MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 2
Liberty University
Question 1
Question
Prove the trigonometric identity:
cos3(x)
1−sin(x)+sin3(x)
1−cos(x)= tan(x)
Solution
We start with the expression: cos3(x)
1−sin(x)+sin3(x)
1−cos(x)
We can write: =cos4(x)
cos(x)(1 −sin(x)) +sin4(x)
sin(x)(1 −cos(x)) [Multiply by cos(x)
cos(x)and sin(x)
sin(x)]
Expand: =cos4(x)
cos(x)−cos2(x)+sin4(x)
sin(x)−sin2(x)[Expand denominators]
Simplify: =cos3(x)
1−cos(x)+sin3(x)
1−sin(x)
Now, notice that tan(x) = sin(x)
cos(x)=sin3(x)
cos(x) sin2(x)=sin3(x)
1−cos2(x)
and also tan(x) = sin(x)
cos(x)=cos(x) sin(x)
cos2(x)=cos(x)
1−sin2(x)
Therefore, we have: tan(x) = sin3(x)
1−cos2(x)=cos(x)
1−sin2(x)
So, the identity holds: cos3(x)
1−sin(x)+sin3(x)
1−cos(x)= tan(x)
Question 2
Question
Prove the following trigonometric identity:
cos(θ)
1−sin(θ)+1
cos(θ)=2
1 + sin(θ)
Solution
1. We start by obtaining a common denominator on the left side of the
equation:
cos(θ)
1−sin(θ)+1
cos(θ)=cos(θ) cos(θ)
cos(θ)(1 −sin(θ)) +(1 −sin(θ))
cos(θ)(1 −sin(θ))
=cos2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
2. Since cos2(θ) = 1 −sin2(θ)(Pythagorean identity), we can substitute this
into the numerator:
cos2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ)) =1−sin2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
=2−sin(θ)−sin2(θ)
cos(θ)(1 −sin(θ))
3. Next, we use the Pythagorean identity sin2(θ)=1−cos2(θ)to simplify
the numerator further:
2−sin(θ)−sin2(θ)
cos(θ)(1 −sin(θ)) =2−sin(θ)−(1 −cos2(θ))
cos(θ)(1 −sin(θ))
=1−sin(θ) + cos2(θ)
cos(θ)(1 −sin(θ))
4. Finally, using the Pythagorean identity cos2(θ)=1−sin2(θ)for the nu-
2
merator:
1−sin(θ) + cos2(θ)
cos(θ)(1 −sin(θ)) =1−sin(θ)+1−sin2(θ)
cos(θ)(1 −sin(θ))
=2−sin(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
=2−2 sin(θ)
cos(θ)(1 −sin(θ))
=2(1 −sin(θ))
cos(θ)(1 −sin(θ))
=2
cos(θ)
=2
1 + sin(θ)
Therefore, we have shown that cos(θ)
1−sin(θ)+1
cos(θ)=2
1+sin(θ)is a true identity.
Question 3
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left side and
simplify it step by step until we reach the right side.
Step 1: Start with the left side of the identity:
sin4(x)−cos4(x)
Step 2: Recall the trigonometric identity: sin2(x) = 1 −cos2(x).
sin4(x) = (sin2(x))2
= (1 −cos2(x))2
= 1 −2 cos2(x) + (cos2(x))2
= 1 −2 cos2(x) + cos4(x)
3
Step 3: Substitute the above result back into the original expression:
sin4(x)−cos4(x) = 1 −2 cos2(x) + cos4(x)−cos4(x)
= 1 −2 cos2(x)
= 2(1/2−cos2(x))
= 2(sin2(x))
= 2 sin2(x) cos2(x)
Step 4: Thus, we have shown that sin4(x)−cos4(x) = 2 sin2(x) cos2(x),
which completes the proof of the given trigonometric identity.
Question 4
Question
Prove the following trigonometric identity:
sin(2x)
1 + cos(2x)= tan(x)
Solution
Step 1: We will start with the left-hand side of the equation and attempt to
simplify it using trigonometric identities. Step 2: Recall the double angle iden-
tity for sine: sin(2x) = 2 sin(x) cos(x). Step 3: Likewise, recall the double
angle identity for cosine: cos(2x) = cos2(x)−sin2(x). Step 4: Substitute these
identities into the left-hand side of the equation:
sin(2x)
1 + cos(2x)=2 sin(x) cos(x)
1 + (cos2(x)−sin2(x))
Step 5: Simplify the denominator by expanding the square term:
1 + cos2(x)−sin2(x) = cos2(x)+1−sin2(x) = cos2(x) + cos2(x) = 2 cos2(x)
Step 6: Substitute this back into the expression:
2 sin(x) cos(x)
2 cos2(x)=sin(x)
cos(x)= tan(x)
Step 7: Therefore, we have shown that the left-hand side is equal to the right-
hand side, and the trigonometric identity sin(2x)
1+cos(2x)= tan(x)is proven.
4
Question 5
Question
Prove the following trigonometric identity:
cos(x)
1−sin(x)+1
cos(x) + sin(x)=1
cos(x)
Solution
Step 1: Start by manipulating each side individually.
cos(x)
1−sin(x)+1
cos(x) + sin(x)=cos(x)
1−sin(x)+1
cos(x) + sin(x)·1 + sin(x)
1 + sin(x)
=cos(x)
1−sin(x)+1 + sin(x)
cos(x) + sin(x)
=cos(x)(1 + sin(x)) + (1 −sin(x))
(1 −sin(x))(1 + sin(x))
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−sin2(x)
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−(1 −cos2(x))
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−1 + cos2(x)
=cos(x) + cos(x) sin(x)+1−sin(x)
2 cos2(x)−1
Step 2: Simplify the expression obtained in Step 1.
cos(x) + cos(x) sin(x)+1−sin(x)
2 cos2(x)−1=cos(x) (1 + sin(x)) + 1 −sin(x)
2 cos2(x)−1
=1 + sin(x)+1−sin(x)
2 cos2(x)−1
=2
2 cos2(x)−1
=1
cos2(x)−sin2(x)
=1
cos(x)(Using trigonometric identity: cos2(x)−sin2(x) = cos(x))
Therefore, cos(x)
1−sin(x)+1
cos(x)+sin(x)=1
cos(x)is true, and the identity is proven.
5
Question 6
Question
Prove the trigonometric identity:
1−cos 2θ
sin2θ= csc2θ−cot2θ
Solution
Start with the left-hand side: 1−cos 2θ
sin2θ=1−cos2θ+ sin2θ
sin2θ
=sin2θ
sin2θ
= 1
Now, simplify the right-hand side: csc2θ−cot2θ=(1
sin θ)2
−(cos θ
sin θ)2
=1
sin2θ−cos2θ
sin2θ
=1−cos2θ
sin2θ
=sin2θ
sin2θ
= 1
Thus, we have shown that the left-hand side is equal to the right-hand side,
and the trigonometric identity is proven.
Question 7
Question
Solve the trigonometric equation sin2(x) + sin(x)−6=0for xin the interval
[0,2π).
Solution
Step 1: Let u= sin(x). Then the equation becomes u2+u−6 = 0.
Step 2: Factor the quadratic equation: (u+ 3)(u−2) = 0.
Step 3: Set each factor to zero and solve for u:u+ 3 = 0 =⇒u=−3or
u−2 = 0 =⇒u= 2.
6
Step 4: Recall that u= sin(x). So, sin(x) = −3and sin(x) = 2 are not valid
solutions since the range of sin is [−1,1].
Step 5: Therefore, u= sin(x) = −3or u= sin(x) = 2 have no solutions in
the interval [0,2π).
Step 6: The only valid solution comes from u= sin(x) = 2, which has no
solutions in the interval [0,2π).
Step 7: Thus, the original trigonometric equation sin2(x) + sin(x)−6 = 0
has no solutions in the interval [0,2π).
Question 8
Question
Prove the following trigonometric identity:
1
sin(θ) + cos(θ)=√2
2·1
sin(θ+ 45◦)
Solution
Step 1: We will start by converting 45◦to radians. Recall that πradians is
equivalent to 180◦, so 1radian is approximately 57.2958◦. Therefore, 45◦is
equivalent to 45
57.2958 ≈0.7854 radians.
