MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 6
Liberty University
Question 1
Question
Prove the following trigonometric identity:
tan(x)·cot(x) = sec(x)·csc(x)
Solution
To prove the given trigonometric identity, we will start by expressing all the
trigonometric functions in terms of sine and cosine.
Step 1: Express tan(x)and cot(x)in terms of sine and cosine:
tan(x) = sin(x)
cos(x)and cot(x) = cos(x)
sin(x)
Step 2: Substitute the expressions for tan(x)and cot(x)into the left-hand
side of the identity:
tan(x)·cot(x) = (sin(x)
cos(x))(cos(x)
sin(x))= 1
Step 3: Express sec(x)and csc(x)in terms of sine and cosine:
sec(x) = 1
cos(x)and csc(x) = 1
sin(x)
Step 4: Substitute the expressions for sec(x)and csc(x)into the right-hand
side of the identity:
sec(x)·csc(x) = (1
cos(x))( 1
sin(x))=1
cos(x) sin(x)
Step 5: Recall the trigonometric identity sin(2θ) = 2 sin(θ) cos(θ), then:
cos(x) sin(x) = 1
2sin(2x)
Step 6: Substitute 1
2sin(2x)back into the expression for the right-hand side
of the identity: 1
cos(x) sin(x)=1
1
2sin(2x)= 2 csc(2x)
Therefore, tan(x)·cot(x) = sec(x)·csc(x)is proven to be true.
Question 2
Question
Prove the trigonometric identity:
1
sin(θ)−sin(θ)
cos2(θ)=cos(θ)
sin(θ)
Solution
Step 1: Start with the left-hand side of the given trigonometric identity:
1
sin(θ)−sin(θ)
cos2(θ)
Step 2: Find a common denominator for the two fractions:
cos2(θ)
sin(θ) cos2(θ)−sin2(θ)
sin(θ) cos2(θ)
Step 3: Combine the fractions:
cos2(θ)−sin2(θ)
sin(θ) cos2(θ)
Step 4: Recall the Pythagorean identity: cos2(θ)−sin2(θ) = 1.
1
sin(θ) cos2(θ)
Step 5: Simplify the expression:
1
sin(θ) cos(θ)
Step 6: Use the reciprocal identity 1
sin(θ)= csc(θ):
csc(θ) csc(θ) = csc2(θ)
2
Step 7: Therefore, the left-hand side simplifies to:
csc2(θ)
Step 8: Since csc(θ) = 1
sin(θ), we have:
1
sin(θ)−sin(θ)
cos2(θ)= csc2(θ) = 1
sin(θ)=cos(θ)
sin(θ)
Thus, the trigonometric identity is proved.
Question 3
Question
Solve the trigonometric equation sin2(x)−2 sin(x) + 1 = 0 for xin the interval
[0,2π).
Solution
To solve the trigonometric equation sin2(x)−2 sin(x) + 1 = 0, we can treat it
as a quadratic equation in terms of sin(x). Let y= sin(x), then the equation
becomes y2−2y+ 1 = 0.
Step 1: Solve the quadratic equation y2−2y+ 1 = 0.
y2−2y+ 1 = 0
(y−1)2= 0
y= 1
Since y= sin(x), we have sin(x) = 1. As sin(x) = 1 occurs when x=π
2, we
now have one solution.
Step 2: Check for additional solutions within the interval [0,2π). Since the
sine function has a period of 2π, we can find a second solution by adding 2πto
the first solution. π
2+ 2π=5π
2
However, 5π
2is not in the interval [0,2π). Therefore, the only solution to
the equation sin2(x)−2 sin(x) + 1 = 0 in the interval [0,2π)is x=π
2.
Question 4
Question
Prove the following trigonometric identity:
1−cos(x)
sin(x)=cot(x)−csc(x)
cot(x) + csc(x)
3
Solution
To prove the identity, we will work with the left-hand side (LHS) and the right-
hand side (RHS) separately, and show that they are equal.
Step 1: Start with the LHS:
LHS =1−cos(x)
sin(x)
=1
sin(x)−cos(x)
sin(x)
= csc(x)−cot(x)
=csc(x)
1−cot(x)
1
=csc(x)
1−cot(x)
1·csc(x)
csc(x)
=csc(x)−cot(x) csc(x)
csc(x)
=cot(x)−csc(x)
csc(x)(by rearranging terms)
Step 2: Show that LHS = RHS: Since we have shown that LHS is equal
to cot(x)−csc(x)
csc(x), and cot(x)−csc(x)
csc(x)=cot(x)−csc(x)
cot(x)+csc(x)=RHS, we conclude that the
identity is true.
Therefore, the trigonometric identity
1−cos(x)
sin(x)=cot(x)−csc(x)
cot(x) + csc(x)
is proven.
Question 5
Question
Prove the identity:
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
Solution
To prove the given identity, we will first express all terms in terms of sin(θ)and
cos(θ)using the Pythagorean trigonometric identity: sin2(θ) + cos2(θ) = 1.
Step 1: Rewrite the identity. We rewrite the identity to express all terms
in terms of sin(θ)and cos(θ):
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
4
Step 2: Express sin4(θ)and cos4(θ)in terms of sin2(θ)and cos2(θ). Using
the square of a trigonometric function, we have:
sin4(θ) = (sin2(θ))2= sin2(θ) sin2(θ) = sin2(θ)(1 −cos2(θ))
cos4(θ) = (cos2(θ))2= cos2(θ) cos2(θ) = cos2(θ)(1 −sin2(θ))
Step 3: Substitute the expressions back into the original identity. Substitute
the expressions for sin4(θ)and cos4(θ)back into the original identity:
sin2(θ)(1 −cos2(θ)) −cos2(θ)(1 −sin2(θ)) = 2 sin2(θ) cos2(θ)
Step 4: Simplify the expression. Expand and simplify the left side of the
equation:
sin2(θ)−sin2(θ) cos2(θ)−cos2(θ) + cos2(θ) sin2(θ) = 2 sin2(θ) cos2(θ)
sin2(θ)−cos2(θ) = 2 sin2(θ) cos2(θ)
Step 5: Use the Pythagorean trigonometric identity to finish the proof.
Since sin2(θ) + cos2(θ) = 1, we have:
sin2(θ)−cos2(θ) = 1 −2 cos2(θ) = 2 sin2(θ) cos2(θ)
Therefore, the identity sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)is proven.
Question 6
Question
Prove the following trigonometric identity:
(tan(x) + cot(x))2−2 sin(x) cos(x) = sec2(x)−1
Solution
• We will work on one side of the equation at a time to prove the identity.
• Let’s start with the left side:
Step 1: Simplify (tan(x) + cot(x))2−2 sin(x) cos(x).
(tan(x) + cot(x))2−2 sin(x) cos(x) = tan2(x) + 2 tan(x) cot(x) + cot2(x)−2 sin(x) cos(x)
=sin2(x)
cos2(x)+ 2 sin(x)
cos(x)
cos(x)
sin(x)+cos2(x)
sin2(x)−2 sin(x) cos(x)
=sin2(x) + 2 cos(x) sin(x) + cos2(x)
cos2(x)−2 sin(x) cos(x)
=1
cos2(x)−2 sin(x) cos(x)
= sec2(x)−2 sin(x) cos(x)
5
• Now we have arrived at a form that is similar to what we want to prove.
Let’s simplify this further to match the expression sec2(x)−1.
Step 2: Simplify sec2(x)−2 sin(x) cos(x).
sec2(x)−2 sin(x) cos(x) = sec2(x)−1+1−2 sin(x) cos(x)
= sec2(x)−1 + (sin2(x) + cos2(x)) −2 sin(x) cos(x)
= (tan(x) + cot(x))2−2 sin(x) cos(x) + (sin2(x) + cos2(x)) −2 sin(x) cos(x)
= (tan(x) + cot(x))2−2 sin(x) cos(x)
• The left side equals the right side, therefore, the trigonometric identity
(tan(x) + cot(x))2−2 sin(x) cos(x) = sec2(x)−1is verified.
Question 7
Question
Prove the following trigonometric identity:
cos(2x)
sin x−sin x
cos(2x)= 2 cot x
Solution
To prove the given trigonometric identity, we will manipulate both sides of the
equation separately until they are equal.
Step 1: Rewrite the left side of the equation using angle addition/subtraction
trigonometric formulas.
cos(2x)
sin x−sin x
cos(2x)=cos2x−sin2x
sin xcos(2x)−sin x
cos2x−sin2x
=cos2x−sin2x
sin xcos2x−sin xsin2x−sin x
cos2x−sin2x
=cos2x−sin2x
sin xcos2x−sin2x−sin x
cos2x−sin2x
Step 2: Use the Pythagorean identity (sin2x+ cos2x= 1) to simplify.
cos(2x)
sin x−sin x
cos(2x)=cos2x−sin2x
sin xcos2x−sin2x−sin x
cos2x−sin2x
=1−2 sin2x
sin x(1 −2 sin2x)−sin2x−sin x
1−2 sin2x
=1−2 sin2x
sin x−2 sin3x−sin2x−sin x
1−2 sin2x
6
Step 3: Simplify the expression by factoring and canceling terms.
cos(2x)
sin x−sin x
cos(2x)=1−2 sin2x
sin x−2 sin3x−sin2x−sin x
1−2 sin2x
=1−2 sin2x
sin x−sin2x−2 sin3x−sin x
1−2 sin2x
=1−2 sin2x
sin x(1 −sin x−2 sin2x)−sin x
1−2 sin2x
=1−2 sin2x
−sin xcos x−sin x
1−2 sin2x
=−1
sin x+2 sin x
sin x
=−csc x+ 2 = 2 cot x
Therefore, cos(2x)
sin x−sin x
cos(2x)= 2 cot x, and the given trigonometric identity
has been proven.
