2x−π
2))to simplify the equation.
1 + 2 sin xcos x= 1 + cos(2x−π
2)
Step 3: Simplify the equation.
cos(2x−π
2)= 2 sin xcos x
Step 4: Use the double angle formula cos(2θ) = 2 cos2θ−1.
2 cos2(x−π
4)−1 = 2 sin xcos x
Step 5: Apply the Pythagorean identity cos2(x−π
4) = 1 −sin2(x−π
4)and
simplify.
2(1 −sin2(x−π
4)) −1 = 2 sin xcos x
Step 6: Expand and simplify the equation.
2−2 sin2(x−π
4)−1 = 2 sin xcos x
1−2 sin2(x−π
4) = 2 sin xcos x
Step 7: Use the double angle formula for sine to expand sin(x−π
4)
1−2(1
2sin xcos π
4−1
2cos xsin π
4)2
= 2 sin xcos x
Step 8: Simplify and solve the equation.
1−1
2sin2x−1
2cos2x= 2 sin xcos x
1−1
2= 2 sin xcos x
1
2= 2 sin xcos x
Step 9: Use the double angle formula for sine to rewrite sin 2x
sin 2x= 2 sin xcos x
24
Step 10: Substitute sin 2x=1
2and solve for 2x.
1
2=1
2
2x=π
6
Step 11: Find the solutions for xin the interval [0,2π].
x=π
12,5π
12
Question 32
Question
Prove the trigonometric identity:
sin2(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= 1
Solution
Step 1: Start by expressing sin2(θ)
1+cos(θ)as a compound fraction:
sin2(θ)
1 + cos(θ)=sin2(θ)
1 + cos(θ)×1−cos(θ)
1−cos(θ)=sin2(θ)(1 −cos(θ))
1−cos2(θ)
Step 2: Simplify the expression from Step 1:
sin2(θ)(1 −cos(θ))
1−cos2(θ)=sin2(θ)−sin2(θ) cos(θ)
sin2(θ)= 1 −cos(θ)
Step 3: Express cos(θ)
1+sin(θ)as a compound fraction:
cos(θ)
1 + sin(θ)=cos(θ)
1 + sin(θ)×1−sin(θ)
1−sin(θ)=cos(θ)(1 −sin(θ))
1−sin2(θ)
Step 4: Simplify the expression from Step 3:
cos(θ)(1 −sin(θ))
1−sin2(θ)=cos(θ)−sin(θ) cos(θ)
cos2(θ)= 1 −sin(θ)
Step 5: Adding the simplified expressions from Step 2 and Step 4:
1−cos(θ)+1−sin(θ) = 2 −(cos(θ) + sin(θ))
Step 6: Recall that cos(θ)+sin(θ) = √2 sin (θ+π
4). Therefore, substituting
this into the expression from Step 5:
2−√2 sin (θ+π
4)
Thus, the original identity sin2(θ)
1+cos(θ)+cos(θ)
1+sin(θ)= 1 is proven.
25
Question 33
Question
Solve the equation sin(3x) + sin(2x) = 0 for 0≤x≤2π.
Solution
Step 1: Utilize the sum-to-product trigonometric identity to simplify the equa-
tion. Step 2: Apply trigonometric identities to solve for x.
Step 1: Using the sum-to-product trigonometric identity
sin(a) + sin(b) = 2 sin (a+b
2)cos (a−b
2),
we can rewrite the equation as
2 sin (3x+ 2x
2)cos (3x−2x
2)= 0.
Simplifying this expression gives
sin(2.5x) cos(0.5x) = 0.
Step 2: Now, we have the equation sin(2.5x) cos(0.5x)=0. To solve this
equation, we need to consider two cases where either sin(2.5x) = 0 or cos(0.5x) =
0.
Case 1: sin(2.5x)=0This occurs when 2.5x=nπ for n∈Z. Solving for x
gives x=nπ
2.5.
Case 2: cos(0.5x)=0This occurs when 0.5x=π
2+nπ for n∈Z. Solving
for xgives x=π
2+ 2nπ.
Therefore, the solutions to the equation sin(3x) + sin(2x) = 0 for 0≤x≤2π
are x=nπ
2.5and x=π
2+ 2nπ, where nis an integer such that 0≤x≤2π.
Question 34
Question
Prove the following trigonometric identity:
sin4x
cos2x+cos4x
sin2x= tan2x+ cot2x
Solution
To prove the given trigonometric identity, we will start with one side of the
equation and manipulate it to be equal to the other side.
26
Step 1: Rewrite the terms using basic trigonometric identities:
sin4x
cos2x+cos4x
sin2x
Step 2: Express sin4xand cos4xin terms of sin 2xand cos 2x:
(sin2x)2
cos2x+(cos2x)2
sin2x
Step 3: Express sin2xand cos2xin terms of sin 2xand cos 2x:
(1−cos 2x
2)2
cos2x+(1+cos 2x
2)2
sin2x
Step 4: Simplify the expressions by expanding the squares:
1−2 cos 2x+cos22x
4
cos2x+
1+2 cos 2x+cos22x
4
sin2x
Step 5: Simplify further by multiplying the fractions and combining like
terms: 1−2 cos 2x+ cos22x
4 cos2x+1 + 2 cos 2x+ cos22x
4 sin2x
Step 6: Use the double angle identities cos 2x= 1 −2 sin2xand sin 2x=
2 sin xcos x:
1−2(1 −2 sin2x) + (1 −2 sin2x)2
4(1 −sin2x)+1 + 2(1 −2 sin2x) + (1 −2 sin2x)2
4 sin2x
Step 7: Simplify and factorize the expressions:
1 + 4 sin2x−2 + 4 sin2x+ 1 −4 sin2x+ 4 sin4x
4 cos2x
Step 8: Simplify the expression further:
4 sin4x
4 cos2x= tan2x
Step 9: Since the right-hand side of the identity is tan2x+ cot2x, we
have shown that the left-hand side is equal to the right-hand side. Thus, the
trigonometric identity is proven.