Step 2: Let’s rewrite the right-hand side of the given identity. Using the
angle addition formula for sine, we have:
sin(θ+ 45◦) = sin(θ) cos(45◦) + cos(θ) sin(45◦)
Plugging in the values for cos(45◦)and sin(45◦)(which are √2
2), we get:
sin(θ+ 45◦) = √2
2sin(θ) + √2
2cos(θ)
Step 3: Now, let’s simplify the right-hand side of the given identity using
this expression:
√2
2·1
sin(θ+ 45◦)=√2
2·1
√2
2sin(θ) + √2
2cos(θ)
=√2
2·1
√2(sin(θ) + cos(θ))
=1
sin(θ) + cos(θ)
Step 4: Since the right-hand side simplifies to the left-hand side of the given
identity, the identity is proven.
7
Question 9
Question
Prove the following trigonometric identity:
tan(θ)−sin(θ)
tan(θ) + sin(θ)=1−cos(θ)
1 + cos(θ)
Solution
Step 1: Start with the left side of the equation and manipulate to get the right
side:
tan(θ)−sin(θ)
tan(θ) + sin(θ)=
sin(θ)
cos(θ)−sin(θ)
sin(θ)
cos(θ)+ sin(θ)
=
sin(θ)−cos(θ) sin(θ)
cos(θ)
sin(θ)+cos(θ) sin(θ)
cos(θ)
=sin(θ)(1 −cos(θ))
sin(θ)(1 + cos(θ))
=1−cos(θ)
1 + cos(θ)
Therefore, the identity is proved.
Question 10
Question
Prove the following trigonometric identity:
1−cos θ
sin θ=tan θ
1 + cot θ
Solution
To prove the given trigonometric identity, we will simplify the left-hand side
(LHS) and the right-hand side (RHS) and show that they are equal.
Step 1: Simplify the LHS.
1−cos θ
sin θ=1−cos θ
sin θ·1 + cos θ
1 + cos θ=1−cos2θ
sin θ(1 + cos θ)
Using the Pythagorean identity sin2θ+ cos2θ= 1, we have cos2θ= 1 −sin2θ.
Substituting this into the expression above:
1−cos2θ
sin θ(1 + cos θ)=sin2θ
sin θ(1 + cos θ)=sin θ
1 + cos θ
8
Step 2: Simplify the RHS.
tan θ
1 + cot θ=sin θ/ cos θ
1 + cos θ/ sin θ=sin θ
cos θ·sin θ
sin θ+ cos θ=sin2θ
sin θ+ cos θ
Step 3: Compare the LHS and RHS. Since both the LHS and RHS simplify
to sin θ
1+cos θ, the given trigonometric identity is proven.
Question 11
Question
Prove the identity cot2(θ)−sin2(θ) = csc2(θ)for all θwhere the expressions are
defined.
Solution
To prove the given identity, we will start with the left-hand side and manipulate
it to obtain the right-hand side.
Step 1: Express cot2(θ)in terms of sin(θ)and cos(θ)using the definitions
of cotangent, secant, and cosecant:
cot2(θ)−sin2(θ) = cos2(θ)
sin2(θ)−sin2(θ) = cos2(θ)−sin4(θ)
sin2(θ)
Step 2: Simplify the expression by factoring out a sin2(θ)in the numerator:
cos2(θ)−sin4(θ)
sin2(θ)=(1 −sin2(θ)) cos2(θ)
sin2(θ)=cos2(θ)
sin2(θ)−sin2(θ) cos2(θ)
sin2(θ)
Step 3: Simplify further by canceling out sin2(θ)in the last term:
cos2(θ)
sin2(θ)−sin2(θ) cos2(θ)
sin2(θ)= csc2(θ)−sin2(θ)
Step 4: Hence, we have shown that cot2(θ)−sin2(θ) = csc2(θ)for all θ
where the expressions are defined.
Question 12
Question
Prove the trigonometric identity:
sin(x)
1−cos(x)= csc(x) + cot(x)
9
Solution
Step 1: Rewrite the right-hand side of the equation in terms of sine and cosine
functions.
csc(x) + cot(x) = 1
sin(x)+cos(x)
sin(x)=sin(x) + cos(x)
sin(x)
Step 2: Simplify the expression on the right-hand side.
sin(x) + cos(x)
sin(x)=sin(x)
sin(x)+cos(x)
sin(x)= 1 + cos(x)
sin(x)= 1 + cot(x)
Step 3: Now, we need to show that the two sides are equal.
Left-hand side =sin(x)
1−cos(x)
Step 4: Rationalize the denominator of the left-hand side.
sin(x)
1−cos(x)=sin(x)(1 + cos(x))
(1 −cos(x))(1 + cos(x)) =sin(x) + sin(x) cos(x)
1−cos2(x)
Step 5: Recall that sin2(x) + cos2(x) = 1.
sin(x) + sin(x) cos(x)
1−cos2(x)=sin(x) + sin(x) cos(x)
sin2(x)=sin(x)(1 + cos(x))
sin2(x)= csc(x)+cot(x)
Therefore, the given trigonometric identity sin(x)
1−cos(x)= csc(x) + cot(x)is
proven to be true.
Question 13
Question
Prove the trigonometric identity:
1
sin(A) + cot(A)= cos(A)
Solution
To prove the given trigonometric identity, we will use the definitions of the
trigonometric functions and algebraic manipulations.
Step 1: Rewrite in terms of sine and cosine Let’s rewrite cot(A)in
terms of sin(A)and cos(A). Recall that cot(A) = 1
tan(A)=1
sin(A)
cos(A)
=cos(A)
sin(A).
Now, rewrite the given expression as:
1
sin(A) + cos(A)
sin(A)
=1
sin2(A)+cos(A)
sin(A)
10
Step 2: Simplify the expression Now, simplify the expression further:
1
sin2(A)+cos(A)
sin(A)
=sin(A)
sin2(A) + cos(A)
Step 3: Use Pythagorean Identity Next, use the Pythagorean identity
sin2(A) + cos2(A) = 1 to simplify the expression:
sin(A)
sin2(A) + cos(A)=sin(A)
1 + cos(A)
Step 4: Distribute and Simplify Distribute the sin(A)in the denomina-
tor: sin(A)
1 + cos(A)=sin(A)
1·1
1 + cos(A)= sin(A)·1
1 + cos(A)
Step 5: Revisit the given expression Now, recall the given expression
we want to prove: 1
sin(A) + cot(A)= cos(A)
Step 6: Substitute back Substitute back the expression we simplified:
sin(A)·1
1 + cos(A)= cos(A)
Step 7: Refine the expression Further simplify the expression:
sin(A)
1 + cos(A)= cos(A)
Step 8: Use the Reciprocal Identity Finally, notice that cos(A) =
1
sec(A)=sin(A)
1+cos(A). Hence, we have proved the given trigonometric identity:
1
sin(A) + cot(A)= cos(A)
Question 14
Question
Prove the following trigonometric identity:
1 + tan(θ)
1−tan(θ)= sec2(θ)
11
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until we achieve the right-hand side.
Step 1: Start with the left-hand side of the equation:
1 + tan(θ)
1−tan(θ)
Step 2: Use the definition of tangent:
1 + sin(θ)
cos(θ)
1−sin(θ)
cos(θ)
Step 3: Simplify the expression further:
cos(θ) + sin(θ)
cos(θ)−sin(θ)
Step 4: Multiply both the numerator and the denominator by cos(θ):
(cos(θ) + sin(θ)) cos(θ)
(cos(θ)−sin(θ)) cos(θ)
Step 5: Expand the numerators and denominators:
cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)
Step 6: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1:
cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)=cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)·1
1=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)
Step 7: Use the double-angle identity sin(2θ) = 2 sin(θ) cos(θ):
1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)=1 + 1
2sin(2θ)
1−1
2sin(2θ)=1 + 1
2sin(2θ)
1−1
2sin(2θ)
Step 8: Use the double-angle identity for sine: sin(2θ) = 2 sin(θ) cos(θ):
1 + 1
2sin(2θ)
1−1
2sin(2θ)=1 + 1
2·2 sin(θ) cos(θ)
1−1
2·2 sin(θ) cos(θ)=1 + sin(θ)
1−sin(θ)= sec2(θ)
Therefore, we have shown that 1+tan(θ)
1−tan(θ)= sec2(θ), thus proving the trigono-
metric identity.