Question 8
Question
Prove the trigonometric identity:
sin(θ)−cos(θ)
sin(θ) + cos(θ)= tan (θ
2)
Solution
Step 1: Let’s start by expressing tan (θ
2)in terms of sin(θ)and cos(θ).
tan (θ
2)=sin (θ
2)
cos (θ
2)
=
2 sin(θ
2)cos(θ
2)
2 cos(θ
2)
cos2(θ
2)−sin2(θ
2)
=2 sin (θ
2)cos (θ
2)
2 cos2(θ
2)−2 sin2(θ
2)
=2 sin (θ
2)cos (θ
2)
2(cos2(θ
2)−sin2(θ
2))
=2 sin (θ
2)cos (θ
2)
2(cos(θ)−sin(θ))
=sin(θ)
sin(θ)−cos(θ)(using double angle identities)
7
Step 2: Now, we can substitute this expression back into the original identity
and simplify.
sin(θ)−cos(θ)
sin(θ) + cos(θ)=sin(θ)
sin(θ)−cos(θ)
= tan (θ
2)
Therefore, we have successfully proved the trigonometric identity:
sin(θ)−cos(θ)
sin(θ) + cos(θ)= tan (θ
2)
Question 9
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the given identity, we will make use of the angle addition formula for
sin(3x)and then simplify the expression step by step.
Step 1: Apply the angle addition formula for sin(3x)to get:
sin(3x) = sin(2x+x)
Step 2: Expand sin(2x+x)using the angle addition formula:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 3: Simplify sin(2x)and cos(2x)using double angle formulas:
sin(3x) = (2 sin(x) cos(x))(cos(x)) + (1 −2 sin2(x))(sin(x))
Step 4: Distribute and simplify:
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin2(x) sin(x)
Step 5: Simplify further:
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
Step 6: Simplify and rearrange terms to obtain the desired identity:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, we have successfully proven the trigonometric identity sin(3x) =
3 sin(x)−4 sin3(x).
8
Question 10
Question
Prove the following trigonometric identity:
sin x
1 + cos x+1 + cos x
sin x= 2 csc xcsc (x
2+π
4)
Solution
Step 1: Recall the definition of cosecant and simplify the right side.
Right side = 2 csc xcsc (x
2+π
4)
= 2 1
sin x·1
sin (x
2+π
4)
= 2 1
sin x·1
(sin x
2cos π
4+ cos x
2sin π
4)
= 2 1
sin x·1
(sin x
2·1
√2+ cos x
2·1
√2)
= 2 1
sin x·1
1
√2(sin x
2+ cos x
2)
= 2 √2
sin x(sin x
2+ cos x
2)
= 2 √2
sin x·sin (x
2+π
4)
= 2 (1
sin x+1
sin (x
2+π
4))
= 2 (csc x+ csc (x
2+π
4))
9
Step 2: Expand the left side and simplify.
Left side =sin x
1 + cos x+1 + cos x
sin x
=sin x
1 + cos x+1
sin x+ cos x
=sin2x+ 1 + cos xsin x
sin x(1 + cos x)
=(1 + sin x)(1 + cos x)
sin x(1 + cos x)
= csc x+ csc x
= 2 csc x
Step 3: Conclude that the original identity is true. Since the right side equals
2(csc x+ csc (x
2+π
4))and the left side equals 2 csc x, we have shown that the
given trigonometric identity is true.
Question 11
Question
Solve the equation sin(2x) cos(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Rewrite the given equation using double angle formulas.
sin(2x) cos(x) = 1
2
2 sin(x) cos(x)·cos(x) = 1
2
2 sin(x) cos2(x) = 1
2
Step 2: Rewrite cos2(x)in terms of sin(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1.
2 sin(x)(1 −sin2(x)) = 1
2
2 sin(x)−2 sin3(x) = 1
2
4 sin3(x)−2 sin(x) + 1 = 0
Step 3: Let y= sin(x), then the equation becomes a cubic equation.
4y3−2y+ 1 = 0
10
Step 4: Find the roots of the cubic equation. The roots of this cubic equation
might not have simple expressions in terms of radicals.
Step 5: Solve for xby finding the corresponding values of y. Finally, find the
values of xby solving for y= sin(x)using either numerical methods or graphing
techniques.
Therefore, the solutions to the equation sin(2x) cos(x) = 1
2in the interval
[0,2π)involve finding the roots of a cubic equation and then solving for xby
finding the corresponding values of y= sin(x).
Question 12
Question
Prove the following trigonometric identity:
cos(θ)−sin(θ) tan(θ) = cos(θ)
Solution
To prove the trigonometric identity cos(θ)−sin(θ) tan(θ) = cos(θ), we will
manipulate the left-hand side of the equation using trigonometric identities.
Step 1: Recall the definitions of tangent and sine functions:
tan(θ) = sin(θ)
cos(θ)
Step 2: Replace tan(θ)with sin(θ)
cos(θ)in the equation:
cos(θ)−sin(θ)·sin(θ)
cos(θ)
Step 3: Simplify the expression:
cos(θ)−sin2(θ)
cos(θ)
Step 4: Use the Pythagorean identity sin2(θ) + cos2(θ) = 1:
cos(θ)−1−cos2(θ)
cos(θ)
Step 5: Simplify further:
cos(θ)−1
cos(θ)+ cos(θ)
Step 6: Combine like terms:
cos(θ)−1
cos(θ)+ cos(θ) = cos(θ)
Therefore, cos(θ)−sin(θ) tan(θ) = cos(θ), and the identity is proved.
11
Question 13
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine, sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the equation: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides of the equation by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x≤2πare x=π
6
and x=5π
6.
Therefore, the solutions to the trigonometric equation sin(2x) = cos(x)in
the interval 0≤x≤2πare x=π
6and x=5π
6.
Question 14
Question
Prove the identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
formula for sine which states that sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
Step 1: Let’s start by applying the angle addition formula for sine to sin(3x):
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, we need to express sin(2x)and cos(2x)in terms of sin(x)and
cos(x)using the double angle identities:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
Step 3: Substitute in the expressions for sin(2x)and cos(2x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + cos2(x) sin(x)−sin3(x)
= 2 sin(x)−3 sin3(x) + (1 −sin2(x)) sin(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
12
Therefore, we have shown that sin(3x) = 3 sin(x)−4 sin3(x), completing the
proof.
Question 15
Question
Prove the trigonometric identity:
2 sin xcos xcos(2x) = sin(2x)
where xis a real number.
Solution
We will start by expanding the left-hand side of the equation using trigonometric
identities.
Step 1: Expand the left-hand side using trigonometric identities:
2 sin xcos xcos(2x) = 2 sin xcos x(cos2x−sin2x)
= 2 sin xcos x(cos2x−(1 −cos2x))
= 2 sin xcos x(cos2x−1 + cos2x)
= 2 sin xcos x(2 cos2x−1)
= 4 sin xcos2x−2 sin xcos x.
Step 2: Next, simplify the expression further:
4 sin xcos2x−2 sin xcos x= 2 sin x(2 cos2x)−2 sin xcos x
= 2 sin x(2 cos2x−cos x)
= 2 sin x(cos x)(2 cos x−1)
= sin xcos(2x).
Therefore, the left-hand side is equal to the right-hand side, and we have
proven the trigonometric identity:
2 sin xcos xcos(2x) = sin(2x).
Question 16
Question
Prove the trigonometric identity:
cot(θ)·sec(θ) = csc(θ)
13
Solution
Step 1: Rewrite all trigonometric functions in terms of sine and cosine.
cot(θ)·sec(θ) = cos(θ)
sin(θ)·1
cos(θ)
=cos(θ)
sin(θ)·cos(θ)
Step 2: Use the trigonometric identity sin(θ)·cos(θ) = 1
2·sin(2θ).
cos(θ)
sin(θ)·cos(θ)=cos(θ)
1
2·sin(2θ)
=2 cos(θ)
sin(2θ)
Step 3: Apply the double-angle identity for sine: sin(2θ) = 2 sin(θ) cos(θ).
2 cos(θ)
sin(2θ)=2 cos(θ)
2 sin(θ) cos(θ)
=2 cos(θ)
2 sin(θ) cos(θ)
=cos(θ)
sin(θ) cos(θ)
=1
sin(θ)
= csc(θ)
Therefore, cot(θ)·sec(θ) = csc(θ)is proven.
Question 17
Question
Solve the equation sin2(x)−√3 sin(x) + 1 = 0 for xin the interval [0,2π].
Solution
Step 1: Let u= sin(x), so the equation becomes u2−√3u+ 1 = 0.
Step 2: To solve for u, we can use the quadratic formula: u=−b±√b2−4ac
2a
with a= 1,b=−√3, and c= 1.
Step 3: Plugging in the values of a,b, and c, we get u=√3±√(−√3)2−4∗1∗1
2∗1.
Step 4: Simplifying under the square root, we have u=√3±√3−4
2.
Step 5: Further simplifying, we get u=√3±√−1
2. Since √−1 = i, we have
u=√3±i
2.
14
Step 6: Therefore, the solutions for uare u=√3+i
2and u=√3−i
2.
Step 7: Now, we need to find the corresponding values of xusing u= sin(x).
Step 8: For u=√3+i
2, we have sin(x) = √3+i
2. This implies x= sin−1(√3+i
2).
Step 9: Similarly, for u=√3−i
2, we have x= sin−1(√3−i
2).
Step 10: Finally, we need to find the values of xin the interval [0,2π].
Step 11: Therefore, the solutions to the equation sin2(x)−√3 sin(x) + 1 = 0
for xin the interval [0,2π]are x= sin−1(√3+i
2)and x= sin−1(√3−i
2).
Question 18
Question
Prove the following trigonometric identity:
(cos x+ sin x)2= 1 + sin 2x
Solution
To prove the trigonometric identity
(cos x+ sin x)2= 1 + sin 2x,
we will expand the left side and simplify to match the right side of the equation.
Step 1: Expand the left side using the distributive property.