Question 35
Question
Solve the trigonometric equation sin(3x) = cos(2x)for xin the interval [0,2π].
27
Solution
Step 1: Recall the trigonometric identities:
•sin(2θ) = 2 sin(θ) cos(θ)
•cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Rewrite the equation using these identities:
sin(3x) = cos(2x)
sin(2x+x) = cos2(x)−sin2(x)
Step 3: Use the double angle identity for sin to simplify:
2 sin(x) cos(x)·cos(x)−(1 −cos2(x)) = 0
2 sin(x) cos2(x)−1 + cos2(x) = 0
Step 4: Substitute sin(x) = cos(x)
2:
2(cos(x)
2)cos2(x)−1 + cos2(x) = 0
cos3(x)−2 cos(x) + 1 = 0
Step 5: Let u= cos(x)to simplify the equation:
u3−2u+ 1 = 0
Step 6: Factor the equation using synthetic division, graphical methods, or
numerical methods. The roots are u= 1, u =−0.618, u = 0.618.
Step 7: Convert uback to cos(x)to find the corresponding values of x:
cos(x) = 1 =⇒x= 0
cos(x) = −0.618 =⇒x≈2.223
cos(x) = 0.618 =⇒x≈0.941
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π]are x= 0,0.941,2.223.
28
Question 3
Question
Prove the following trigonometric identity:
cos x−sin x
cos x+ sin x= tan (π
4−x)
Solution
To prove the given trigonometric identity, we will simplify the left-hand side
until it is equivalent to the right-hand side.
Step 1: Start with the given expression:
cos x−sin x
cos x+ sin x
Step 2: Multiply both the numerator and denominator by cos x:
(cos x−sin x) cos x
(cos x+ sin x) cos x
Step 3: Expand the numerator and denominator:
cos2x−cos xsin x
cos2x+ sin xcos x
Step 4: Recognize that cos2x−sin2x= 1 (a Pythagorean identity) and
simplify: 1−cos xsin x
1 + cos xsin x
Step 5: Use the angle subtraction formula for tangent: tan(a−b) = tan a−tan b
1+tan atan b
with a=π
4and b=x:
tan(π
4)−tan(x)
1 + tan(π
4)tan(x)
Step 6: Simplify to get the right-hand side:
tan (π
4−x)
Therefore, we have proved that cos x−sin x
cos x+sin x= tan (π
4−x).
Question 4
Question
Prove the trigonometric identity:
cos2(x)−sin2(x)
cos(x)−sin(x)= cos(x) + sin(x)
2
Solution
To prove the given trigonometric identity, we will start by simplifying the left-
hand side of the equation.
Step 1: Rewrite the left-hand side using basic trigonometric identities.
cos2(x)−sin2(x)
cos(x)−sin(x)=(cos(x)−sin(x))(cos(x) + sin(x))
cos(x)−sin(x)
Step 2: Cancel out the common factor of cos(x)−sin(x)in the numerator
and denominator.
cos(x) + sin(x)
Step 3: Therefore, the left-hand side equals the right-hand side, which
proves the trigonometric identity.
cos2(x)−sin2(x)
cos(x)−sin(x)= cos(x) + sin(x)
Thus, we have successfully proved the given trigonometric identity.
Question 5
Question
Solve the trigonometric equation sin(3x) = cos(2x)for xin the interval [0,2π).
Solution
Step 1: We know the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x). Substituting these into the given equation, we get:
3 sin(x)−4 sin3(x) = 1 −2 sin2(x)
Step 2: Rearranging the terms, we have:
4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0
Step 3: This is a cubic equation in sin(x). Let’s denote sin(x) = y. Therefore,
the equation becomes:
4y3−2y2−3y+ 1 = 0
Step 4: By trial and error, we find that one of the solutions to this cubic
equation is y= 1, which corresponds to sin(x)=1. However, since sin(x)
cannot exceed 1, this is an extraneous solution and must be discarded.
Step 5: To solve the cubic equation, we need to find the other two roots. We
can factorize the cubic as follows:
(y−1)(4y2+ 2y−1) = 0
3
Step 6: Setting each factor to zero, we get two possible solutions:
y= 1 or 4y2+ 2y−1 = 0
Step 7: We solve the quadratic equation 4y2+2y−1 = 0. Using the quadratic
formula, we find:
y=−2±√22−4∗4∗(−1)
8
y=−2±√20
8
Step 8: Simplifying the root further, we get:
y=−1±√5
4
Step 9: Now, we substitute back sin(x) = yto find the corresponding angles
xin the interval [0,2π):
x= sin−1(−1 + √5
4)
x= sin−1(−1−√5
4)
So, the solutions to the equation sin(3x) = cos(2x)in the interval [0,2π)are
x= sin−1(−1+√5
4)and x= sin−1(−1−√5
4).
Question 6
Question
Prove the trigonometric identity:
1−sin θ
cos θ+1 + sin θ
cos θ= 2 csc θcot θ
Solution
Start with the left-hand side (LHS) of the equation:
1−sin θ
cos θ+1 + sin θ
cos θ=1
cos θ−sin θ
cos θ+1
cos θ+sin θ
cos θ
=1
cos θ+1
cos θ
= 2 csc θ(Recall that csc θ=1
sin θand cot θ=1
tan θ)
= 2 csc θcot θ(Since cot θ=cos θ
sin θ)
Therefore, the given trigonometric identity is verified.
4
Question 7
Question
Solve the trigonometric equation sin2(x) + sin(x)−2=0for xin the interval
[0,2π).
Solution
Step 1: Let u= sin(x). Then the equation becomes u2+u−2 = 0.
Step 2: Solve the quadratic equation u2+u−2=0by factoring as (u+
2)(u−1) = 0. Thus, u=−2or u= 1.