12
Question 15
Question
Prove the following trigonometric identity:
2 sin2(x)−1 = cos(2x)
Solution
Step 1: We will start with the right-hand side of the equation, which is cos(2x).
Using the double angle identity for cosine, we have:
cos(2x) = cos2(x)−sin2(x)
Step 2: Now, we will replace cos2(x)in the above expression using the
Pythagorean identity cos2(x) = 1 −sin2(x). Substituting this in, we get:
cos(2x) = (1 −sin2(x)) −sin2(x) = 1 −2 sin2(x)
Step 3: Simplifying the expression 1−2 sin2(x), we have:
cos(2x) = 1 −2 sin2(x) = −2 sin2(x)+1
Step 4: Comparing the expression for cos(2x)from Step 3 with the left-hand
side of the original equation, 2 sin2(x)−1, we see that:
2 sin2(x)−1 = −2 sin2(x)+1
Step 5: Hence, we have proved that:
2 sin2(x)−1 = cos(2x)
Therefore, the trigonometric identity is verified.
Question 16
Question
Prove the following trigonometric identity:
1−tan x
1 + tan x=sin x−cos x
sin x+ cos x
13
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation until it matches the right-hand side.
Step 1: Start with the left-hand side of the equation:
1−tan x
1 + tan x
Step 2: Use the definition of tangent:
1−sin x
cos x
1 + sin x
cos x
Step 3: Find a common denominator:
cos x−sin x
cos x+ sin x
Step 4: Multiply both the numerator and the denominator by 1
cos x:
cos x−sin x
cos x+ sin x×
1
cos x
1
cos x
=1−tan x
1 + tan x=cos x−sin x
cos x+ sin x
Step 5: Therefore, we have shown that:
1−tan x
1 + tan x=sin x−cos x
sin x+ cos x
And the given trigonometric identity is proved.
Question 18
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: Start with the angle addition formula for sine:
sin(3x) = sin(2x+x)
Step 2: Apply the angle addition formula:
sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
14
Step 3: Use the double angle identities to simplify the expression:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 4: Substitute these into the expression from Step 2:
sin(2x) cos(x) + cos(2x) sin(x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
Step 5: Simplify the expression from Step 4:
2 sin(x) cos2(x) + sin(x)−2 sin3(x)
Step 6: Factor out a sin(x)term:
sin(x)(2 cos2(x)+1−2 sin2(x))
Step 7: Use the Pythagorean identities:
cos2(x) = 1 −sin2(x)
2 cos2(x)+1−2 sin2(x) = 2(1 −sin2(x)) + 1 −2 sin2(x) = 3 −4 sin2(x)
Step 8: Substitute back into the expression:
sin(x)(3 −4 sin2(x)) = 3 sin(x)−4 sin3(x)
Step 9: Therefore, we have proved that sin(3x) = 3 sin(x)−4 sin3(x).
Question 19
Question
Prove the following trigonometric identity:
2 sin(θ) cos(θ)−2 sin2(θ) = cos(2θ)
Solution
To prove the identity 2 sin(θ) cos(θ)−2 sin2(θ) = cos(2θ), we will use trigono-
metric identities along with algebraic manipulations.
Step 1: Rewriting the right-hand side Since cos(2θ) = cos2(θ)−sin2(θ),
we rewrite the right-hand side as:
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Expanding the left-hand side Expand the left-hand side using
the double angle identity sin(2θ) = 2 sin(θ) cos(θ):
2 sin(θ) cos(θ)−2 sin2(θ)
15
= 2 sin(θ) cos(θ)−(1 −cos2(θ))
= 2 sin(θ) cos(θ)−1 + cos2(θ)
Step 3: Rearranging terms Rearrange the terms on the left-hand side to
match the right-hand side:
2 sin(θ) cos(θ)−1 + cos2(θ)
Step 4: Completing the proof Since 2 sin(θ) cos(θ)−1 + cos2(θ) =
cos(2θ), the identity is proved.
Question 20
Question
Prove the following trigonometric identity:
tan(θ)−sin(θ)
1 + tan(θ) sin(θ)= tan(θ)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
until it simplifies to the right-hand side.
tan(θ)−sin(θ)
1 + tan(θ) sin(θ)=
sin(θ)
cos(θ)−sin(θ)
1 + sin(θ)
cos(θ)·sin(θ)
=
sin(θ)−sin(θ) cos(θ)
cos(θ)
1 + sin2(θ)
cos(θ)
=sin(θ)(1 −cos(θ))
cos(θ) + sin2(θ)
=sin2(θ)
1−cos(θ)·1−cos(θ)
1−cos(θ)
=sin2(θ)
1−cos2(θ)
=sin2(θ)
sin2(θ)
= tan(θ)
Therefore, we have successfully proven the trigonometric identity tan(θ)−sin(θ)
1+tan(θ) sin(θ)=
tan(θ).
16
Question 21
Question
Simplify the following expression:
cos2xtan x+ sin xcsc x
Solution
To simplify the given expression, we will express all terms in sine and cosine
functions, and then simplify by using trigonometric identities.
Step 1: Recall the definitions of trigonometric functions:
tan x=sin x
cos x,csc x=1
sin x
Step 2: Substitute the expressions for tan xand csc xinto the given expres-
sion:
cos2x·sin x
cos x+ sin x·1
sin x
Step 3: Simplify the expression by canceling out common factors:
cos x·sin x+ sin x·cos x
Step 4: Combine the terms:
sin x+ cos x
Step 5: The simplified form of the given expression is sin x+ cos x.
Question 22
Question
Prove the trigonometric identity:
1−tan θ
1 + tan θ= cos θ−sin θ
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 identity.
1−tan θ
1 + tan θ
17
Step 2: Use the definitions of tan θ=sin θ
cos θand simplify.
1−sin θ
cos θ
1 + sin θ
cos θ
Step 3: Combine the fractions by finding a common denominator.
cos θ−sin θ
cos θ+ sin θ
Step 4: Multiply the numerator and denominator by the conjugate of the
denominator to simplify.
(cos θ−sin θ)(cos θ−sin θ)
(cos θ+ sin θ)(cos θ−sin θ)
cos2θ−2 cos θsin θ+ sin2θ
cos2θ−sin2θ
Step 5: Apply the Pythagorean identity sin2θ+ cos2θ= 1 to simplify the
numerator. 1−2 cos θsin θ
cos2θ−sin2θ
Step 6: Use the Pythagorean identity again to simplify the denominator.
1−2 cos θsin θ
cos2θ−(1 −cos2θ)
1−2 cos θsin θ
2 cos2θ−1
Step 7: Recognize that 2 cos2θ−1 = cos 2θ(double angle formula for
cosine).
1−2 cos θsin θ
cos 2θ
Step 8: Use the double angle formula for sine, sin 2θ= 2 sin θcos θ.
1−2 sin θcos θ= 1 −sin 2θ= cos 2θ
Hence, the left-hand side is equal to the right-hand side, thus proving the
trigonometric identity.
Question 23
Question
Prove the trigonometric identity: tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y).
18
Solution
To prove the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y), we can start
by expressing tan(x)and tan(y)in terms of sine and cosine functions and then
simplify the left-hand side to match the right-hand side of the equation.
Step 1: Recall that tan(x) = sin(x)
cos(x)and tan(y) = sin(y)
cos(y).
Step 2: Substitute the expressions for tan(x)and tan(y)into the left-hand
side of the equation:
tan(x) tan(y) = (sin(x)
cos(x))(sin(y)
cos(y))
Step 3: Simplify the expression:
tan(x) tan(y) = sin(x) sin(y)
cos(x) cos(y)
Step 4: Notice that cos(x) cos(y)−sin(x) sin(y) = cos(x+y)using the
angle addition identity.
Step 5: Substitute the above formula into the original expression to obtain:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Therefore, the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)is proven.