(cos x+ sin x)2= (cos x+ sin x)(cos x+ sin x)
= cos2x+ 2 cos xsin x+ sin2x
Step 2: Use the Pythagorean Identity cos2x+ sin2x= 1.
cos2x+ 2 cos xsin x+ sin2x= 1 + 2 cos xsin x
Step 3: Use the double angle identity sin 2x= 2 sin xcos x.
1 + 2 cos xsin x= 1 + sin 2x
Hence, we have shown that (cos x+ sin x)2= 1 + sin 2x, which proves the
trigonometric identity.
Question 19
Question
Prove the following trigonometric identity:
√2 cos(θ) sin(θ) = sin(2θ)
15
Solution
To prove the trigonometric identity √2 cos(θ) sin(θ) = sin(2θ), we will use the
double angle formula for sine.
Step 1: Recall the double angle formula for sine:
sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Begin with the left side of the equation and simplify:
√2 cos(θ) sin(θ) = √2·1
2(sin(2θ))
=√2
2sin(2θ)
= sin(2θ)·√2
2
= sin(2θ)(since √2
2= 1)
= sin(2θ)
Step 3: Therefore, we have shown that √2 cos(θ) sin(θ) = sin(2θ), which
proves the given trigonometric identity.
Question 20
Question
Prove the identity:
cos(θ)
1−sin(θ)+1
cos(θ)=2
1 + sin(θ)
Solution
Step 1: Find a common denominator To add the fractions on the left-hand
side, we need to find a common denominator. The common denominator for
1−sin(θ)and cos(θ)is cos(θ)(1−sin(θ)). Rewrite the fractions with this common
denominator:
cos(θ)2
cos(θ)(1 −sin(θ)) +1(1 −sin(θ))
cos(θ)(1 −sin(θ))
Step 2: Combine the fractions Combine the fractions over the common
denominator:
cos(θ)2+ 1 −sin(θ)
cos(θ)(1 −sin(θ))
16
Step 3: Expand the numerator Expand the numerator using the Pythagorean
identity cos(θ)2= 1 −sin(θ)2:
1−sin(θ)2+ 1 −sin(θ)
cos(θ)(1 −sin(θ))
Step 4: Simplify the numerator Simplify the numerator:
2−sin(θ)−sin(θ)2
cos(θ)(1 −sin(θ))
Step 5: Use the Pythagorean identity Since sin(θ)2= 1 −cos(θ)2, sub-
stitute this into the numerator:
2−sin(θ)−(1 −cos(θ)2)
cos(θ)(1 −sin(θ))
Step 6: Further simplify the numerator Simplify the numerator:
2−sin(θ)−1 + cos(θ)2
cos(θ)(1 −sin(θ)) =1 + cos(θ)2−sin(θ)
cos(θ)(1 −sin(θ))
Step 7: Use the Pythagorean identity again Since cos(θ)2= 1−sin(θ)2,
substitute this into the numerator:
1 + (1 −sin(θ)2)−sin(θ)
cos(θ)(1 −sin(θ))
Step 8: Simplify the numerator Simplify the numerator:
1+1−sin(θ)2−sin(θ)
cos(θ)(1 −sin(θ)) =2−sin(θ)2−sin(θ)
cos(θ)(1 −sin(θ))
Step 9: Simplify the numerator further Since sin(θ)2+sin(θ) = sin(θ)(1+
sin(θ)), substitute this into the numerator:
2−sin(θ)(1 + sin(θ))
cos(θ)(1 −sin(θ)) =2
1 + sin(θ)
Therefore, the given identity is true.
Question 21
Question
Prove the trigonometric identity:
sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= tan (θ
2)
17
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation into the form of the right-hand side.
Step 1: Start with the left-hand side of the equation and combine the
fractions: sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)
Step 2: To simplify the expression, multiply the first term by 1−cos(θ)
1−cos(θ)and
the second term by 1−sin(θ)
1−sin(θ):
sin(θ)(1 −cos(θ))
(1 + cos(θ))(1 −cos(θ)) +cos(θ)(1 −sin(θ))
(1 + sin(θ))(1 −sin(θ))
Step 3: Expand and simplify the expression:
sin(θ)−sin(θ) cos(θ)
1−cos2(θ)+cos(θ)−cos(θ) sin(θ)
1−sin2(θ)
Step 4: Recall the Pythagorean identities sin2(θ) + cos2(θ)=1and 1−
sin2(θ) = cos2(θ):
sin(θ)−sin(θ) cos(θ)
sin2(θ)+cos(θ)−cos(θ) sin(θ)
cos2(θ)
Step 5: Simplify further to obtain a common denominator:
sin(θ)(1 −cos(θ))
sin2(θ)+cos(θ)(1 −sin(θ))
cos2(θ)
Step 6: Use double angle formula for tangent: tan(2α) = 2 tan(α)
1−tan2(α):
2 tan (θ
2)
1−tan2(θ
2)= tan (θ
2)
Therefore, we have successfully proven the trigonometric identity:
sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= tan (θ
2)
Question 22
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
18
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to match the right-hand side.
Step 1: Start with the left-hand side of the equation:
sin4(x)−cos4(x)
Step 2: Rewrite sin4(x)as (sin2(x))2and cos4(x)as (cos2(x))2:
(sin2(x))2−(cos2(x))2
Step 3: Use the difference of squares identity: a2−b2= (a+b)(a−b):
(sin2(x) + cos2(x))(sin2(x)−cos2(x))
Step 4: Recall that sin2(x) + cos2(x) = 1 (Pythagorean identity):
1(sin2(x)−cos2(x))
Step 5: Rewrite sin2(x)−cos2(x)as sin(2x)using the double angle identity
sin(2θ) = 2 sin(θ) cos(θ):
1·sin(2x)
Step 6: Now, simplify the expression on the right-hand side:
sin(2x) = sin(2x)·1 = sin(2x)
Step 7: Hence, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x), as
required.
Question 23
Question
Prove the following trigonometric identity:
cos(θ) cot(θ)−sin(θ) = 1−sin2(θ)
sin(θ)
Solution
Step 1: Rewrite the left-hand side of the equation using trigonometric identi-
ties.
cos(θ) cot(θ)−sin(θ) = cos(θ)
sin(θ)−sin(θ)
Step 2: Find a common denominator for the terms.
cos(θ)
sin(θ)−sin(θ) = cos(θ)−sin2(θ)
sin(θ)
19
Step 3: Use the Pythagorean identity cos2(θ)=1−sin2(θ)to simplify the
numerator. cos(θ)−sin2(θ)
sin(θ)=1−sin2(θ)
sin(θ)
Step 4: Thus, we have shown that
cos(θ) cot(θ)−sin(θ) = 1−sin2(θ)
sin(θ)
And the trigonometric identity is proven.
Question 24
Question
Solve the equation 2 sin2(x)−5 cos(x) + 1 = 0 for xin the interval [0,2π].
Solution
Step 1: Start by using the Pythagorean identity for sine and cosine: sin2(x) =
1−cos2(x). Step 2: Rewrite the equation using the Pythagorean identity:
2(1 −cos2(x)) −5 cos(x) + 1 = 0.
Step 3: Simplify the equation:
2−2 cos2(x)−5 cos(x) + 1 = 0.
Step 4: Rearrange the terms to form a quadratic equation in terms of cosine:
2 cos2(x) + 5 cos(x)−3 = 0.
Step 5: Solve this quadratic equation by factoring or using the quadratic for-
mula. The solutions are:
cos(x) = −5±√52−4(2)(−3)
2(2) =−5±√49
4=−5±7
4.
Step 6: Thus, we have two possible values for cos(x):cos(x) = 1 or cos(x) = −3
2.
Step 7: Since cosine values lie between -1 and 1, the second solution is extraneous
and we must discard it. Therefore, we have cos(x) = 1. Step 8: Solve for x
using the fact that cos(x) = 1 implies x= 2nπ where nis an integer. Step 9:
Checking the interval [0,2π], the solution is x= 0.
Therefore, the solution to the equation 2 sin2(x)−5 cos(x) + 1 = 0 for xin
the interval [0,2π]is x= 0.
20
Question 25
Question
Prove the following trigonometric identity:
cot(θ)−tan(θ) = 2
sin(2θ)
Solution
We will start by expressing cot(θ)and tan(θ)in terms of sine and cosine:
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Subtracting tan(θ)from cot(θ)
cot(θ)−tan(θ) = cos(θ)
sin(θ)−sin(θ)
cos(θ)
Step 3: Finding a common denominator
cot(θ)−tan(θ) = cos2(θ)
cos(θ) sin(θ)−sin2(θ)
cos(θ) sin(θ)
Step 4: Combining the fractions
cot(θ)−tan(θ) = cos2(θ)−sin2(θ)
cos(θ) sin(θ)
Step 5: Using the double angle identity sin(2θ) = 2 sin(θ) cos(θ)
cot(θ)−tan(θ) = cos(2θ)
sin(2θ)
Step 6: Recall that cos(2θ) = 2 cos2(θ)−1
cot(θ)−tan(θ) = 2 cos2(θ)−1
sin(2θ)
Step 7: Simplifying further
cot(θ)−tan(θ) = 2 cos2(θ)−1
sin(2θ)=2−2 sin2(θ)
sin(2θ)
21
Step 8: Using the Pythagorean identity sin2(θ) = 1 −cos2(θ)
cot(θ)−tan(θ) = 2−2(1 −cos2(θ))
sin(2θ)
cot(θ)−tan(θ) = 2−2 + 2 cos2(θ)
sin(2θ)
cot(θ)−tan(θ) = 2 cos2(θ)
sin(2θ)
Step 9: Finally, we see that cot(θ)−tan(θ) = 2 cos2(θ)
sin(2θ)=2
sin(2θ), which
proves the given trigonometric identity.
Question 26
Question
Prove the following trigonometric identity:
cot(x) csc(x)−tan(x) sec(x) = cos(x)−sin(x)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation using basic trigonometric identities until we obtain the right-
hand side.