Step 3: Recall that u= sin(x). Therefore, we have two cases to solve: Case
1: u= sin(x) = −2This case has no solutions since the sine function is bounded
between -1 and 1.
Case 2: u= sin(x)=1This case implies x=π
2since sine of π
2is 1 and
0≤x < 2π.
Step 4: Therefore, the solution to the trigonometric equation sin2(x) +
sin(x)−2 = 0 in the interval [0,2π)is x=π
2.
Question 8
Question
Solve the equation cos2(x) + cos(x)−2 = 0 for xin the interval [0,2π).
Solution
Step 1: Let u= cos(x), then the equation becomes u2+u−2 = 0. Step 2: We
can factor the quadratic equation as (u+ 2)(u−1) = 0, which gives u=−2
or u= 1. Step 3: Recall that u= cos(x), so cos(x) = −2is not possible as
cos(x)ranges from -1 to 1. Step 4: Therefore, we have cos(x) = 1. Step 5: The
solutions for cos(x)=1in the interval [0,2π)are x= 0 and x= 2π. Step 6:
Thus, the solutions to the given equation are x= 0 and x= 2π.
Question 9
Question
Solve the trigonometric equation sin2(x) + sin(x)−6=0for xin the interval
[0,2π].
5
Solution
Step 1: Let’s rewrite the given equation as a quadratic equation in terms of
sin(x):
sin2(x) + sin(x)−6 = 0
⇒(sin(x)−2)(sin(x) + 3) = 0
Step 2: We now have the equations sin(x)−2 = 0 or sin(x) + 3 = 0.
Step 3: Solve sin(x)−2 = 0:
sin(x) = 2
This equation has no real solutions since sin(x)is bounded by −1and 1.
Step 4: Solve sin(x) + 3 = 0:
sin(x) = −3
This equation also has no real solutions since sin(x)is bounded by −1and 1.
Step 5: Since the equation sin2(x)+sin(x)−6 = 0 has no real solutions, there
are no solutions to the original trigonometric equation in the interval [0,2π].
Question 10
Question
Solve the trigonometric equation sin(2x)−cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: We’ll start by using the double angle formula for sin(2x):
sin(2x) = 2 sin(x) cos(x)
Step 2: Substituting sin(2x) = 2 sin(x) cos(x)into the given equation gives
us:
2 sin(x) cos(x)−cos(x) = 0
Step 3: Factoring out cos(x), we get:
cos(x)(2 sin(x)−1) = 0
Step 4: Setting each factor to zero gives us two equations:
cos(x) = 0 or 2 sin(x)−1 = 0
Step 5: For cos(x) = 0, the solutions are x=π
2,3π
2.
Step 6: For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which occurs at x=π
6,5π
6.
Step 7: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0≤x≤2πare x=π
6,π
2,5π
6,3π
2.
6
Question 11
Question
Solve the trigonometric equation sin(2x) = cos2(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substituting this identity into the given equation, we get 2 sin(x) cos(x) =
cos2(x).
Step 3: Dividing both sides by cos(x), we obtain 2 sin(x) = cos(x).
Step 4: Remember the Pythagorean identity: sin2(x) + cos2(x) = 1.
Step 5: Squaring both sides of 2 sin(x) = cos(x), we get 4 sin2(x) = cos2(x).
Step 6: Using the Pythagorean identity in 4 sin2(x) = cos2(x), we have
4(1 −cos2(x)) = cos2(x).
Step 7: Distributing and simplifying, we get 4−4 cos2(x) = cos2(x).
Step 8: Rearranging terms, we have 5 cos2(x) = 4.
Step 9: Solving for cos(x), we find cos(x) = ±2
√5.
Step 10: Since we are looking for solutions in the interval [0,2π), we need to
check for extraneous solutions.
Step 11: The cosine function is positive in the first and fourth quadrants.
Step 12: Therefore, the solutions for cos(x) = 2
√5are x= arccos (2
√5).
Step 13: Similarly, the solutions for cos(x) = −2
√5are x= arccos (−2
√5).
Step 14: Evaluating the arccosines, we find x=π
3for cos(x) = 2
√5.
Step 15: And x=5π
3for cos(x) = −2
√5.
Therefore, the solutions to the equation sin(2x) = cos2(x)for xin the inter-
val [0,2π)are x=π
3and x=5π
3.
Question 12
Question
Prove the trigonometric identity:
cot(θ) csc(θ) = csc(θ)−cot(θ)
Solution
To prove the trigonometric identity cot(θ) csc(θ) = csc(θ)−cot(θ), we will ma-
nipulate the left-hand side (LHS) to obtain the right-hand side (RHS).
7
Step 1: Rewrite the left-hand side using the definitions of cotangent and
cosecant.
LHS = cot(θ) csc(θ)
=cos(θ)
sin(θ)·1
sin(θ)
=cos(θ)
sin2(θ)
Step 2: Rewrite the right-hand side using the definitions of cosecant and
cotangent.
RHS = csc(θ)−cot(θ)
=1
sin(θ)−cos(θ)
sin(θ)
=1−cos(θ)
sin(θ)
Step 3: Simplify the expressions in Steps 1 and 2 to see if they are equal.
LHS =cos(θ)
sin2(θ)
=cos(θ)
1−cos2(θ)(using the Pythagorean identity)
=cos(θ)
sin2(θ)·1 + cos(θ)
1 + cos(θ)
=cos(θ)(1 + cos(θ))
sin2(θ)(1 + cos(θ))
=cos(θ) + cos2(θ)
sin2(θ) + cos(θ) sin2(θ)
=cos(θ) + cos2(θ)
sin2(θ) + cos2(θ)
=cos(θ) + cos2(θ)
1
= cos(θ) + cos2(θ)
Therefore, the left-hand side does not simplify to the right-hand side csc(θ)−
cot(θ). Thus, the trigonometric identity cot(θ) csc(θ) = csc(θ)−cot(θ)is not
true.