Question 24
Question
Prove the following trigonometric identity:
cos3(x)
sin(x)+1
cos2(x)=1
sin(x)
Solution
Step 1: Recall that sin2(x) + cos2(x) = 1. Step 2: Start with the left side of the
given identity:
cos3(x)
sin(x)+1
cos2(x)
Step 3: Rewrite 1
cos2(x)as sin2(x)
sin(x) cos2(x). Step 4: So, the left side becomes
cos3(x)
sin(x)+sin2(x)
sin(x) cos2(x)
19
Step 5: Factor out cos(x)from the first term and simplify:
=cos(x)·cos2(x)
sin(x)+sin2(x)
sin(x) cos2(x)
=cos(x) cos2(x) + sin2(x)
sin(x) cos2(x)
Step 6: Now, use the Pythagorean identity sin2(x) = 1 −cos2(x)to simplify the
numerator:
=cos(x) cos2(x)+1−cos2(x)
sin(x) cos2(x)
Step 7: Combine like terms in the numerator:
=1
sin(x) cos2(x)
Step 8: Now, notice that 1 = sin2(x) + cos2(x)can be written as cos2(x) =
1
sin2(x). Step 9: Substitute cos2(x) = 1
sin2(x)into the expression:
=1
sin(x)·1
sin2(x)
=1
1
sin(x)
=1
sin(x)
Step 10: Therefore, the left side equals the right side, proving the given trigono-
metric identity.
Question 25
Question
Prove the following trigonometric identity:
1 + sin x
cos x+1 + cos x
sin x= 2 csc xcot x
20
Solution
We start by manipulating the left-hand side of the equation:
Step 1: 1 + sin x
cos x+1 + cos x
sin x
=1
cos x+sin x
cos x+1
sin x+cos x
sin x
=1
cos x+sin x
cos x+1
sin x+cos x
sin x×cos x
cos x(rationalizing)
=1
cos x+sin x
cos x+cos x
sin x+cos2x
sin xcos x
=1 + sin x+ cos x+ cos2x
sin xcos x(combining terms)
=(sin x+ 1)(cos x+ 1)
sin xcos x
=sin xcos x+ sin x+ cos x+ 1
sin xcos x
=sin xcos x+ sin x+ cos x
sin xcos x+1
sin xcos x
=sin x(cos x+ 1) + cos x(sin x+ 1)
sin xcos x+1
sin xcos x
=sin x+ cos x
sin xcos x+ csc xcot x(using trigonometric identities)
=1
sin x+1
cos x+ csc xcot x
= csc x+ sec x+ csc xcot x
= csc x(1 + cot x) + sec x
= 2 csc xcot x+ sec x
= 2 csc xcot x(since sec x = 1/cos x)
Therefore, we have proved the given trigonometric identity.
Question 26
Question
Prove the following trigonometric identity:
sin6θ−cos6θ= sin 2θsin 4θ
21
Solution
To prove the given trigonometric identity, we will first express sin6θand cos6θ
in terms of sin 2θand cos 2θusing double angle formulas.
sin 2θ= 2 sin θcos θ
cos 2θ= cos2θ−sin2θ
Step 1: Express sin6θin terms of sin 2θand cos 2θ.
sin6θ= (sin2θ)3
= (sin2θ)2·sin2θ
= (1 −cos2θ)2·sin2θ
= (1 −cos2θ)2·(1 −cos2θ)
= (1 −2 cos2θ+ cos4θ)·(1 −cos2θ)
= 1 −3 cos2θ+ 2 cos4θ
Step 2: Express cos6θin terms of sin 2θand cos 2θ.
cos6θ= (cos2θ)3
= (cos2θ)2·cos2θ
= (1 −sin2θ)2·cos2θ
= (1 −sin2θ)2·(1 −sin2θ)
= (1 −2 sin2θ+ sin4θ)·(1 −sin2θ)
= 1 −3 sin2θ+ 2 sin4θ
Step 3: Substitute the expressions for sin6θand cos6θinto the original
identity and simplify.
sin6θ−cos6θ= (1 −3 cos2θ+ 2 cos4θ)−(1 −3 sin2θ+ 2 sin4θ)
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2 sin4θ
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2(1 −cos2θ)2
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2(1 −2 cos2θ+ cos4θ)
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2 + 4 cos2θ−2 cos4θ
= 4 cos2θ−3 cos2θ+ 3 sin2θ−2
= cos2θ+ 3 sin2θ−2
= (1 −sin2θ) + 3 sin2θ−2
= 1 + 2 sin2θ−2
= 2 sin2θ
Finally, using the double angle formula for sine, sin 2θ= 2
22
Question 27
Question
Solve the trigonometric equation sin4(x)−cos4(x) = 1
2for 0◦≤x≤360◦.
Solution
Step 1: Start by applying the Pythagorean identity sin2(x) + cos2(x) = 1 to
obtain an expression involving only one trigonometric function.
sin4(x)−(1 −sin2(x))2=1
2
Step 2: Expand and simplify the equation.
sin4(x)−(1 −2 sin2(x) + sin4(x)) = 1
2
2 sin2(x)−1 = 1
2
Step 3: Simplify the equation by moving all terms to one side.
2 sin2(x)−3
2= 0
Step 4: Factor the quadratic equation.
2 sin2(x)−3
2= 0
4 sin2(x)−3 = 0
(2 sin(x)−√3)(2 sin(x) + √3) = 0
Step 5: Solve for sin(x).
sin(x) = √3
2or sin(x) = −√3
2
Step 6: Find the corresponding angles within the given interval. For sin(x) =
√3
2, the solutions are x= 60◦and x= 120◦. For sin(x) = −√3
2, there are no
solutions within the given interval.
Therefore, the solutions to the trigonometric equation sin4(x)−cos4(x) = 1
2
for 0◦≤x≤360◦are x= 60◦and x= 120◦.
Question 28
Question
Prove the following trigonometric identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
23
Solution
Step 1: We will start by using the angle addition formula for sine: sin(A+B) =
sin Acos B+ cos Asin B.
sin(3θ) = sin(2θ+θ)
Step 2: Now, expand sin(2θ+θ)using the angle addition formula:
sin(3θ) = sin(2θ) cos(θ) + cos(2θ) sin(θ)
Step 3: Use the double angle formulas for sine and cosine:
sin(2θ) = 2 sin(θ) cos(θ)and cos(2θ) = 2 cos2(θ)−1
Step 4: Substitute the double angle formulas into the expression for sin(3θ):
sin(3θ) = 2(2 sin(θ) cos(θ)) cos(θ) + (2 cos2(θ)−1) sin(θ)
Step 5: Simplify the expression:
sin(3θ) = 4 sin(θ) cos2(θ) + 2 cos2(θ) sin(θ)−sin(θ)
Step 6: Recall that cos2(θ) = 1 −sin2(θ):
sin(3θ) = 4 sin(θ)(1 −sin2(θ)) + 2(1 −sin2(θ)) sin(θ)−sin(θ)
Step 7: Distribute and simplify the expression:
sin(3θ) = 4 sin(θ)−4 sin3(θ) + 2 sin(θ)−2 sin3(θ)−sin(θ)
Step 8: Combine like terms to obtain the desired identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Therefore, the trigonometric identity sin(3θ) = 3 sin(θ)−4 sin3(θ)is proven.
Question 29
Question
Prove the trigonometric identity: cot2(θ)−1 = csc2(θ).
Solution
To prove the trigonometric identity cot2(θ)−1 = csc2(θ), we will start with the
left-hand side of the equation and manipulate it step by step until we reach the
right-hand side of the equation.
24
Step 1: Start with the left-hand side of the equation.
cot2(θ)−1
Step 2: Rewrite cot2(θ)in terms of sine and cosine.
cos2(θ)
sin2(θ)−1
Step 3: Find a common denominator to combine the fractions.
cos2(θ)−sin2(θ)
sin2(θ)
Step 4: Recall the Pythagorean identity: cos2(θ)−sin2(θ) = 1.
1
sin2(θ)
Step 5: Rewrite 1
sin2(θ)in terms of csc2(θ).
csc2(θ)
Therefore, we have shown that cot2(θ)−1 = csc2(θ), completing the proof
of the trigonometric identity.
Question 30
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: Start with the left side of the identity and find a common denominator.