22
LHS = cot(x) csc(x)−tan(x) sec(x)
=cos(x)
sin(x)·1
sin(x)−sin(x)
cos(x)·1
cos(x)
=cos(x)
sin(x)2−sin(x)
cos(x)2
=cos(x) cos(x)
sin(x)2cos(x)−sin(x) sin(x)
cos(x)2sin(x)
=cos2(x)
sin2(x)−sin2(x)
cos2(x)
=cos2(x)−sin2(x)
sin2(x) cos2(x)
=cos(2x)
sin(2x)
=2 cos(x) sin(x)
2 sin(x) cos(x)
= 1
= cos(x)−sin(x)
=RHS
Therefore, we have shown that the left-hand side (LHS) is equal to the right-
hand side (RHS), and the trigonometric identity is proved.
Question 27
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
1. We can start with the triple angle identity: sin(3x) = 3 sin(x)−4 sin3(x).
2. Let’s rewrite the triple angle identity using the double angle identity:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x).
3. Recall the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x).
23
4. Substituting these expressions into our rewritten triple angle identity:
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x).
5. Simplifying the expression:
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x).
6. Factoring out a sin(x)from the first two terms gives:
sin(3x) = sin(x)(2 cos2(x) + cos2(x)) −sin3(x).
7. Further simplifying:
sin(3x) = sin(x)(3 cos2(x)) −sin3(x).
8. Using the Pythagorean identity sin2(x) + cos2(x) = 1:
3 cos2(x) = 3(1 −sin2(x)) = 3 −3 sin2(x).
9. Substituting this back into our expression:
sin(3x) = sin(x)(3 −3 sin2(x)) −sin3(x).
10. Factorizing 3−3 sin2(x)gives 3(1 −sin2(x)) = 3 cos2(x) = 3:
sin(3x) = 3 sin(x)−3 sin3(x)−sin3(x).
11. Simplifying, we arrive at the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x).
Question 28
Question
Solve the trigonometric equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x < 2π.
Solution
Step 1: Let’s rewrite the equation in terms of a single variable, say u= sin(x).
u2−3u+ 2 = 0
Step 2: We can factor the quadratic equation to find the roots.
(u−2)(u−1) = 0
24
Step 3: Setting each factor to zero gives us two possible values for u.
u= 2 or u= 1
Step 4: Recall that u= sin(x), so we have two possibilities for sin(x).
sin(x) = 2 or sin(x) = 1
Step 5: Since the range of sin is [−1,1], the first possibility is not valid.
Therefore, we focus on solving sin(x) = 1.
Step 6: The equation sin(x)=1has solutions at x=π
2+ 2πk, where kis
an integer.
Step 7: Therefore, the solutions to the trigonometric equation sin2(x)−
3 sin(x) + 2 = 0 for 0≤x < 2πare x=π
2and x=3π
2.
Question 29
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: We will start by using the angle addition formula for sine: sin(A+B) =
sin(A) cos(B) + cos(A) sin(B). Applying this formula to sin(3x) = sin(2x+x),
we get:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, we’ll express sin(2x)and cos(2x)in terms of sin(x)and
cos(x). Using the double-angle formulas, we have:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 3: Now, substitute the expressions for sin(2x)and cos(2x)back into
the equation from Step 1:
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
Step 4: Simplify the expression by expanding terms and combining like
terms:
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin3(x)
Step 5: Recall the double-angle identity: cos2(x) = 1 −sin2(x). Substitute
cos2(x) = 1 −sin2(x)into the expression:
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
25
Step 6: Simplify the expression further:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
Step 7: Combine like terms to obtain the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)has been
proven.
Question 30
Question
Prove the following trigonometric identity:
tan2(x)−sin2(x) = sec2(x)−1
Solution
To prove the trigonometric identity tan2(x)−sin2(x) = sec2(x)−1, we will first
rewrite all trigonometric functions in terms of sin(x)and cos(x).
Step 1: Rewrite tan xin terms of sin xand cos x.
tan(x) = sin(x)
cos(x)
Step 2: Square the expression for tan(x).
tan2(x) = (sin(x)
cos(x))2
=sin2(x)
cos2(x)
Step 3: Rewrite sin2(x)in terms of cos2(x).
sin2(x) = 1 −cos2(x)
Step 4: Substitute the expressions for tan2(x)and sin2(x)back into the
original identity.
sin2(x)
cos2(x)−(1 −cos2(x)) = sec2(x)−1
Step 5: Simplify the left-hand side of the equation.
sin2(x)
cos2(x)−1 + cos2(x) = sec2(x)−1
26
Step 6: Combine like terms on the left-hand side.
sin2(x)
cos2(x)+ cos2(x)−1 = sec2(x)−1
Step 7: Use the Pythagorean identity sin2(x)+cos2(x) = 1 on the left-hand
side. 1−cos2(x)
cos2(x)+ cos2(x)−1 = sec2(x)−1
Step 8: Simplify the left-hand side.
1
cos2(x)−1 = sec2(x)−1
Step 9: Recognize that sec2(x) = 1
cos2(x).
sec2(x)−1 = sec2(x)−1
Therefore, the trigonometric identity tan2(x)−sin2(x) = sec2(x)−1has
been proven.
Question 31
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for xin the interval [0,2π].
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
u2−3u+ 2 = 0
Step 2: Solve the quadratic equation u2−3u+ 2 = 0 by factoring or using
the quadratic formula:
u2−2u−u+ 2 = 0
u(u−2) −1(u−2) = 0
(u−2)(u−1) = 0
So, u= 2 or u= 1.
Step 3: Since u= sin(x), we have two cases to consider: Case 1: u= 1 =⇒
sin(x) = 1. The solutions in [0,2π]for this case are x=π
2, and 5π
2. Case 2:
u= 2 =⇒sin(x) = 2, which is not possible.
Step 4: Thus, the solutions to the equation sin2(x)−3 sin(x) + 2 = 0 in the
interval [0,2π]are x=π
2, and 5π
2.
27
Question 32
Question
Prove the trigonometric identity:
1−sin x
cos x= tan (x
2)
Solution
To prove the given trigonometric identity, we will start by expressing tan (x
2)in
terms of sin xand cos x.
Step 1: Express tan (x
2)in terms of sin xand cos x. We know that tan (x
2)=
sin(x
2)
cos(x
2). By half-angle identities, we have:
sin (x
2)=±√1−cos x
2
cos (x
2)=±√1 + cos x
2
Since xlies in the first quadrant, both sine and cosine values are positive. Thus,
sin (x
2)=√1−cos x
2
cos (x
2)=√1 + cos x
2
Step 2: Substitute sin (x
2)and cos (x
2)into tan (x
2).
tan (x
2)=sin (x
2)
cos (x
2)=√1−cos x
2
√1+cos x
2
=√1−cos x
1 + cos x
Step 3: Manipulate the right side of the identity to match with tan (x
2).
We can simplify the expression on the right side of the identity in the following
way:
tan (x
2)=√1−cos x
1 + cos x=√(1 −cos x)(1 −cos x)
(1 + cos x)(1 −cos x)=√1−2 cos x+ cos2x
1−cos2x
=√1−2 cos x+ cos2x
sin2x=√(1 −cos x)2
sin2x=1−cos x
sin x=1−sin x
cos x
Therefore, we have proven the trigonometric identity 1−sin x
cos x= tan (x
2).
28
Question 33
Question
Solve the equation sin(3x) = cos(2x)for xin the interval [0,2π).
Solution
Step 1: Recall the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x). Rewriting the equation in terms of sin(x), we have:
3 sin(x)−4 sin3(x) = 1 −2 sin2(x)
4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0
Step 2: Let y= sin(x), then the equation becomes 4y3−2y2−3y+ 1 = 0.
We can factor this equation as:
(y−1)(4y2+ 2y−1) = 0
Step 3: Solving the quadratic factor, we have 4y2+ 2y−1 = 0 ⇒y=
−2±√22−4(4)(−1)
2(4) . Thus, y=−1±√3
4.
Step 4: Since y= sin(x), we have two possible values for yin the interval
[0,2π):sin(x) = −1+√3
4or sin(x) = −1−√3
4.
Step 5: Solving for x, we find sin(x) = −1+√3
4corresponds to x=5π
6and
sin(x) = −1−√3
4corresponds to x=7π
6.
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π)are x=5π
6and x=7π
6.
Question 34
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
and triple angle formulas for sine.
Step 1: Write sin(3x)using the angle addition formula By the angle
addition formula for sine, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
29
Step 2: Express sin(2x)and cos(2x)using double angle formulas
Recall the double angle formulas:
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ)
Therefore, sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x).
Step 3: Substitute sin(2x)and cos(2x)back into the expression
for sin(3x)Substituting these expressions back into sin(3x) = sin(2x) cos(x) +
cos(2x) sin(x)gives:
sin(3x) = (2 sin(x) cos(x)) cos(x) + (cos2(x)−sin2(x)) sin(x)
Step 4: Simplify the expression
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x)
Now, since sin2(x) + cos2(x) = 1:
sin(3x) = 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Hence, we have proved the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 35
Question
Prove the following trigonometric identity:
tan(x)−sin(x)
1 + tan(x) sin(x)=cos(x)
1 + sin(x)
Solution
To prove the given trigonometric identity, we will manipulate the left side of the
equation until it matches the right side.
Step 1: Rewrite the left side using trigonometric identities.
tan(x)−sin(x)
1 + tan(x) sin(x)=
sin(x)
cos(x)−sin(x)
1 + sin(x)
cos(x)sin(x)
30
Step 2: Simplify the expression by finding a common denominator.
=
sin(x)−sin(x) cos(x)
cos(x)
1 + sin2(x)
cos(x)
Step 3: Combine the fractions.
=sin(x)−sin(x) cos(x)
cos(x) + sin2(x)
Step 4: Factor out a cos(x)in the numerator.
=sin(x)(1 −cos(x))
cos(x)(1 + sin(x))
Step 5: Cancel out a sin(x)term.
=1−cos(x)
1 + sin(x)
Step 6: Use the trigonometric identity cos(x) = 1 −sin2(x).