8
Question 13
Question
Prove the following trigonometric identity:
sin(x)
1−cos(x)+1 + cos(x)
sin(x)= csc(x)
Solution
To prove the given trigonometric identity, we will work on the left-hand side of
the equation and simplify it step by step.
Step 1: Start with the left-hand side of the equation.
sin(x)
1−cos(x)+1 + cos(x)
sin(x)
Step 2: Find a common denominator.
sin(x) sin(x)
sin(x)(1 −cos(x)) +(1 + cos(x))(1 −cos(x))
sin(x)(1 −cos(x))
Step 3: Combine the fractions.
sin2(x) + (1 + cos(x))(1 −cos(x))
sin(x)(1 −cos(x))
Step 4: Expand the numerator.
sin2(x) + (1 −cos2(x))
sin(x)(1 −cos(x))
Step 5: Simplify further using trigonometric identities.
sin2(x) + sin2(x)
sin(x)(1 −cos(x))
Step 6: Simplify the numerator.
2 sin2(x)
sin(x)(1 −cos(x))
Step 7: Simplify the expression.
2 sin(x) sin(x)
sin(x)(1 −cos(x)) =2 sin(x)
1−cos(x)
Step 8: Use the reciprocal identity for sine.
2 csc(x) = csc(x)
Therefore, the given trigonometric identity is proven.
9
Question 14
Question
Prove the following trigonometric identity:
1 + tan θ
1−tan θ= sec2θ
Solution
We know that tan θ=sin θ
cos θand sec θ=1
cos θ
Rewriting the identity, we have 1 + tan θ
1−tan θ=1 + sin θ
cos θ
1−sin θ
cos θ
=
cos θ+sin θ
cos θ
cos θ−sin θ
cos θ
=cos θ+ sin θ
cos θ−sin θ·cos θ
cos θ
=cos θ(cos θ+ sin θ)
cos θ(cos θ−sin θ)
=cos2θ+ cos θsin θ
cos2θ−cos θsin θ
=cos2θ(1 + tan θ)
cos2θ(1 −tan θ)
=1 + tan θ
1−tan θ
= sec2θ
Therefore, the trigonometric identity 1+tan θ
1−tan θ= sec2θis proven.
Question 15
Question
Prove the trigonometric identity:
cos6(x)−sin6(x) = cos(2x)
Solution
To prove the trigonometric identity cos6(x)−sin6(x) = cos(2x), we will use the
trigonometric identity for the cos of double angle.
10
Step 1: Start with the left side of the given identity.
cos6(x)−sin6(x) = (cos2(x))3−(sin2(x))3
= (cos2(x)−sin2(x))(cos4(x) + cos2(x) sin2(x) + sin4(x))
= (cos(2x))((cos2(x) + sin2(x))2−2 cos2(x) sin2(x))
= (cos(2x))(1 −2(cos(2x)))
= cos(2x)−2 cos2(2x)
Step 2: Use the double angle formula for cosine to simplify the expression.
cos(2x)−2 cos2(2x) = 2 cos2(x)−1−2(2 cos2(x)−1)2
= 2 cos2(x)−1−8 cos4(x) + 8 cos2(x)−4
=−8 cos4(x) + 10 cos2(x)−5
Step 3: Simplify the expression further.
−8 cos4(x) + 10 cos2(x)−5 = −(8 cos4(x)−10 cos2(x) + 5)
=−cos(4x)
=−cos(2(2x))
=−cos(2x)
Since cos6(x)−sin6(x) = −cos(2x)and cos(2x) = −cos(2x), we have suc-
cessfully proved the trigonometric identity cos6(x)−sin6(x) = cos(2x).
Question 16
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x)−3 sin(x)−1 = 0
Solution
Step 1: Rewrite the equation using the Pythagorean identity: cos2(x)=1−
sin2(x).
2(1 −sin2(x)) −3 sin(x)−1 = 0
2−2 sin2(x)−3 sin(x)−1 = 0
−2 sin2(x)−3 sin(x) + 1 = 0
Step 2: Rearrange the equation and substitute u= sin(x):
2u2+ 3u−1 = 0
11
Step 3: Solve the quadratic equation by factoring or using the quadratic
formula. The solutions are u=1
2and u=−1.
Step 4: Substitute back to find the values of x: For u=1
2:
sin(x) = 1
2
x=π
6,5π
6
For u=−1:
sin(x) = −1
x=3π
2
Therefore, the solutions to the trigonometric equation are x=π
6,5π
6,3π
2in
the interval [0,2π].
Question 17
Question
Solve the trigonometric equation sin2(x)−√3 cos(x) = 1 for x∈[0,2π].
Solution
Step 1: Rewrite the given equation using the Pythagorean identity for sine and
cosine:
sin2(x)−√3 cos(x) = 1
⇒1−cos2(x)−√3 cos(x) = 1
Step 2: Rearrange the terms and simplify:
cos2(x) + √3 cos(x) = 0
Step 3: Rewrite the equation in terms of cosine:
cos(x)(cos(x) + √3) = 0
Step 4: Find the values of xthat satisfy the equation:
cos(x) = 0 or cos(x) = −√3
Step 5: Solve for xwhen cos(x) = 0: For cos(x) = 0,x=π
2,3π
2
Step 6: Solve for xwhen cos(x) = −√3: Since cos(x)cannot be greater
than 1or less than −1, there are no solutions for cos(x) = −√3in the interval
[0,2π].
Step 7: Therefore, the solutions to the equation sin2(x)−√3 cos(x)=1in
the interval [0,2π]are x=π
2,3π
2.