Step 2: Use trigonometric identities to simplify the expression. Step 3: Simplify
the expression further to show that it equals 1.
Step 1: Start with the left side of the identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)
Finding a common denominator, we have:
sin2(x)(1 + sin(x))
(1 + cos(x))(1 + sin(x)) +cos2(x)(1 + cos(x))
(1 + cos(x))(1 + sin(x))
25
Step 2: Now, use trigonometric identities to simplify the expression:
sin2(x) + sin3(x) + cos2(x) + cos3(x)
1 + sin(x) + cos(x) + sin(x) cos(x)
Since sin2(x) + cos2(x) = 1, the expression becomes:
1 + (sin3(x) + cos3(x))
1 + sin(x) + cos(x) + sin(x) cos(x)
Step 3: Now, simplify the expression further:
1 + (sin(x) + cos(x))(1 −sin(x) cos(x))
1 + sin(x) + cos(x) + sin(x) cos(x)
=1 + sin(x) + cos(x)−sin(x) cos(x)
1 + sin(x) + cos(x) + sin(x) cos(x)
Since 1 + sin(x) + cos(x) + sin(x) cos(x) = 1 + (sin(x) + cos(x))(1 + sin(x)) =
1 + sin(x) + cos(x), the expression simplifies to:
1 + sin(x) + cos(x)−sin(x) cos(x)
1 + sin(x) + cos(x)= 1
Therefore, we have proved the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Question 31
Question
Solve the trigonometric equation cos(θ) + cos(2θ) = 1 for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity cos(2θ) = 2 cos2(θ)−1to substitute
cos(2θ)in the equation, giving us:
cos(θ) + 2 cos2(θ)−1 = 1
Step 2: Rearrange the equation to get all terms on one side of the equation:
2 cos2(θ) + cos(θ)−2 = 0
Step 3: Now, let x= cos(θ). Substituting xinto the quadratic equation, we
get:
2x2+x−2 = 0
26
Step 4: Solve for xusing the quadratic formula x=−b±√b2−4ac
2awhere a= 2,
b= 1, and c=−2.
Step 5: Plugging in the values and simplifying, we get:
x=−1±√1 + 16
4=−1±√17
4
Step 6: Since −1≤cos(θ)≤1, we have to discard the extraneous solution
where x > 1. Thus, the two valid solutions for cos(θ)are −1−√17
4and −1+√17
4.
Step 7: For each valid solution of cos(θ), find the corresponding values of θby
taking the inverse cosine. Hence, the solutions for θare θ= cos−1(−1−√17
4andθ =
cos−1(−1+√17
4for0≤θ≤2π.
Question 32
Question
Prove the trigonometric identity:
cot(x) + csc(x) = sin(x)
cos(x)
Solution
To prove the trigonometric identity, we will manipulate the left-hand side until
it matches the right-hand side.
Step 1: Rewrite cot(x)and csc(x)in terms of sine and cosine:
cot(x) = cos(x)
sin(x)
csc(x) = 1
sin(x)
Step 2: Add cot(x)and csc(x)together:
cot(x) + csc(x) = cos(x)
sin(x)+1
sin(x)
Step 3: Find a common denominator and combine the fractions:
cot(x) + csc(x) = cos(x)+1
sin(x)
Step 4: Multiply the numerator and denominator by cos(x)to rationalize
the expression:
cot(x) + csc(x) = (cos(x) + 1) ·cos(x)
sin(x)·cos(x)
27
Step 5: Simplify the expression:
cot(x) + csc(x) = cos2(x) + cos(x)
sin(x) cos(x)
Step 6: Recall the Pythagorean identity sin2(x) + cos2(x) = 1:
cos2(x) = 1 −sin2(x)
Step 7: Substitute cos2(x)in terms of sin(x):
cot(x) + csc(x) = 1−sin2(x) + cos(x)
sin(x) cos(x)
Step 8: Simplify the numerator:
cot(x) + csc(x) = 1 + cos(x)
sin(x) cos(x)
Step 9: Recall the double angle identity cos(2x) = cos2(x)−sin2(x):
cos(2x) = 2 cos2(x)−1
cos(x) = cos(2x)+1
2
Step 10: Substitute cos(x)in terms of cos(2x):
cot(x) + csc(x) = 1 + cos(2x)+1
2
sin(x) cos(x)
Step 11: Simplify further:
cot(x) + csc(x) =
cos(2x)+3
2
sin(x) cos(x)
Step 12: Simplify:
cot(x) + csc(x) = cos(2x)+3
2 sin(x) cos(x)
Step 13: Recall the double angle identity sin(2x) = 2 sin(x) cos(x):
cot(x) + csc(x) = cos(2x)+3
2 sin(x) cos(x)=sin(x)
cos(x)
Therefore, the trigonometric identity is proven:
cot(x) + csc(x) = sin(x)
cos(x)
28
Question 33
Question
Prove the following trigonometric identity:
1 + tan(θ)
1−tan(θ)= sec2(θ)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it using trigonometric identities until we arrive at the right-hand
side.
Step 1: Rewrite tan(θ)in terms of sin(θ)and cos(θ)using the definition of
tangent:
tan(θ) = sin(θ)
cos(θ)
Step 2: Substitute sin(θ)
cos(θ)for tan(θ)in 1+tan(θ)
1−tan(θ):
1 + sin(θ)
cos(θ)
1−sin(θ)
cos(θ)
Step 3: Simplify the expression by combining the fractions:
cos(θ) + sin(θ)
cos(θ)−sin(θ)
Step 4: Multiply the numerator and denominator by (cos(θ) + sin(θ)) to
rationalize the expression:
(cos(θ) + sin(θ))(cos(θ) + sin(θ))
(cos(θ)−sin(θ))(cos(θ) + sin(θ))
Step 5: Expand the numerator and the denominator:
cos2(θ) + 2 sin(θ) cos(θ) + sin2(θ)
cos2(θ)−sin2(θ)
Step 6: Recall the Pythagorean identities: cos2(θ) = 1 −sin2(θ)and
sin2(θ) = 1 −cos2(θ). Substitute them into the expression:
1−cos2(θ) + 2 sin(θ) cos(θ)+1−cos2(θ)
1−cos2(θ)−1 + cos2(θ)
Step 7: Simplify the expression further:
2−2 cos2(θ) + 2 sin(θ) cos(θ)
−2 sin(θ) cos(θ)
29
Step 8: Simplify by factoring out a negative sign and canceling out common
factors:
−2(1 −cos2(θ)−sin(θ) cos(θ))
2 sin(θ) cos(θ)
Step 9: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to simplify
further:
−2(2 −1)
2 sin(θ) cos(θ)
Step 10: Simplify the expression to obtain the right-hand side:
−2
2 sin(θ) cos(θ)=−1
sin(θ) cos(θ)=−csc(θ) cot(θ) = −1
sin(θ)·1
cos(θ)=−sec(θ) csc(θ) = −sec(θ) cot(θ) = −sec(θ) cot(θ)
Therefore, we have proven the identity 1+tan(θ)
1−tan(θ)= sec2(θ).
Question 34
Question
Solve the equation sin4(x) + cos4(x) = 5
8for 0≤x≤2π.
Solution
Step 1: Use the Pythagorean identity sin2(x)+cos2(x) = 1 to simplify the given
equation:
sin4(x)+cos4(x) = (sin2(x))2+(cos2(x))2= sin2(x)+cos2(x)−2 sin2(x) cos2(x)
Step 2: Substitute sin2(x) + cos2(x) = 1 into the simplified equation:
sin4(x) + cos4(x) = 1 −2 sin2(x) cos2(x)
Step 3: Rearrange the equation:
2 sin2(x) cos2(x) = 1 −sin4(x)−cos4(x)
Step 4: Using the given equation sin4(x) + cos4(x) = 5
8, substitute into the
rearranged equation:
2 sin2(x) cos2(x) = 1 −5
8=3
8
Step 5: Simplify the equation further:
sin2(x) cos2(x) = 3
16
30
Step 6: Use the double angle identity sin(2x) = 2 sin(x) cos(x)to rewrite the
equation: 1
4sin(2x) = 3
16
Step 7: Solve for sin(2x):
sin(2x) = 3
4
Step 8: Find the solutions for 2x: Since sin(2x) = 3
4is positive in both the
first and second quadrants, we have:
2x=π
3,2π
3
Therefore, the solutions for xare:
x=π
6,π
3
Question 35
Question
Prove the trigonometric identity:
sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
for all x∈R.