=sin2(x)
1−sin2(x)
Step 7: Simplify using the Pythagorean identity sin2(x) + cos2(x) = 1.
=sin2(x)
cos2(x)
Step 8: Simplify the expression to match the right side.
=cos(x)
1 + sin(x)
Therefore, the left side of the given trigonometric identity is equal to the
right side, and the identity has been proven.
31
Step 5: Recall the trigonometric identity sin(2θ) = 2 sin(θ) cos(θ), then:
cos(x) sin(x) = 1
2sin(2x)
Step 6: Substitute 1
2sin(2x)back into the expression for the right-hand side
of the identity: 1
cos(x) sin(x)=1
1
2sin(2x)= 2 csc(2x)
Therefore, tan(x)·cot(x) = sec(x)·csc(x)is proven to be true.
Question 2
Question
Prove the trigonometric identity:
1
sin(θ)−sin(θ)
cos2(θ)=cos(θ)
sin(θ)
Solution
Step 1: Start with the left-hand side of the given trigonometric identity:
1
sin(θ)−sin(θ)
cos2(θ)
Step 2: Find a common denominator for the two fractions:
cos2(θ)
sin(θ) cos2(θ)−sin2(θ)
sin(θ) cos2(θ)
Step 3: Combine the fractions:
cos2(θ)−sin2(θ)
sin(θ) cos2(θ)
Step 4: Recall the Pythagorean identity: cos2(θ)−sin2(θ) = 1.
1
sin(θ) cos2(θ)
Step 5: Simplify the expression:
1
sin(θ) cos(θ)
Step 6: Use the reciprocal identity 1
sin(θ)= csc(θ):
csc(θ) csc(θ) = csc2(θ)
2
Step 7: Therefore, the left-hand side simplifies to:
csc2(θ)
Step 8: Since csc(θ) = 1
sin(θ), we have:
1
sin(θ)−sin(θ)
cos2(θ)= csc2(θ) = 1
sin(θ)=cos(θ)
sin(θ)
Thus, the trigonometric identity is proved.
Question 3
Question
Solve the trigonometric equation sin2(x)−2 sin(x) + 1 = 0 for xin the interval
[0,2π).
Solution
To solve the trigonometric equation sin2(x)−2 sin(x) + 1 = 0, we can treat it
as a quadratic equation in terms of sin(x). Let y= sin(x), then the equation
becomes y2−2y+ 1 = 0.
Step 1: Solve the quadratic equation y2−2y+ 1 = 0.
y2−2y+ 1 = 0
(y−1)2= 0
y= 1
Since y= sin(x), we have sin(x) = 1. As sin(x) = 1 occurs when x=π
2, we
now have one solution.
Step 2: Check for additional solutions within the interval [0,2π). Since the
sine function has a period of 2π, we can find a second solution by adding 2πto
the first solution. π
2+ 2π=5π
2
However, 5π
2is not in the interval [0,2π). Therefore, the only solution to
the equation sin2(x)−2 sin(x) + 1 = 0 in the interval [0,2π)is x=π
2.
Question 4
Question
Prove the following trigonometric identity:
1−cos(x)
sin(x)=cot(x)−csc(x)
cot(x) + csc(x)
3
Solution
To prove the identity, we will work with the left-hand side (LHS) and the right-
hand side (RHS) separately, and show that they are equal.
Step 1: Start with the LHS:
LHS =1−cos(x)
sin(x)
=1
sin(x)−cos(x)
sin(x)
= csc(x)−cot(x)
=csc(x)
1−cot(x)
1
=csc(x)
1−cot(x)
1·csc(x)
csc(x)
=csc(x)−cot(x) csc(x)
csc(x)
=cot(x)−csc(x)
csc(x)(by rearranging terms)
Step 2: Show that LHS = RHS: Since we have shown that LHS is equal
to cot(x)−csc(x)
csc(x), and cot(x)−csc(x)
csc(x)=cot(x)−csc(x)
cot(x)+csc(x)=RHS, we conclude that the
identity is true.
Therefore, the trigonometric identity
1−cos(x)
sin(x)=cot(x)−csc(x)
cot(x) + csc(x)
is proven.
Question 5
Question
Prove the identity:
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
Solution
To prove the given identity, we will first express all terms in terms of sin(θ)and
cos(θ)using the Pythagorean trigonometric identity: sin2(θ) + cos2(θ) = 1.
Step 1: Rewrite the identity. We rewrite the identity to express all terms
in terms of sin(θ)and cos(θ):
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
4
Step 2: Express sin4(θ)and cos4(θ)in terms of sin2(θ)and cos2(θ). Using
the square of a trigonometric function, we have:
sin4(θ) = (sin2(θ))2= sin2(θ) sin2(θ) = sin2(θ)(1 −cos2(θ))
cos4(θ) = (cos2(θ))2= cos2(θ) cos2(θ) = cos2(θ)(1 −sin2(θ))
Step 3: Substitute the expressions back into the original identity. Substitute
the expressions for sin4(θ)and cos4(θ)back into the original identity:
sin2(θ)(1 −cos2(θ)) −cos2(θ)(1 −sin2(θ)) = 2 sin2(θ) cos2(θ)
Step 4: Simplify the expression. Expand and simplify the left side of the
equation:
sin2(θ)−sin2(θ) cos2(θ)−cos2(θ) + cos2(θ) sin2(θ) = 2 sin2(θ) cos2(θ)
sin2(θ)−cos2(θ) = 2 sin2(θ) cos2(θ)
Step 5: Use the Pythagorean trigonometric identity to finish the proof.
Since sin2(θ) + cos2(θ) = 1, we have:
sin2(θ)−cos2(θ) = 1 −2 cos2(θ) = 2 sin2(θ) cos2(θ)
Therefore, the identity sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)is proven.
Question 6
Question
Prove the following trigonometric identity:
(tan(x) + cot(x))2−2 sin(x) cos(x) = sec2(x)−1
Solution
• We will work on one side of the equation at a time to prove the identity.
• Let’s start with the left side:
Step 1: Simplify (tan(x) + cot(x))2−2 sin(x) cos(x).
(tan(x) + cot(x))2−2 sin(x) cos(x) = tan2(x) + 2 tan(x) cot(x) + cot2(x)−2 sin(x) cos(x)
=sin2(x)
cos2(x)+ 2 sin(x)
cos(x)
cos(x)
sin(x)+cos2(x)
sin2(x)−2 sin(x) cos(x)
=sin2(x) + 2 cos(x) sin(x) + cos2(x)
cos2(x)−2 sin(x) cos(x)
=1
cos2(x)−2 sin(x) cos(x)
= sec2(x)−2 sin(x) cos(x)
5
• Now we have arrived at a form that is similar to what we want to prove.
Let’s simplify this further to match the expression sec2(x)−1.
Step 2: Simplify sec2(x)−2 sin(x) cos(x).
sec2(x)−2 sin(x) cos(x) = sec2(x)−1+1−2 sin(x) cos(x)
= sec2(x)−1 + (sin2(x) + cos2(x)) −2 sin(x) cos(x)
= (tan(x) + cot(x))2−2 sin(x) cos(x) + (sin2(x) + cos2(x)) −2 sin(x) cos(x)
= (tan(x) + cot(x))2−2 sin(x) cos(x)
• The left side equals the right side, therefore, the trigonometric identity
(tan(x) + cot(x))2−2 sin(x) cos(x) = sec2(x)−1is verified.
Question 7
Question
Prove the following trigonometric identity:
cos(2x)
sin x−sin x
cos(2x)= 2 cot x
Solution
To prove the given trigonometric identity, we will manipulate both sides of the
equation separately until they are equal.
Step 1: Rewrite the left side of the equation using angle addition/subtraction
trigonometric formulas.
cos(2x)
sin x−sin x
cos(2x)=cos2x−sin2x
sin xcos(2x)−sin x
cos2x−sin2x
=cos2x−sin2x
sin xcos2x−sin xsin2x−sin x
cos2x−sin2x
=cos2x−sin2x
sin xcos2x−sin2x−sin x
cos2x−sin2x
Step 2: Use the Pythagorean identity (sin2x+ cos2x= 1) to simplify.
cos(2x)
sin x−sin x
cos(2x)=cos2x−sin2x
sin xcos2x−sin2x−sin x
cos2x−sin2x
=1−2 sin2x
sin x(1 −2 sin2x)−sin2x−sin x
1−2 sin2x
=1−2 sin2x
sin x−2 sin3x−sin2x−sin x
1−2 sin2x
6
Step 3: Simplify the expression by factoring and canceling terms.
cos(2x)
sin x−sin x
cos(2x)=1−2 sin2x
sin x−2 sin3x−sin2x−sin x
1−2 sin2x
=1−2 sin2x
sin x−sin2x−2 sin3x−sin x
1−2 sin2x
=1−2 sin2x
sin x(1 −sin x−2 sin2x)−sin x
1−2 sin2x
=1−2 sin2x
−sin xcos x−sin x
1−2 sin2x
=−1
sin x+2 sin x
sin x
=−csc x+ 2 = 2 cot x
Therefore, cos(2x)
sin x−sin x
cos(2x)= 2 cot x, and the given trigonometric identity
has been proven.
Question 8
Question
Prove the trigonometric identity:
sin(θ)−cos(θ)
sin(θ) + cos(θ)= tan (θ
2)
Solution
Step 1: Let’s start by expressing tan (θ
2)in terms of sin(θ)and cos(θ).
tan (θ
2)=sin (θ
2)
cos (θ
2)
=
2 sin(θ
2)cos(θ
2)
2 cos(θ
2)
cos2(θ
2)−sin2(θ
2)
=2 sin (θ
2)cos (θ
2)
2 cos2(θ
2)−2 sin2(θ
2)
=2 sin (θ
2)cos (θ
2)
2(cos2(θ
2)−sin2(θ
2))
=2 sin (θ
2)cos (θ
2)
2(cos(θ)−sin(θ))
=sin(θ)
sin(θ)−cos(θ)(using double angle identities)
7
Step 2: Now, we can substitute this expression back into the original identity
and simplify.
sin(θ)−cos(θ)
sin(θ) + cos(θ)=sin(θ)
sin(θ)−cos(θ)
= tan (θ
2)
Therefore, we have successfully proved the trigonometric identity:
sin(θ)−cos(θ)
sin(θ) + cos(θ)= tan (θ
2)
Question 9
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the given identity, we will make use of the angle addition formula for
sin(3x)and then simplify the expression step by step.