12
Question 18
Question
Prove the following trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ)
Solution
To prove the trigonometric identity cos4(θ)−sin4(θ) = cos(2θ), we will use
various trigonometric identities and algebraic manipulations.
cos4(θ)−sin4(θ) = (cos2(θ) + sin2(θ))(cos2(θ)−sin2(θ)) (Difference of squares)
= (cos2(θ)−sin2(θ))
= (cos(θ) + sin(θ))(cos(θ)−sin(θ))(cos(θ)−sin(θ)) (Difference of squares again)
= cos(2θ)·(cos(θ)−sin(θ)) (Double angle formula)
= cos(2θ)·cos(θ)−cos(2θ)·sin(θ)
= cos(θ+ 2θ)(Angle sum formula)
= cos(3θ)
Therefore, cos4(θ)−sin4(θ) = cos(3θ)= cos(2θ).
Question 19
Question
Prove the trigonometric identity:
sin2(A)−sin2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Solution
Step 1: First, let’s express sin2(A)and sin2(B)in terms of cos(A)and cos(B)
using the Pythagorean identity sin2(θ) = 1 −cos2(θ).
We have:
sin2(A) = 1 −cos2(A)
sin2(B) = 1 −cos2(B)
Step 2: Substitute the expressions for sin2(A)and sin2(B)into the given
equation to get:
1−cos2(A)−(1 −cos2(B))
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
13
Step 3: Simplify the left side of the equation:
−cos2(A) + cos2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 4: Notice that we can factor out a negative sign from the numerator
of the left side:
−cos2(A)−cos2(B)
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 5: Since cos2(θ)−cos2(ϕ) = (cos(θ) + cos(ϕ))(cos(θ)−cos(ϕ)), we can
rewrite the numerator on the left side:
−(cos(A) + cos(B))(cos(A)−cos(B))
cos(A) + cos(B)=tan(A)−tan(B)
1 + tan(A) tan(B)
Step 6: Cancel out the common factor of cos(A) + cos(B)from the numer-
ator and denominator on the left side:
−(cos(A)−cos(B)) = tan(A)−tan(B)
1 + tan(A) tan(B)
Step 7: Finally, use the identity cos(θ)−cos(ϕ) = −2 sin (θ+ϕ
2)sin (θ−ϕ
2)
to simplify the left side:
2 sin (A+B
2)sin (A−B
2)=tan(A)−tan(B)
1 + tan(A) tan(B)
Therefore, we have proven the given trigonometric identity.
Question 20
Question
Solve the trigonometric equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x < 2π.
Solution
Step 1: Let u= sin(x), then we have u2−3u+ 2 = 0.
Step 2: Factor the quadratic equation in uto get (u−1)(u−2) = 0.
Step 3: Set each factor to zero to solve for u:
u−1 = 0 or u−2 = 0
Step 4: Solve for uto get u= 1 or u= 2.
Step 5: Recall that u= sin(x), so we have sin(x) = 1 or sin(x) = 2.
Step 6: Since sin(x)≤1, the equation sin(x) = 2 has no solution. Therefore,
we focus on solving sin(x) = 1.
Step 7: Recall that sin(x) = 1 when x=π
2+ 2πk, where kis an integer.
Step 8: Therefore, the solutions to the trigonometric equation are x=π
2.
14
Question 21
Question
Prove the trigonometric identity:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Solution
To prove the given trigonometric identity, we will begin by expressing tan(x)
and tan(y)in terms of sine and cosine functions. Then, we will manipulate the
expressions to arrive at the right-hand side of the equation.
Step 1: Express tan(x)and tan(y)in terms of sine and cosine:
tan(x) = sin(x)
cos(x),tan(y) = sin(y)
cos(y)
Step 2: Multiply tan(x)and tan(y)together:
tan(x) tan(y) = sin(x) sin(y)
cos(x) cos(y)
Step 3: Use the trigonometric identity cos(a+b) = cos(a) cos(b)−sin(a) sin(b):
cos(x+y) = cos(x) cos(y)−sin(x) sin(y)
Step 4: Rearrange the equation to solve for cos(x) cos(y):
cos(x) cos(y) = cos(x+y) + sin(x) sin(y)
Step 5: Substitute cos(x) cos(y)back into the original expression:
tan(x) tan(y) = sin(x) sin(y)
1−(cos(x+y) + sin(x) sin(y))
Step 6: Combine like terms and simplify:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)−sin2(x)
Step 7: Apply the trigonometric identity sin2(x) = 1 −cos2(x):
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)−(1 −cos2(x))
Step 8: Simplify further:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y) + cos2(x)
15
Step 9: Use the Pythagorean identity cos2(x) = 1 −sin2(x):
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)+1−sin2(x)
Step 10: Simplify and rearrange terms:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y)+1−(1 −cos(x) cos(y))
Step 11: Combine like terms and simplify:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x+y) + cos(x) cos(y)
Step 12: Use the trigonometric identity cos(x+y) = cos(x) cos(y)−sin(x) sin(y)
one more time:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Therefore, we have proved the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y).
Question 22
Question
Solve the trigonometric equation cos2(x)−sin(x) = 0 for x∈[0,2π].
Solution
Step 1: We can rewrite the equation cos2(x)−sin(x) = 0 using the Pythagorean
identity cos2(x)=1−sin2(x). Substituting this into the equation, we get
1−sin2(x)−sin(x) = 0.
Step 2: Rearranging the equation, we have sin2(x) + sin(x)−1 = 0. This is
a quadratic equation in terms of sin(x).
Step 3: To solve the quadratic equation sin2(x) + sin(x)−1 = 0, we can use
the quadratic formula: sin(x) = −b±√b2−4ac
2a, where a= 1,b= 1, and c=−1.
Step 4: Plugging in the values of a,b, and c, we get sin(x) = −1±√1+4
2,
which simplifies to sin(x) = −1±√5
2.
Step 5: The solutions for sin(x)are sin(x) = −1+√5
2and sin(x) = −1−√5
2.
However, we need to check if these solutions are within the interval [0,2π].