Solution
To prove the trigonometric identity sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x),
we will use the trigonometric identity sin(A+B) = sin(A) cos(B)+cos(A) sin(B)
repeatedly.
Step 1: Apply the angle addition formula:
sin(5x) = sin(4x+x) = sin(4x) cos(x) + cos(4x) sin(x)
Step 2: Express sin(4x)using the double angle identity:
sin(4x) = 2 sin(2x) cos(2x) = 2(2 sin(x) cos(x))(2 cos2(x)−1)
= 8 sin(x) cos(x)(2 cos2(x)−1) = 8 sin(x) cos(x)(2 −2 sin2(x)−1)
= 8 sin(x) cos(x)(1 −2 sin2(x)) = 8 sin(x) cos(x)−16 sin3(x) cos(x)
Step 3: Substitute sin(4x)back into the expression for sin(5x):
31
sin(5x) = (8 sin(x) cos(x)−16 sin3(x) cos(x)) cos(x) + cos(4x) sin(x)
= 8 sin(x) cos2(x)−16 sin3(x) cos2(x) + cos(4x) sin(x)
= 8 sin(x)(1 −sin2(x)) −16 sin3(x)(1 −sin2(x)) + cos(4x) sin(x)
= 8 sin(x)−8 sin3(x)−16 sin3(x) + 16 sin5(x) + cos(4x) sin(x)
= 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + cos(4x) sin(x)
Step 4: Use the double angle identity to express cos(4x):
cos(4x) = 2 cos2(2x)−1 = 2(2 cos2(x)−1)2−1
= 2(4 cos4(x)−4 cos2(x) + 1) −1 = 8 cos4(x)−8 cos2(x)+1−1
= 8 cos4(x)−8 cos2(x)
= 8(1 −sin2(x))2−8(1 −sin2(x)) = 8(1 −2 sin2(x) + sin4(x)) −8 + 8 sin2(x)
= 8 −16 sin2(x) + 8 sin4(x)−8 + 8 sin2(x) = 8 sin4(x)−8 sin2(x)
Step 5: Substitute cos(4x)back into the expression for sin(5x):
sin(5x) = 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + cos(4x) sin(x)
= 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + (8 sin4(x)−8 sin2(x)) sin(x)
=
32
Question 2
Question
Prove the following trigonometric identity:
cos(θ)
1−sin(θ)+1
cos(θ)=2
1 + sin(θ)
Solution
1. We start by obtaining a common denominator on the left side of the
equation:
cos(θ)
1−sin(θ)+1
cos(θ)=cos(θ) cos(θ)
cos(θ)(1 −sin(θ)) +(1 −sin(θ))
cos(θ)(1 −sin(θ))
=cos2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
2. Since cos2(θ) = 1 −sin2(θ)(Pythagorean identity), we can substitute this
into the numerator:
cos2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ)) =1−sin2(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
=2−sin(θ)−sin2(θ)
cos(θ)(1 −sin(θ))
3. Next, we use the Pythagorean identity sin2(θ)=1−cos2(θ)to simplify
the numerator further:
2−sin(θ)−sin2(θ)
cos(θ)(1 −sin(θ)) =2−sin(θ)−(1 −cos2(θ))
cos(θ)(1 −sin(θ))
=1−sin(θ) + cos2(θ)
cos(θ)(1 −sin(θ))
4. Finally, using the Pythagorean identity cos2(θ)=1−sin2(θ)for the nu-
2
merator:
1−sin(θ) + cos2(θ)
cos(θ)(1 −sin(θ)) =1−sin(θ)+1−sin2(θ)
cos(θ)(1 −sin(θ))
=2−sin(θ)+1−sin(θ)
cos(θ)(1 −sin(θ))
=2−2 sin(θ)
cos(θ)(1 −sin(θ))
=2(1 −sin(θ))
cos(θ)(1 −sin(θ))
=2
cos(θ)
=2
1 + sin(θ)
Therefore, we have shown that cos(θ)
1−sin(θ)+1
cos(θ)=2
1+sin(θ)is a true identity.
Question 3
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left side and
simplify it step by step until we reach the right side.
Step 1: Start with the left side of the identity:
sin4(x)−cos4(x)
Step 2: Recall the trigonometric identity: sin2(x) = 1 −cos2(x).
sin4(x) = (sin2(x))2
= (1 −cos2(x))2
= 1 −2 cos2(x) + (cos2(x))2
= 1 −2 cos2(x) + cos4(x)
3
Step 3: Substitute the above result back into the original expression:
sin4(x)−cos4(x) = 1 −2 cos2(x) + cos4(x)−cos4(x)
= 1 −2 cos2(x)
= 2(1/2−cos2(x))
= 2(sin2(x))
= 2 sin2(x) cos2(x)
Step 4: Thus, we have shown that sin4(x)−cos4(x) = 2 sin2(x) cos2(x),
which completes the proof of the given trigonometric identity.
Question 4
Question
Prove the following trigonometric identity:
sin(2x)
1 + cos(2x)= tan(x)
Solution
Step 1: We will start with the left-hand side of the equation and attempt to
simplify it using trigonometric identities. Step 2: Recall the double angle iden-
tity for sine: sin(2x) = 2 sin(x) cos(x). Step 3: Likewise, recall the double
angle identity for cosine: cos(2x) = cos2(x)−sin2(x). Step 4: Substitute these
identities into the left-hand side of the equation:
sin(2x)
1 + cos(2x)=2 sin(x) cos(x)
1 + (cos2(x)−sin2(x))
Step 5: Simplify the denominator by expanding the square term:
1 + cos2(x)−sin2(x) = cos2(x)+1−sin2(x) = cos2(x) + cos2(x) = 2 cos2(x)
Step 6: Substitute this back into the expression:
2 sin(x) cos(x)
2 cos2(x)=sin(x)
cos(x)= tan(x)
Step 7: Therefore, we have shown that the left-hand side is equal to the right-
hand side, and the trigonometric identity sin(2x)
1+cos(2x)= tan(x)is proven.
4
Question 5
Question
Prove the following trigonometric identity:
cos(x)
1−sin(x)+1
cos(x) + sin(x)=1
cos(x)
Solution
Step 1: Start by manipulating each side individually.
cos(x)
1−sin(x)+1
cos(x) + sin(x)=cos(x)
1−sin(x)+1
cos(x) + sin(x)·1 + sin(x)
1 + sin(x)
=cos(x)
1−sin(x)+1 + sin(x)
cos(x) + sin(x)
=cos(x)(1 + sin(x)) + (1 −sin(x))
(1 −sin(x))(1 + sin(x))
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−sin2(x)
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−(1 −cos2(x))
=cos(x) + cos(x) sin(x)+1−sin(x)
cos2(x)−1 + cos2(x)
=cos(x) + cos(x) sin(x)+1−sin(x)
2 cos2(x)−1
Step 2: Simplify the expression obtained in Step 1.
cos(x) + cos(x) sin(x)+1−sin(x)
2 cos2(x)−1=cos(x) (1 + sin(x)) + 1 −sin(x)
2 cos2(x)−1
=1 + sin(x)+1−sin(x)
2 cos2(x)−1
=2
2 cos2(x)−1
=1
cos2(x)−sin2(x)
=1
cos(x)(Using trigonometric identity: cos2(x)−sin2(x) = cos(x))
Therefore, cos(x)
1−sin(x)+1
cos(x)+sin(x)=1
cos(x)is true, and the identity is proven.
5
Question 6
Question
Prove the trigonometric identity:
1−cos 2θ
sin2θ= csc2θ−cot2θ
Solution
Start with the left-hand side: 1−cos 2θ
sin2θ=1−cos2θ+ sin2θ
sin2θ
=sin2θ
sin2θ
= 1
Now, simplify the right-hand side: csc2θ−cot2θ=(1
sin θ)2
−(cos θ
sin θ)2
=1
sin2θ−cos2θ
sin2θ
=1−cos2θ
sin2θ
=sin2θ
sin2θ
= 1
Thus, we have shown that the left-hand side is equal to the right-hand side,
and the trigonometric identity is proven.