Step 1: Apply the angle addition formula for sin(3x)to get:
sin(3x) = sin(2x+x)
Step 2: Expand sin(2x+x)using the angle addition formula:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 3: Simplify sin(2x)and cos(2x)using double angle formulas:
sin(3x) = (2 sin(x) cos(x))(cos(x)) + (1 −2 sin2(x))(sin(x))
Step 4: Distribute and simplify:
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin2(x) sin(x)
Step 5: Simplify further:
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
Step 6: Simplify and rearrange terms to obtain the desired identity:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, we have successfully proven the trigonometric identity sin(3x) =
3 sin(x)−4 sin3(x).
8
Question 10
Question
Prove the following trigonometric identity:
sin x
1 + cos x+1 + cos x
sin x= 2 csc xcsc (x
2+π
4)
Solution
Step 1: Recall the definition of cosecant and simplify the right side.
Right side = 2 csc xcsc (x
2+π
4)
= 2 1
sin x·1
sin (x
2+π
4)
= 2 1
sin x·1
(sin x
2cos π
4+ cos x
2sin π
4)
= 2 1
sin x·1
(sin x
2·1
√2+ cos x
2·1
√2)
= 2 1
sin x·1
1
√2(sin x
2+ cos x
2)
= 2 √2
sin x(sin x
2+ cos x
2)
= 2 √2
sin x·sin (x
2+π
4)
= 2 (1
sin x+1
sin (x
2+π
4))
= 2 (csc x+ csc (x
2+π
4))
9
Step 2: Expand the left side and simplify.
Left side =sin x
1 + cos x+1 + cos x
sin x
=sin x
1 + cos x+1
sin x+ cos x
=sin2x+ 1 + cos xsin x
sin x(1 + cos x)
=(1 + sin x)(1 + cos x)
sin x(1 + cos x)
= csc x+ csc x
= 2 csc x
Step 3: Conclude that the original identity is true. Since the right side equals
2(csc x+ csc (x
2+π
4))and the left side equals 2 csc x, we have shown that the
given trigonometric identity is true.
Question 11
Question
Solve the equation sin(2x) cos(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Rewrite the given equation using double angle formulas.
sin(2x) cos(x) = 1
2
2 sin(x) cos(x)·cos(x) = 1
2
2 sin(x) cos2(x) = 1
2
Step 2: Rewrite cos2(x)in terms of sin(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1.
2 sin(x)(1 −sin2(x)) = 1
2
2 sin(x)−2 sin3(x) = 1
2
4 sin3(x)−2 sin(x) + 1 = 0
Step 3: Let y= sin(x), then the equation becomes a cubic equation.
4y3−2y+ 1 = 0
10
Step 4: Find the roots of the cubic equation. The roots of this cubic equation
might not have simple expressions in terms of radicals.
Step 5: Solve for xby finding the corresponding values of y. Finally, find the
values of xby solving for y= sin(x)using either numerical methods or graphing
techniques.
Therefore, the solutions to the equation sin(2x) cos(x) = 1
2in the interval
[0,2π)involve finding the roots of a cubic equation and then solving for xby
finding the corresponding values of y= sin(x).
Question 12
Question
Prove the following trigonometric identity:
cos(θ)−sin(θ) tan(θ) = cos(θ)
Solution
To prove the trigonometric identity cos(θ)−sin(θ) tan(θ) = cos(θ), we will
manipulate the left-hand side of the equation using trigonometric identities.
Step 1: Recall the definitions of tangent and sine functions:
tan(θ) = sin(θ)
cos(θ)
Step 2: Replace tan(θ)with sin(θ)
cos(θ)in the equation:
cos(θ)−sin(θ)·sin(θ)
cos(θ)
Step 3: Simplify the expression:
cos(θ)−sin2(θ)
cos(θ)
Step 4: Use the Pythagorean identity sin2(θ) + cos2(θ) = 1:
cos(θ)−1−cos2(θ)
cos(θ)
Step 5: Simplify further:
cos(θ)−1
cos(θ)+ cos(θ)
Step 6: Combine like terms:
cos(θ)−1
cos(θ)+ cos(θ) = cos(θ)
Therefore, cos(θ)−sin(θ) tan(θ) = cos(θ), and the identity is proved.
11
Question 13
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine, sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the equation: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides of the equation by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x≤2πare x=π
6
and x=5π
6.
Therefore, the solutions to the trigonometric equation sin(2x) = cos(x)in
the interval 0≤x≤2πare x=π
6and x=5π
6.
Question 14
Question
Prove the identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
formula for sine which states that sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
Step 1: Let’s start by applying the angle addition formula for sine to sin(3x):
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, we need to express sin(2x)and cos(2x)in terms of sin(x)and
cos(x)using the double angle identities:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
Step 3: Substitute in the expressions for sin(2x)and cos(2x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + cos2(x) sin(x)−sin3(x)
= 2 sin(x)−3 sin3(x) + (1 −sin2(x)) sin(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
12
Therefore, we have shown that sin(3x) = 3 sin(x)−4 sin3(x), completing the
proof.
Question 15
Question
Prove the trigonometric identity:
2 sin xcos xcos(2x) = sin(2x)
where xis a real number.
Solution
We will start by expanding the left-hand side of the equation using trigonometric
identities.
Step 1: Expand the left-hand side using trigonometric identities:
2 sin xcos xcos(2x) = 2 sin xcos x(cos2x−sin2x)
= 2 sin xcos x(cos2x−(1 −cos2x))
= 2 sin xcos x(cos2x−1 + cos2x)
= 2 sin xcos x(2 cos2x−1)
= 4 sin xcos2x−2 sin xcos x.
Step 2: Next, simplify the expression further:
4 sin xcos2x−2 sin xcos x= 2 sin x(2 cos2x)−2 sin xcos x
= 2 sin x(2 cos2x−cos x)
= 2 sin x(cos x)(2 cos x−1)
= sin xcos(2x).
Therefore, the left-hand side is equal to the right-hand side, and we have
proven the trigonometric identity:
2 sin xcos xcos(2x) = sin(2x).
Question 16
Question
Prove the trigonometric identity:
cot(θ)·sec(θ) = csc(θ)
13
Solution
Step 1: Rewrite all trigonometric functions in terms of sine and cosine.
cot(θ)·sec(θ) = cos(θ)
sin(θ)·1
cos(θ)
=cos(θ)
sin(θ)·cos(θ)
Step 2: Use the trigonometric identity sin(θ)·cos(θ) = 1
2·sin(2θ).
cos(θ)
sin(θ)·cos(θ)=cos(θ)
1
2·sin(2θ)
=2 cos(θ)
sin(2θ)
Step 3: Apply the double-angle identity for sine: sin(2θ) = 2 sin(θ) cos(θ).
2 cos(θ)
sin(2θ)=2 cos(θ)
2 sin(θ) cos(θ)
=2 cos(θ)
2 sin(θ) cos(θ)
=cos(θ)
sin(θ) cos(θ)
=1
sin(θ)
= csc(θ)
Therefore, cot(θ)·sec(θ) = csc(θ)is proven.
Question 17
Question
Solve the equation sin2(x)−√3 sin(x) + 1 = 0 for xin the interval [0,2π].
Solution
Step 1: Let u= sin(x), so the equation becomes u2−√3u+ 1 = 0.
Step 2: To solve for u, we can use the quadratic formula: u=−b±√b2−4ac
2a
with a= 1,b=−√3, and c= 1.
Step 3: Plugging in the values of a,b, and c, we get u=√3±√(−√3)2−4∗1∗1
2∗1.
Step 4: Simplifying under the square root, we have u=√3±√3−4
2.
Step 5: Further simplifying, we get u=√3±√−1
2. Since √−1 = i, we have
u=√3±i
2.
14
Step 6: Therefore, the solutions for uare u=√3+i
2and u=√3−i
2.
Step 7: Now, we need to find the corresponding values of xusing u= sin(x).
Step 8: For u=√3+i
2, we have sin(x) = √3+i
2. This implies x= sin−1(√3+i
2).
Step 9: Similarly, for u=√3−i
2, we have x= sin−1(√3−i
2).
Step 10: Finally, we need to find the values of xin the interval [0,2π].
Step 11: Therefore, the solutions to the equation sin2(x)−√3 sin(x) + 1 = 0
for xin the interval [0,2π]are x= sin−1(√3+i
2)and x= sin−1(√3−i
2).
Question 18
Question
Prove the following trigonometric identity:
(cos x+ sin x)2= 1 + sin 2x
Solution
To prove the trigonometric identity
(cos x+ sin x)2= 1 + sin 2x,
we will expand the left side and simplify to match the right side of the equation.
Step 1: Expand the left side using the distributive property.
(cos x+ sin x)2= (cos x+ sin x)(cos x+ sin x)
= cos2x+ 2 cos xsin x+ sin2x
Step 2: Use the Pythagorean Identity cos2x+ sin2x= 1.
cos2x+ 2 cos xsin x+ sin2x= 1 + 2 cos xsin x
Step 3: Use the double angle identity sin 2x= 2 sin xcos x.
1 + 2 cos xsin x= 1 + sin 2x
Hence, we have shown that (cos x+ sin x)2= 1 + sin 2x, which proves the
trigonometric identity.
Question 19
Question
Prove the following trigonometric identity:
√2 cos(θ) sin(θ) = sin(2θ)
15
Solution
To prove the trigonometric identity √2 cos(θ) sin(θ) = sin(2θ), we will use the
double angle formula for sine.