Step 6: Since sin is negative for angles between πand 2π, the solution
sin(x) = −1−√5
2is not in the interval [0,2π]. So, we only consider sin(x) =
−1+√5
2.
Step 7: To find the corresponding values of x, we take the arcsin of −1+√5
2
and 2π−arcsin (−1+√5
2).
16
Step 8: Using a calculator, we find that x≈0.6283 radians and x≈5.6549
radians are the solutions in the interval [0,2π].
Therefore, the solutions to the trigonometric equation cos2(x)−sin(x) = 0
for x∈[0,2π]are x≈0.6283 and x≈5.6549 radians.
Question 23
Question
Simplify the expression √3 cos(θ)−sin(θ).
Solution
Step 1: Recall the trigonometric identity sin(π
3)=√3
2and cos(π
3)=1
2.
Step 2: Let’s express √3 cos(θ)−sin(θ)in terms of sines and cosines with
θ=π
3.√3 cos(θ)−sin(θ) = √3 cos (π
3)−sin (π
3)
Step 3: Substitute the values of cos(π
3)and sin(π
3).
=√3·1
2−√3
2
Step 4: Simplify the expression.
=√3
2−√3
2= 0
Therefore, √3 cos(θ)−sin(θ) = 0.
Question 24
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 2 csc(x) sec(x)
Solution
To prove the trigonometric identity, we will start by working with the left-hand
side (LHS) of the equation and manipulate it until it matches the right-hand
side (RHS) of the equation.
Step 1: Start with the LHS of the equation.
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)
17
Step 2: Rewrite sin2(x)and cos2(x)using the Pythagorean identity: sin2(x) =
1−cos2(x)and cos2(x) = 1 −sin2(x).
1−cos2(x)
1−cos(x)+1−sin2(x)
1 + sin(x)
Step 3: Simplify each fraction separately.
1−cos2(x)
1−cos(x)=(1 + cos(x))(1 −cos(x))
1−cos(x)= 1 + cos(x)
1−sin2(x)
1 + sin(x)=(1 + sin(x))(1 −sin(x))
1 + sin(x)= 1 −sin(x)
Step 4: Combine the simplified fractions.
1 + cos(x)+1−sin(x)
Step 5: Simplify the expression further.
2−sin(x) + cos(x)
Step 6: Use trigonometric identities to simplify the expression further.
sin(x) = 1
csc(x)and cos(x) = 1
sec(x)
2−1
csc(x)+1
sec(x)= 2 csc(x) sec(x)
Therefore, we have shown that the LHS is equal to the RHS, and the trigono-
metric identity is proven.
Question 25
Question
Prove the following trigonometric identity:
1
sin(α) cos(α)=2
sin(2α)
Solution
Step 1: Begin with the right-hand side of the equation and apply the double-
angle formula for sine.
RHS =2
sin(2α)=2
2 sin(α) cos(α)=1
sin(α) cos(α)
Therefore, the given trigonometric identity is proved.
18
Question 26
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will first rewrite
the left side in terms of sines and cosines using trigonometric identities.
Step 1: Rewrite the left side in terms of sines and cosines
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
= (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= 1 ·(sin2(x)−cos2(x))
= sin2(x)−cos2(x)
Step 2: Use the double angle formula for sine Recall the double angle
formula for sine: sin(2θ) = 2 sin(θ) cos(θ). We can use this formula with θ=x
to rewrite the right side.
sin2(x)−cos2(x) = −cos(2x)
= sin(2x+π
2)(using the angle sum identity for cosine)
Step 3: Use the double angle formula for sine again Now, apply the
double angle formula for sine again: sin(2θ) = 2 sin(θ) cos(θ). This time, let
θ= 2x+π
4.
sin(2x+π
2)= sin(2(2x+π
4))
= 2 sin(2x+π
4)cos(2x+π
4)
Step 4: Simplify the expression Now, we expand and simplify the ex-
pression.
2 sin(2x+π
4)cos(2x+π
4)= 2(sin(2x) cos(π
4)+ cos(2x) sin(π
4))
= 2(sin(2x)√2
2+ cos(2x)√2
2)
=√2(sin(2x) + cos(2x))
=√2 sin(2x+π
4)
Therefore, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x).
19
Question 27
Question
Prove the following trigonometric identity:
1 + cos θ
sin θ= csc θ+ cot θ
Solution
Step 1: Start with the left-hand side of the identity.
1 + cos θ
sin θ
Step 2: Use the reciprocal identities to rewrite csc θand cot θ:
csc θ=1
sin θand cot θ=cos θ
sin θ
Step 3: Substitute the rewritten forms of csc θand cot θinto the expression:
1
sin θ+cos θ
sin θ
Step 4: Simplify by combining the fractions:
1 + cos θ
sin θ
Step 5: Notice that the simplified expression is equal to the original left-hand
side: 1 + cos θ
sin θ=1 + cos θ
sin θ
Step 6: Therefore, the trigonometric identity
1 + cos θ
sin θ= csc θ+ cot θ
is proven.
Question 28
Question
Prove the identity:
cos3(x)
1−sin(x)=1 + sin(x) + sin2(x)
cos2(x)
20
Solution
Step 1: Rewrite the expression using trigonometric identities.
cos3(x)
1−sin(x)=cos2(x)·cos(x)
1−sin(x)=cos2(x)·cos(x)
1−sin(x)×1 + sin(x)
1 + sin(x)
Step 2: Simplify the expression using the difference of squares formula and
Pythagorean identity.
cos2(x)·cos(x)·(1 + sin(x))
1−sin2(x)=cos2(x)·cos(x)·(1 + sin(x))
cos2(x)
Step 3: Cancel out the common factor of cos2(x).
cos(x)·(1 + sin(x))
1= cos(x) + cos(x) sin(x)
Step 4: Rewrite the expression on the right hand side using trigonometric
identities.
cos(x) + cos(x) sin(x) = cos(x) + sin(x) cos(x)
cos(x)= cos(x) + tan(x)
Step 5: Recall that tan(x) = sin(x)
cos(x).
cos(x) + tan(x) = cos(x) + sin(x)
cos(x)=cos2(x) + sin(x)
cos(x)
Step 6: Since cos3(x)
1−sin(x)=cos2(x)+sin(x)
cos(x), we have proven the identity.