Question 7
Question
Solve the trigonometric equation sin2(x) + sin(x)−6=0for xin the interval
[0,2π).
Solution
Step 1: Let u= sin(x). Then the equation becomes u2+u−6 = 0.
Step 2: Factor the quadratic equation: (u+ 3)(u−2) = 0.
Step 3: Set each factor to zero and solve for u:u+ 3 = 0 =⇒u=−3or
u−2 = 0 =⇒u= 2.
6
Step 4: Recall that u= sin(x). So, sin(x) = −3and sin(x) = 2 are not valid
solutions since the range of sin is [−1,1].
Step 5: Therefore, u= sin(x) = −3or u= sin(x) = 2 have no solutions in
the interval [0,2π).
Step 6: The only valid solution comes from u= sin(x) = 2, which has no
solutions in the interval [0,2π).
Step 7: Thus, the original trigonometric equation sin2(x) + sin(x)−6 = 0
has no solutions in the interval [0,2π).
Question 8
Question
Prove the following trigonometric identity:
1
sin(θ) + cos(θ)=√2
2·1
sin(θ+ 45◦)
Solution
Step 1: We will start by converting 45◦to radians. Recall that πradians is
equivalent to 180◦, so 1radian is approximately 57.2958◦. Therefore, 45◦is
equivalent to 45
57.2958 ≈0.7854 radians.
Step 2: Let’s rewrite the right-hand side of the given identity. Using the
angle addition formula for sine, we have:
sin(θ+ 45◦) = sin(θ) cos(45◦) + cos(θ) sin(45◦)
Plugging in the values for cos(45◦)and sin(45◦)(which are √2
2), we get:
sin(θ+ 45◦) = √2
2sin(θ) + √2
2cos(θ)
Step 3: Now, let’s simplify the right-hand side of the given identity using
this expression:
√2
2·1
sin(θ+ 45◦)=√2
2·1
√2
2sin(θ) + √2
2cos(θ)
=√2
2·1
√2(sin(θ) + cos(θ))
=1
sin(θ) + cos(θ)
Step 4: Since the right-hand side simplifies to the left-hand side of the given
identity, the identity is proven.
7
Question 9
Question
Prove the following trigonometric identity:
tan(θ)−sin(θ)
tan(θ) + sin(θ)=1−cos(θ)
1 + cos(θ)
Solution
Step 1: Start with the left side of the equation and manipulate to get the right
side:
tan(θ)−sin(θ)
tan(θ) + sin(θ)=
sin(θ)
cos(θ)−sin(θ)
sin(θ)
cos(θ)+ sin(θ)
=
sin(θ)−cos(θ) sin(θ)
cos(θ)
sin(θ)+cos(θ) sin(θ)
cos(θ)
=sin(θ)(1 −cos(θ))
sin(θ)(1 + cos(θ))
=1−cos(θ)
1 + cos(θ)
Therefore, the identity is proved.
Question 10
Question
Prove the following trigonometric identity:
1−cos θ
sin θ=tan θ
1 + cot θ
Solution
To prove the given trigonometric identity, we will simplify the left-hand side
(LHS) and the right-hand side (RHS) and show that they are equal.
Step 1: Simplify the LHS.
1−cos θ
sin θ=1−cos θ
sin θ·1 + cos θ
1 + cos θ=1−cos2θ
sin θ(1 + cos θ)
Using the Pythagorean identity sin2θ+ cos2θ= 1, we have cos2θ= 1 −sin2θ.
Substituting this into the expression above:
1−cos2θ
sin θ(1 + cos θ)=sin2θ
sin θ(1 + cos θ)=sin θ
1 + cos θ
8
Step 2: Simplify the RHS.
tan θ
1 + cot θ=sin θ/ cos θ
1 + cos θ/ sin θ=sin θ
cos θ·sin θ
sin θ+ cos θ=sin2θ
sin θ+ cos θ
Step 3: Compare the LHS and RHS. Since both the LHS and RHS simplify
to sin θ
1+cos θ, the given trigonometric identity is proven.
Question 11
Question
Prove the identity cot2(θ)−sin2(θ) = csc2(θ)for all θwhere the expressions are
defined.
Solution
To prove the given identity, we will start with the left-hand side and manipulate
it to obtain the right-hand side.
Step 1: Express cot2(θ)in terms of sin(θ)and cos(θ)using the definitions
of cotangent, secant, and cosecant:
cot2(θ)−sin2(θ) = cos2(θ)
sin2(θ)−sin2(θ) = cos2(θ)−sin4(θ)
sin2(θ)
Step 2: Simplify the expression by factoring out a sin2(θ)in the numerator:
cos2(θ)−sin4(θ)
sin2(θ)=(1 −sin2(θ)) cos2(θ)
sin2(θ)=cos2(θ)
sin2(θ)−sin2(θ) cos2(θ)
sin2(θ)
Step 3: Simplify further by canceling out sin2(θ)in the last term:
cos2(θ)
sin2(θ)−sin2(θ) cos2(θ)
sin2(θ)= csc2(θ)−sin2(θ)
Step 4: Hence, we have shown that cot2(θ)−sin2(θ) = csc2(θ)for all θ
where the expressions are defined.
Question 12
Question
Prove the trigonometric identity:
sin(x)
1−cos(x)= csc(x) + cot(x)
9
Solution
Step 1: Rewrite the right-hand side of the equation in terms of sine and cosine
functions.
csc(x) + cot(x) = 1
sin(x)+cos(x)
sin(x)=sin(x) + cos(x)
sin(x)
Step 2: Simplify the expression on the right-hand side.
sin(x) + cos(x)
sin(x)=sin(x)
sin(x)+cos(x)
sin(x)= 1 + cos(x)
sin(x)= 1 + cot(x)
Step 3: Now, we need to show that the two sides are equal.
Left-hand side =sin(x)
1−cos(x)
Step 4: Rationalize the denominator of the left-hand side.
sin(x)
1−cos(x)=sin(x)(1 + cos(x))
(1 −cos(x))(1 + cos(x)) =sin(x) + sin(x) cos(x)
1−cos2(x)
Step 5: Recall that sin2(x) + cos2(x) = 1.
sin(x) + sin(x) cos(x)
1−cos2(x)=sin(x) + sin(x) cos(x)
sin2(x)=sin(x)(1 + cos(x))
sin2(x)= csc(x)+cot(x)
Therefore, the given trigonometric identity sin(x)
1−cos(x)= csc(x) + cot(x)is
proven to be true.
Question 13
Question
Prove the trigonometric identity:
1
sin(A) + cot(A)= cos(A)
Solution
To prove the given trigonometric identity, we will use the definitions of the
trigonometric functions and algebraic manipulations.
Step 1: Rewrite in terms of sine and cosine Let’s rewrite cot(A)in
terms of sin(A)and cos(A). Recall that cot(A) = 1
tan(A)=1
sin(A)
cos(A)
=cos(A)
sin(A).
Now, rewrite the given expression as:
1
sin(A) + cos(A)
sin(A)
=1
sin2(A)+cos(A)
sin(A)
10
Step 2: Simplify the expression Now, simplify the expression further:
1
sin2(A)+cos(A)
sin(A)
=sin(A)
sin2(A) + cos(A)
Step 3: Use Pythagorean Identity Next, use the Pythagorean identity
sin2(A) + cos2(A) = 1 to simplify the expression:
sin(A)
sin2(A) + cos(A)=sin(A)
1 + cos(A)
Step 4: Distribute and Simplify Distribute the sin(A)in the denomina-
tor: sin(A)
1 + cos(A)=sin(A)
1·1
1 + cos(A)= sin(A)·1
1 + cos(A)
Step 5: Revisit the given expression Now, recall the given expression
we want to prove: 1
sin(A) + cot(A)= cos(A)
Step 6: Substitute back Substitute back the expression we simplified:
sin(A)·1
1 + cos(A)= cos(A)
Step 7: Refine the expression Further simplify the expression:
sin(A)
1 + cos(A)= cos(A)
Step 8: Use the Reciprocal Identity Finally, notice that cos(A) =
1
sec(A)=sin(A)
1+cos(A). Hence, we have proved the given trigonometric identity:
1
sin(A) + cot(A)= cos(A)
Question 14
Question
Prove the following trigonometric identity:
1 + tan(θ)
1−tan(θ)= sec2(θ)
11
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until we achieve the right-hand side.