Step 1: Recall the double angle formula for sine:
sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Begin with the left side of the equation and simplify:
√2 cos(θ) sin(θ) = √2·1
2(sin(2θ))
=√2
2sin(2θ)
= sin(2θ)·√2
2
= sin(2θ)(since √2
2= 1)
= sin(2θ)
Step 3: Therefore, we have shown that √2 cos(θ) sin(θ) = sin(2θ), which
proves the given trigonometric identity.
Question 20
Question
Prove the identity:
cos(θ)
1−sin(θ)+1
cos(θ)=2
1 + sin(θ)
Solution
Step 1: Find a common denominator To add the fractions on the left-hand
side, we need to find a common denominator. The common denominator for
1−sin(θ)and cos(θ)is cos(θ)(1−sin(θ)). Rewrite the fractions with this common
denominator:
cos(θ)2
cos(θ)(1 −sin(θ)) +1(1 −sin(θ))
cos(θ)(1 −sin(θ))
Step 2: Combine the fractions Combine the fractions over the common
denominator:
cos(θ)2+ 1 −sin(θ)
cos(θ)(1 −sin(θ))
16
Step 3: Expand the numerator Expand the numerator using the Pythagorean
identity cos(θ)2= 1 −sin(θ)2:
1−sin(θ)2+ 1 −sin(θ)
cos(θ)(1 −sin(θ))
Step 4: Simplify the numerator Simplify the numerator:
2−sin(θ)−sin(θ)2
cos(θ)(1 −sin(θ))
Step 5: Use the Pythagorean identity Since sin(θ)2= 1 −cos(θ)2, sub-
stitute this into the numerator:
2−sin(θ)−(1 −cos(θ)2)
cos(θ)(1 −sin(θ))
Step 6: Further simplify the numerator Simplify the numerator:
2−sin(θ)−1 + cos(θ)2
cos(θ)(1 −sin(θ)) =1 + cos(θ)2−sin(θ)
cos(θ)(1 −sin(θ))
Step 7: Use the Pythagorean identity again Since cos(θ)2= 1−sin(θ)2,
substitute this into the numerator:
1 + (1 −sin(θ)2)−sin(θ)
cos(θ)(1 −sin(θ))
Step 8: Simplify the numerator Simplify the numerator:
1+1−sin(θ)2−sin(θ)
cos(θ)(1 −sin(θ)) =2−sin(θ)2−sin(θ)
cos(θ)(1 −sin(θ))
Step 9: Simplify the numerator further Since sin(θ)2+sin(θ) = sin(θ)(1+
sin(θ)), substitute this into the numerator:
2−sin(θ)(1 + sin(θ))
cos(θ)(1 −sin(θ)) =2
1 + sin(θ)
Therefore, the given identity is true.
Question 21
Question
Prove the trigonometric identity:
sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= tan (θ
2)
17
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation into the form of the right-hand side.
Step 1: Start with the left-hand side of the equation and combine the
fractions: sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)
Step 2: To simplify the expression, multiply the first term by 1−cos(θ)
1−cos(θ)and
the second term by 1−sin(θ)
1−sin(θ):
sin(θ)(1 −cos(θ))
(1 + cos(θ))(1 −cos(θ)) +cos(θ)(1 −sin(θ))
(1 + sin(θ))(1 −sin(θ))
Step 3: Expand and simplify the expression:
sin(θ)−sin(θ) cos(θ)
1−cos2(θ)+cos(θ)−cos(θ) sin(θ)
1−sin2(θ)
Step 4: Recall the Pythagorean identities sin2(θ) + cos2(θ)=1and 1−
sin2(θ) = cos2(θ):
sin(θ)−sin(θ) cos(θ)
sin2(θ)+cos(θ)−cos(θ) sin(θ)
cos2(θ)
Step 5: Simplify further to obtain a common denominator:
sin(θ)(1 −cos(θ))
sin2(θ)+cos(θ)(1 −sin(θ))
cos2(θ)
Step 6: Use double angle formula for tangent: tan(2α) = 2 tan(α)
1−tan2(α):
2 tan (θ
2)
1−tan2(θ
2)= tan (θ
2)
Therefore, we have successfully proven the trigonometric identity:
sin(θ)
1 + cos(θ)+cos(θ)
1 + sin(θ)= tan (θ
2)
Question 22
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
18
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to match the right-hand side.
Step 1: Start with the left-hand side of the equation:
sin4(x)−cos4(x)
Step 2: Rewrite sin4(x)as (sin2(x))2and cos4(x)as (cos2(x))2:
(sin2(x))2−(cos2(x))2
Step 3: Use the difference of squares identity: a2−b2= (a+b)(a−b):
(sin2(x) + cos2(x))(sin2(x)−cos2(x))
Step 4: Recall that sin2(x) + cos2(x) = 1 (Pythagorean identity):
1(sin2(x)−cos2(x))
Step 5: Rewrite sin2(x)−cos2(x)as sin(2x)using the double angle identity
sin(2θ) = 2 sin(θ) cos(θ):
1·sin(2x)
Step 6: Now, simplify the expression on the right-hand side:
sin(2x) = sin(2x)·1 = sin(2x)
Step 7: Hence, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x), as
required.
Question 23
Question
Prove the following trigonometric identity:
cos(θ) cot(θ)−sin(θ) = 1−sin2(θ)
sin(θ)
Solution
Step 1: Rewrite the left-hand side of the equation using trigonometric identi-
ties.
cos(θ) cot(θ)−sin(θ) = cos(θ)
sin(θ)−sin(θ)
Step 2: Find a common denominator for the terms.
cos(θ)
sin(θ)−sin(θ) = cos(θ)−sin2(θ)
sin(θ)
19
Step 3: Use the Pythagorean identity cos2(θ)=1−sin2(θ)to simplify the
numerator. cos(θ)−sin2(θ)
sin(θ)=1−sin2(θ)
sin(θ)
Step 4: Thus, we have shown that
cos(θ) cot(θ)−sin(θ) = 1−sin2(θ)
sin(θ)
And the trigonometric identity is proven.
Question 24
Question
Solve the equation 2 sin2(x)−5 cos(x) + 1 = 0 for xin the interval [0,2π].
Solution
Step 1: Start by using the Pythagorean identity for sine and cosine: sin2(x) =
1−cos2(x). Step 2: Rewrite the equation using the Pythagorean identity:
2(1 −cos2(x)) −5 cos(x) + 1 = 0.
Step 3: Simplify the equation:
2−2 cos2(x)−5 cos(x) + 1 = 0.
Step 4: Rearrange the terms to form a quadratic equation in terms of cosine:
2 cos2(x) + 5 cos(x)−3 = 0.
Step 5: Solve this quadratic equation by factoring or using the quadratic for-
mula. The solutions are:
cos(x) = −5±√52−4(2)(−3)
2(2) =−5±√49
4=−5±7
4.
Step 6: Thus, we have two possible values for cos(x):cos(x) = 1 or cos(x) = −3
2.
Step 7: Since cosine values lie between -1 and 1, the second solution is extraneous
and we must discard it. Therefore, we have cos(x) = 1. Step 8: Solve for x
using the fact that cos(x) = 1 implies x= 2nπ where nis an integer. Step 9:
Checking the interval [0,2π], the solution is x= 0.
Therefore, the solution to the equation 2 sin2(x)−5 cos(x) + 1 = 0 for xin
the interval [0,2π]is x= 0.
20
Question 25
Question
Prove the following trigonometric identity:
cot(θ)−tan(θ) = 2
sin(2θ)
Solution
We will start by expressing cot(θ)and tan(θ)in terms of sine and cosine:
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Subtracting tan(θ)from cot(θ)
cot(θ)−tan(θ) = cos(θ)
sin(θ)−sin(θ)
cos(θ)
Step 3: Finding a common denominator
cot(θ)−tan(θ) = cos2(θ)
cos(θ) sin(θ)−sin2(θ)
cos(θ) sin(θ)
Step 4: Combining the fractions
cot(θ)−tan(θ) = cos2(θ)−sin2(θ)
cos(θ) sin(θ)
Step 5: Using the double angle identity sin(2θ) = 2 sin(θ) cos(θ)
cot(θ)−tan(θ) = cos(2θ)
sin(2θ)
Step 6: Recall that cos(2θ) = 2 cos2(θ)−1
cot(θ)−tan(θ) = 2 cos2(θ)−1
sin(2θ)
Step 7: Simplifying further
cot(θ)−tan(θ) = 2 cos2(θ)−1
sin(2θ)=2−2 sin2(θ)
sin(2θ)
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Step 8: Using the Pythagorean identity sin2(θ) = 1 −cos2(θ)
cot(θ)−tan(θ) = 2−2(1 −cos2(θ))
sin(2θ)
cot(θ)−tan(θ) = 2−2 + 2 cos2(θ)
sin(2θ)
cot(θ)−tan(θ) = 2 cos2(θ)
sin(2θ)
Step 9: Finally, we see that cot(θ)−tan(θ) = 2 cos2(θ)
sin(2θ)=2
sin(2θ), which
proves the given trigonometric identity.
Question 26
Question
Prove the following trigonometric identity:
cot(x) csc(x)−tan(x) sec(x) = cos(x)−sin(x)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation using basic trigonometric identities until we obtain the right-
hand side.
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LHS = cot(x) csc(x)−tan(x) sec(x)
=cos(x)
sin(x)·1
sin(x)−sin(x)
cos(x)·1
cos(x)
=cos(x)
sin(x)2−sin(x)
cos(x)2
=cos(x) cos(x)
sin(x)2cos(x)−sin(x) sin(x)
cos(x)2sin(x)
=cos2(x)
sin2(x)−sin2(x)
cos2(x)
=cos2(x)−sin2(x)
sin2(x) cos2(x)
=cos(2x)
sin(2x)
=2 cos(x) sin(x)
2 sin(x) cos(x)
= 1
= cos(x)−sin(x)
=RHS
Therefore, we have shown that the left-hand side (LHS) is equal to the right-
hand side (RHS), and the trigonometric identity is proved.