Question 29
Question
Solve the trigonometric equation tan3(x) = sec(x)for xin the interval [0,2π).
Solution
To solve the trigonometric equation tan3(x) = sec(x), we need to utilize the
trigonometric identities relating tangent and secant functions.
Step 1: Rewrite in terms of sine and cosine First, recall that tan(x) =
sin(x)
cos(x)and sec(x) = 1
cos(x). We rewrite the equation in terms of sine and cosine:
(sin(x)
cos(x))3
=1
cos(x)
21
Step 2: Simplify the equation Simplify the equation by cubing the frac-
tion on the left side: sin3(x)
cos3(x)=1
cos(x)
Step 3: Multiply both sides by cos4(x)To clear the denominators,
multiply both sides by cos4(x):
sin3(x) = cos4(x)
Step 4: Use the Pythagorean Identity We know that sin2(x)+cos2(x) =
1, so cos2(x) = 1 −sin2(x). Substitute this into the equation:
sin3(x) = (1 −sin2(x))2
Step 5: Solve for sin(x)Expand the right side to get a polynomial equation
in terms of sin(x):
sin3(x) = 1 −2 sin2(x) + sin4(x)
Step 6: Rearrange into a quadratic equation Rearrange the equation
to form a quadratic equation in terms of sin(x):
sin4(x)−sin2(x)−sin(x) + 1 = 0
Step 7: Solve the quadratic equation for sin(x)This quadratic equation
may be factored into:
(sin2(x)−sin(x)−1)(sin2(x) + sin(x)−1) = 0
Step 8: Solve for sin(x)using the quadratic formula Solve each
quadratic factor separately using the quadratic formula. The solutions for sin(x)
will lead to solutions for xwhen considering the interval [0,2π). Subsequently,
use the angles whose sine gives those solutions.
Therefore, the solutions for xin the interval [0,2π)are given by the angles
whose sine is the solution for sin(x).
Question 30
Question
Prove the trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x).
22
Solution
Step 1: Start with the triple angle formula for cosine:
cos(3x) = cos(2x+x).
Step 2: Expand the right side using the angle sum formula for cosine:
cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x).
Step 3: Recall the double angle formulas:
cos(2x) = cos2(x)−sin2(x)and sin(2x) = 2 sin(x) cos(x).
Step 4: Substitute the double angle formulas into the expression for cos(2x)
and sin(2x)in the previous step:
cos(2x) cos(x)−sin(2x) sin(x) = (cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x).
Step 5: Simplify the expression:
(cos2(x)−sin2(x)) cos(x)−2 sin(x) cos(x) sin(x) = cos3(x)−sin2(x) cos(x)−2 sin(x) cos2(x).
Step 6: Use the Pythagorean identity sin2(x) + cos2(x)=1to express
sin2(x)in terms of cos(x):
cos3(x)−sin2(x) cos(x)−2 sin(x) cos2(x) = cos3(x)−(1−cos2(x)) cos(x)−2 sin(x) cos2(x).
Step 7: Further simplify the expression:
cos3(x)−(1−cos2(x)) cos(x)−2 sin(x) cos2(x) = cos3(x)−cos(x)+cos3(x)−2 sin(x) cos2(x).
Step 8: Use the Pythagorean identity sin2(x) = 1 −cos2(x)to replace
sin(x) cos2(x)with cos3(x)−cos(x):
cos3(x)−cos(x) + cos3(x)−2 sin(x) cos2(x) = 4 cos3(x)−3 cos(x).
Therefore, cos(3x) = 4 cos3(x)−3 cos(x), which proves the identity.
Question 31
Question
Solve the trigonometric equation sin x+ cos x=√2 cos(x−π
4)for xin the
interval [0,2π].
23
Solution
Step 1: Square both sides of the equation to eliminate the square root.
sin2x+ 2 sin xcos x+ cos2x= 2 cos2(x−π
4)
Step 2: Use the Pythagorean identities sin2x+cos2x= 1 and cos2(x−π
4) =
1
2(1 + cos(2x−π
2))to simplify the equation.
1 + 2 sin xcos x= 1 + cos(2x−π
2)
Step 3: Simplify the equation.
cos(2x−π
2)= 2 sin xcos x
Step 4: Use the double angle formula cos(2θ) = 2 cos2θ−1.
2 cos2(x−π
4)−1 = 2 sin xcos x
Step 5: Apply the Pythagorean identity cos2(x−π
4) = 1 −sin2(x−π
4)and
simplify.
2(1 −sin2(x−π
4)) −1 = 2 sin xcos x
Step 6: Expand and simplify the equation.
2−2 sin2(x−π
4)−1 = 2 sin xcos x
1−2 sin2(x−π
4) = 2 sin xcos x
Step 7: Use the double angle formula for sine to expand sin(x−π
4)
1−2(1
2sin xcos π
4−1
2cos xsin π
4)2
= 2 sin xcos x
Step 8: Simplify and solve the equation.
1−1
2sin2x−1
2cos2x= 2 sin xcos x
1−1
2= 2 sin xcos x
1
2= 2 sin xcos x
Step 9: Use the double angle formula for sine to rewrite sin 2x
sin 2x= 2 sin xcos x
24
Step 10: Substitute sin 2x=1
2and solve for 2x.