Question 27
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
1. We can start with the triple angle identity: sin(3x) = 3 sin(x)−4 sin3(x).
2. Let’s rewrite the triple angle identity using the double angle identity:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x).
3. Recall the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x).
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4. Substituting these expressions into our rewritten triple angle identity:
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x).
5. Simplifying the expression:
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x).
6. Factoring out a sin(x)from the first two terms gives:
sin(3x) = sin(x)(2 cos2(x) + cos2(x)) −sin3(x).
7. Further simplifying:
sin(3x) = sin(x)(3 cos2(x)) −sin3(x).
8. Using the Pythagorean identity sin2(x) + cos2(x) = 1:
3 cos2(x) = 3(1 −sin2(x)) = 3 −3 sin2(x).
9. Substituting this back into our expression:
sin(3x) = sin(x)(3 −3 sin2(x)) −sin3(x).
10. Factorizing 3−3 sin2(x)gives 3(1 −sin2(x)) = 3 cos2(x) = 3:
sin(3x) = 3 sin(x)−3 sin3(x)−sin3(x).
11. Simplifying, we arrive at the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x).
Question 28
Question
Solve the trigonometric equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x < 2π.
Solution
Step 1: Let’s rewrite the equation in terms of a single variable, say u= sin(x).
u2−3u+ 2 = 0
Step 2: We can factor the quadratic equation to find the roots.
(u−2)(u−1) = 0
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Step 3: Setting each factor to zero gives us two possible values for u.
u= 2 or u= 1
Step 4: Recall that u= sin(x), so we have two possibilities for sin(x).
sin(x) = 2 or sin(x) = 1
Step 5: Since the range of sin is [−1,1], the first possibility is not valid.
Therefore, we focus on solving sin(x) = 1.
Step 6: The equation sin(x)=1has solutions at x=π
2+ 2πk, where kis
an integer.
Step 7: Therefore, the solutions to the trigonometric equation sin2(x)−
3 sin(x) + 2 = 0 for 0≤x < 2πare x=π
2and x=3π
2.
Question 29
Question
Prove the following trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: We will start by using the angle addition formula for sine: sin(A+B) =
sin(A) cos(B) + cos(A) sin(B). Applying this formula to sin(3x) = sin(2x+x),
we get:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, we’ll express sin(2x)and cos(2x)in terms of sin(x)and
cos(x). Using the double-angle formulas, we have:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 3: Now, substitute the expressions for sin(2x)and cos(2x)back into
the equation from Step 1:
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
Step 4: Simplify the expression by expanding terms and combining like
terms:
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin3(x)
Step 5: Recall the double-angle identity: cos2(x) = 1 −sin2(x). Substitute
cos2(x) = 1 −sin2(x)into the expression:
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
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Step 6: Simplify the expression further:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x)
Step 7: Combine like terms to obtain the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)has been
proven.
Question 30
Question
Prove the following trigonometric identity:
tan2(x)−sin2(x) = sec2(x)−1
Solution
To prove the trigonometric identity tan2(x)−sin2(x) = sec2(x)−1, we will first
rewrite all trigonometric functions in terms of sin(x)and cos(x).
Step 1: Rewrite tan xin terms of sin xand cos x.
tan(x) = sin(x)
cos(x)
Step 2: Square the expression for tan(x).
tan2(x) = (sin(x)
cos(x))2
=sin2(x)
cos2(x)
Step 3: Rewrite sin2(x)in terms of cos2(x).
sin2(x) = 1 −cos2(x)
Step 4: Substitute the expressions for tan2(x)and sin2(x)back into the
original identity.
sin2(x)
cos2(x)−(1 −cos2(x)) = sec2(x)−1
Step 5: Simplify the left-hand side of the equation.
sin2(x)
cos2(x)−1 + cos2(x) = sec2(x)−1
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Step 6: Combine like terms on the left-hand side.
sin2(x)
cos2(x)+ cos2(x)−1 = sec2(x)−1
Step 7: Use the Pythagorean identity sin2(x)+cos2(x) = 1 on the left-hand
side. 1−cos2(x)
cos2(x)+ cos2(x)−1 = sec2(x)−1
Step 8: Simplify the left-hand side.
1
cos2(x)−1 = sec2(x)−1
Step 9: Recognize that sec2(x) = 1
cos2(x).
sec2(x)−1 = sec2(x)−1
Therefore, the trigonometric identity tan2(x)−sin2(x) = sec2(x)−1has
been proven.
Question 31
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for xin the interval [0,2π].
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
u2−3u+ 2 = 0
Step 2: Solve the quadratic equation u2−3u+ 2 = 0 by factoring or using
the quadratic formula:
u2−2u−u+ 2 = 0
u(u−2) −1(u−2) = 0
(u−2)(u−1) = 0
So, u= 2 or u= 1.
Step 3: Since u= sin(x), we have two cases to consider: Case 1: u= 1 =⇒
sin(x) = 1. The solutions in [0,2π]for this case are x=π
2, and 5π
2. Case 2:
u= 2 =⇒sin(x) = 2, which is not possible.
Step 4: Thus, the solutions to the equation sin2(x)−3 sin(x) + 2 = 0 in the
interval [0,2π]are x=π
2, and 5π
2.
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Question 32
Question
Prove the trigonometric identity:
1−sin x
cos x= tan (x
2)
Solution
To prove the given trigonometric identity, we will start by expressing tan (x
2)in
terms of sin xand cos x.
Step 1: Express tan (x
2)in terms of sin xand cos x. We know that tan (x
2)=
sin(x
2)
cos(x
2). By half-angle identities, we have:
sin (x
2)=±√1−cos x
2
cos (x
2)=±√1 + cos x
2
Since xlies in the first quadrant, both sine and cosine values are positive. Thus,
sin (x
2)=√1−cos x
2
cos (x
2)=√1 + cos x
2
Step 2: Substitute sin (x
2)and cos (x
2)into tan (x
2).
tan (x
2)=sin (x
2)
cos (x
2)=√1−cos x
2
√1+cos x
2
=√1−cos x
1 + cos x
Step 3: Manipulate the right side of the identity to match with tan (x
2).
We can simplify the expression on the right side of the identity in the following
way:
tan (x
2)=√1−cos x
1 + cos x=√(1 −cos x)(1 −cos x)
(1 + cos x)(1 −cos x)=√1−2 cos x+ cos2x
1−cos2x
=√1−2 cos x+ cos2x
sin2x=√(1 −cos x)2
sin2x=1−cos x
sin x=1−sin x
cos x
Therefore, we have proven the trigonometric identity 1−sin x
cos x= tan (x
2).
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Question 33
Question
Solve the equation sin(3x) = cos(2x)for xin the interval [0,2π).
Solution
Step 1: Recall the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x). Rewriting the equation in terms of sin(x), we have:
3 sin(x)−4 sin3(x) = 1 −2 sin2(x)
4 sin3(x)−2 sin2(x)−3 sin(x) + 1 = 0
Step 2: Let y= sin(x), then the equation becomes 4y3−2y2−3y+ 1 = 0.
We can factor this equation as:
(y−1)(4y2+ 2y−1) = 0
Step 3: Solving the quadratic factor, we have 4y2+ 2y−1 = 0 ⇒y=
−2±√22−4(4)(−1)
2(4) . Thus, y=−1±√3
4.
Step 4: Since y= sin(x), we have two possible values for yin the interval
[0,2π):sin(x) = −1+√3
4or sin(x) = −1−√3
4.
Step 5: Solving for x, we find sin(x) = −1+√3
4corresponds to x=5π
6and
sin(x) = −1−√3
4corresponds to x=7π
6.
Therefore, the solutions to the equation sin(3x) = cos(2x)in the interval
[0,2π)are x=5π
6and x=7π
6.
Question 34
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
and triple angle formulas for sine.
Step 1: Write sin(3x)using the angle addition formula By the angle
addition formula for sine, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
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Step 2: Express sin(2x)and cos(2x)using double angle formulas
Recall the double angle formulas:
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ)
Therefore, sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x).
Step 3: Substitute sin(2x)and cos(2x)back into the expression
for sin(3x)Substituting these expressions back into sin(3x) = sin(2x) cos(x) +
cos(2x) sin(x)gives:
sin(3x) = (2 sin(x) cos(x)) cos(x) + (cos2(x)−sin2(x)) sin(x)
Step 4: Simplify the expression
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin3(x)
Now, since sin2(x) + cos2(x) = 1:
sin(3x) = 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Hence, we have proved the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 35
Question
Prove the following trigonometric identity:
tan(x)−sin(x)
1 + tan(x) sin(x)=cos(x)
1 + sin(x)
Solution
To prove the given trigonometric identity, we will manipulate the left side of the
equation until it matches the right side.
Step 1: Rewrite the left side using trigonometric identities.
tan(x)−sin(x)
1 + tan(x) sin(x)=
sin(x)
cos(x)−sin(x)
1 + sin(x)
cos(x)sin(x)
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Step 2: Simplify the expression by finding a common denominator.
=
sin(x)−sin(x) cos(x)
cos(x)
1 + sin2(x)
cos(x)
Step 3: Combine the fractions.
=sin(x)−sin(x) cos(x)
cos(x) + sin2(x)
Step 4: Factor out a cos(x)in the numerator.
=sin(x)(1 −cos(x))
cos(x)(1 + sin(x))
Step 5: Cancel out a sin(x)term.
=1−cos(x)
1 + sin(x)
Step 6: Use the trigonometric identity cos(x) = 1 −sin2(x).
=sin2(x)
1−sin2(x)
Step 7: Simplify using the Pythagorean identity sin2(x) + cos2(x) = 1.
=sin2(x)
cos2(x)
Step 8: Simplify the expression to match the right side.
=cos(x)
1 + sin(x)
Therefore, the left side of the given trigonometric identity is equal to the
right side, and the identity has been proven.
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