1
2=1
2
2x=π
6
Step 11: Find the solutions for xin the interval [0,2π].
x=π
12,5π
12
Question 32
Question
Prove the trigonometric identity:
sin2(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= 1
Solution
Step 1: Start by expressing sin2(θ)
1+cos(θ)as a compound fraction:
sin2(θ)
1 + cos(θ)=sin2(θ)
1 + cos(θ)×1−cos(θ)
1−cos(θ)=sin2(θ)(1 −cos(θ))
1−cos2(θ)
Step 2: Simplify the expression from Step 1:
sin2(θ)(1 −cos(θ))
1−cos2(θ)=sin2(θ)−sin2(θ) cos(θ)
sin2(θ)= 1 −cos(θ)
Step 3: Express cos(θ)
1+sin(θ)as a compound fraction:
cos(θ)
1 + sin(θ)=cos(θ)
1 + sin(θ)×1−sin(θ)
1−sin(θ)=cos(θ)(1 −sin(θ))
1−sin2(θ)
Step 4: Simplify the expression from Step 3:
cos(θ)(1 −sin(θ))
1−sin2(θ)=cos(θ)−sin(θ) cos(θ)
cos2(θ)= 1 −sin(θ)
Step 5: Adding the simplified expressions from Step 2 and Step 4:
1−cos(θ)+1−sin(θ) = 2 −(cos(θ) + sin(θ))
Step 6: Recall that cos(θ)+sin(θ) = √2 sin (θ+π
4). Therefore, substituting
this into the expression from Step 5:
2−√2 sin (θ+π
4)
Thus, the original identity sin2(θ)
1+cos(θ)+cos(θ)
1+sin(θ)= 1 is proven.
25
Question 33
Question
Solve the equation sin(3x) + sin(2x) = 0 for 0≤x≤2π.
Solution
Step 1: Utilize the sum-to-product trigonometric identity to simplify the equa-
tion. Step 2: Apply trigonometric identities to solve for x.
Step 1: Using the sum-to-product trigonometric identity
sin(a) + sin(b) = 2 sin (a+b
2)cos (a−b
2),
we can rewrite the equation as
2 sin (3x+ 2x
2)cos (3x−2x
2)= 0.
Simplifying this expression gives
sin(2.5x) cos(0.5x) = 0.
Step 2: Now, we have the equation sin(2.5x) cos(0.5x)=0. To solve this
equation, we need to consider two cases where either sin(2.5x) = 0 or cos(0.5x) =
0.
Case 1: sin(2.5x)=0This occurs when 2.5x=nπ for n∈Z. Solving for x
gives x=nπ
2.5.
Case 2: cos(0.5x)=0This occurs when 0.5x=π
2+nπ for n∈Z. Solving
for xgives x=π
2+ 2nπ.
Therefore, the solutions to the equation sin(3x) + sin(2x) = 0 for 0≤x≤2π
are x=nπ
2.5and x=π
2+ 2nπ, where nis an integer such that 0≤x≤2π.
Question 34
Question
Prove the following trigonometric identity:
sin4x
cos2x+cos4x
sin2x= tan2x+ cot2x
Solution
To prove the given trigonometric identity, we will start with one side of the
equation and manipulate it to be equal to the other side.
26
Step 1: Rewrite the terms using basic trigonometric identities:
sin4x
cos2x+cos4x
sin2x
Step 2: Express sin4xand cos4xin terms of sin 2xand cos 2x:
(sin2x)2
cos2x+(cos2x)2
sin2x
Step 3: Express sin2xand cos2xin terms of sin 2xand cos 2x:
(1−cos 2x
2)2
cos2x+(1+cos 2x
2)2
sin2x
Step 4: Simplify the expressions by expanding the squares:
1−2 cos 2x+cos22x
4
cos2x+
1+2 cos 2x+cos22x
4
sin2x
Step 5: Simplify further by multiplying the fractions and combining like
terms: 1−2 cos 2x+ cos22x
4 cos2x+1 + 2 cos 2x+ cos22x
4 sin2x
Step 6: Use the double angle identities cos 2x= 1 −2 sin2xand sin 2x=
2 sin xcos x:
1−2(1 −2 sin2x) + (1 −2 sin2x)2
4(1 −sin2x)+1 + 2(1 −2 sin2x) + (1 −2 sin2x)2
4 sin2x
Step 7: Simplify and factorize the expressions:
1 + 4 sin2x−2 + 4 sin2x+ 1 −4 sin2x+ 4 sin4x
4 cos2x
Step 8: Simplify the expression further:
4 sin4x
4 cos2x= tan2x
Step 9: Since the right-hand side of the identity is tan2x+ cot2x, we
have shown that the left-hand side is equal to the right-hand side. Thus, the
trigonometric identity is proven.
Question 35
Question
Solve the trigonometric equation sin(3x) = cos(2x)for xin the interval [0,2π].
27
Solution
Step 1: Recall the trigonometric identities:
•sin(2θ) = 2 sin(θ) cos(θ)
•cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Rewrite the equation using these identities:
sin(3x) = cos(2x)
sin(2x+x) = cos2(x)−sin2(x)
Step 3: Use the double angle identity for sin to simplify:
2 sin(x) cos(x)·cos(x)−(1 −cos2(x)) = 0
2 sin(x) cos2(x)−1 + cos2(x) = 0
Step 4: Substitute sin(x) = cos(x)
2:
2(cos(x)
2)cos2(x)−1 + cos2(x) = 0
cos3(x)−2 cos(x) + 1 = 0
Step 5: Let u= cos(x)to simplify the equation:
u3−2u+ 1 = 0
Step 6: Factor the equation using synthetic division, graphical methods, or
numerical methods. The roots are u= 1, u =−0.618, u = 0.618.
Step 7: Convert uback to cos(x)to find the corresponding values of x:
cos(x) = 1 =⇒x= 0
cos(x) = −0.618 =⇒x≈2.223
cos(x) = 0.618 =⇒x≈0.941
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π]are x= 0,0.941,2.223.
28