Step 1: Start with the left-hand side of the equation:
1 + tan(θ)
1−tan(θ)
Step 2: Use the definition of tangent:
1 + sin(θ)
cos(θ)
1−sin(θ)
cos(θ)
Step 3: Simplify the expression further:
cos(θ) + sin(θ)
cos(θ)−sin(θ)
Step 4: Multiply both the numerator and the denominator by cos(θ):
(cos(θ) + sin(θ)) cos(θ)
(cos(θ)−sin(θ)) cos(θ)
Step 5: Expand the numerators and denominators:
cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)
Step 6: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1:
cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)=cos2(θ) + sin(θ) cos(θ)
cos2(θ)−sin(θ) cos(θ)·1
1=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)
Step 7: Use the double-angle identity sin(2θ) = 2 sin(θ) cos(θ):
1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)=1 + 1
2sin(2θ)
1−1
2sin(2θ)=1 + 1
2sin(2θ)
1−1
2sin(2θ)
Step 8: Use the double-angle identity for sine: sin(2θ) = 2 sin(θ) cos(θ):
1 + 1
2sin(2θ)
1−1
2sin(2θ)=1 + 1
2·2 sin(θ) cos(θ)
1−1
2·2 sin(θ) cos(θ)=1 + sin(θ)
1−sin(θ)= sec2(θ)
Therefore, we have shown that 1+tan(θ)
1−tan(θ)= sec2(θ), thus proving the trigono-
metric identity.
12
Question 15
Question
Prove the following trigonometric identity:
2 sin2(x)−1 = cos(2x)
Solution
Step 1: We will start with the right-hand side of the equation, which is cos(2x).
Using the double angle identity for cosine, we have:
cos(2x) = cos2(x)−sin2(x)
Step 2: Now, we will replace cos2(x)in the above expression using the
Pythagorean identity cos2(x) = 1 −sin2(x). Substituting this in, we get:
cos(2x) = (1 −sin2(x)) −sin2(x) = 1 −2 sin2(x)
Step 3: Simplifying the expression 1−2 sin2(x), we have:
cos(2x) = 1 −2 sin2(x) = −2 sin2(x)+1
Step 4: Comparing the expression for cos(2x)from Step 3 with the left-hand
side of the original equation, 2 sin2(x)−1, we see that:
2 sin2(x)−1 = −2 sin2(x)+1
Step 5: Hence, we have proved that:
2 sin2(x)−1 = cos(2x)
Therefore, the trigonometric identity is verified.
Question 16
Question
Prove the following trigonometric identity:
1−tan x
1 + tan x=sin x−cos x
sin x+ cos x
13
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation until it matches the right-hand side.
Step 1: Start with the left-hand side of the equation:
1−tan x
1 + tan x
Step 2: Use the definition of tangent:
1−sin x
cos x
1 + sin x
cos x
Step 3: Find a common denominator:
cos x−sin x
cos x+ sin x
Step 4: Multiply both the numerator and the denominator by 1
cos x:
cos x−sin x
cos x+ sin x×
1
cos x
1
cos x
=1−tan x
1 + tan x=cos x−sin x
cos x+ sin x
Step 5: Therefore, we have shown that:
1−tan x
1 + tan x=sin x−cos x
sin x+ cos x
And the given trigonometric identity is proved.
Question 18
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: Start with the angle addition formula for sine:
sin(3x) = sin(2x+x)
Step 2: Apply the angle addition formula:
sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
14
Step 3: Use the double angle identities to simplify the expression:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x)
Step 4: Substitute these into the expression from Step 2:
sin(2x) cos(x) + cos(2x) sin(x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x)
Step 5: Simplify the expression from Step 4:
2 sin(x) cos2(x) + sin(x)−2 sin3(x)
Step 6: Factor out a sin(x)term:
sin(x)(2 cos2(x)+1−2 sin2(x))
Step 7: Use the Pythagorean identities:
cos2(x) = 1 −sin2(x)
2 cos2(x)+1−2 sin2(x) = 2(1 −sin2(x)) + 1 −2 sin2(x) = 3 −4 sin2(x)
Step 8: Substitute back into the expression:
sin(x)(3 −4 sin2(x)) = 3 sin(x)−4 sin3(x)
Step 9: Therefore, we have proved that sin(3x) = 3 sin(x)−4 sin3(x).
Question 19
Question
Prove the following trigonometric identity:
2 sin(θ) cos(θ)−2 sin2(θ) = cos(2θ)
Solution
To prove the identity 2 sin(θ) cos(θ)−2 sin2(θ) = cos(2θ), we will use trigono-
metric identities along with algebraic manipulations.
Step 1: Rewriting the right-hand side Since cos(2θ) = cos2(θ)−sin2(θ),
we rewrite the right-hand side as:
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Expanding the left-hand side Expand the left-hand side using
the double angle identity sin(2θ) = 2 sin(θ) cos(θ):
2 sin(θ) cos(θ)−2 sin2(θ)
15
= 2 sin(θ) cos(θ)−(1 −cos2(θ))
= 2 sin(θ) cos(θ)−1 + cos2(θ)
Step 3: Rearranging terms Rearrange the terms on the left-hand side to
match the right-hand side:
2 sin(θ) cos(θ)−1 + cos2(θ)
Step 4: Completing the proof Since 2 sin(θ) cos(θ)−1 + cos2(θ) =
cos(2θ), the identity is proved.
Question 20
Question
Prove the following trigonometric identity:
tan(θ)−sin(θ)
1 + tan(θ) sin(θ)= tan(θ)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
until it simplifies to the right-hand side.
tan(θ)−sin(θ)
1 + tan(θ) sin(θ)=
sin(θ)
cos(θ)−sin(θ)
1 + sin(θ)
cos(θ)·sin(θ)
=
sin(θ)−sin(θ) cos(θ)
cos(θ)
1 + sin2(θ)
cos(θ)
=sin(θ)(1 −cos(θ))
cos(θ) + sin2(θ)
=sin2(θ)
1−cos(θ)·1−cos(θ)
1−cos(θ)
=sin2(θ)
1−cos2(θ)
=sin2(θ)
sin2(θ)
= tan(θ)
Therefore, we have successfully proven the trigonometric identity tan(θ)−sin(θ)
1+tan(θ) sin(θ)=
tan(θ).
16
Question 21
Question
Simplify the following expression:
cos2xtan x+ sin xcsc x
Solution
To simplify the given expression, we will express all terms in sine and cosine
functions, and then simplify by using trigonometric identities.
Step 1: Recall the definitions of trigonometric functions:
tan x=sin x
cos x,csc x=1
sin x
Step 2: Substitute the expressions for tan xand csc xinto the given expres-
sion:
cos2x·sin x
cos x+ sin x·1
sin x
Step 3: Simplify the expression by canceling out common factors:
cos x·sin x+ sin x·cos x
Step 4: Combine the terms:
sin x+ cos x
Step 5: The simplified form of the given expression is sin x+ cos x.
Question 22
Question
Prove the trigonometric identity:
1−tan θ
1 + tan θ= cos θ−sin θ
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 identity.
1−tan θ
1 + tan θ
17
Step 2: Use the definitions of tan θ=sin θ
cos θand simplify.
1−sin θ
cos θ
1 + sin θ
cos θ
Step 3: Combine the fractions by finding a common denominator.
cos θ−sin θ
cos θ+ sin θ
Step 4: Multiply the numerator and denominator by the conjugate of the
denominator to simplify.
(cos θ−sin θ)(cos θ−sin θ)
(cos θ+ sin θ)(cos θ−sin θ)
cos2θ−2 cos θsin θ+ sin2θ
cos2θ−sin2θ
Step 5: Apply the Pythagorean identity sin2θ+ cos2θ= 1 to simplify the
numerator. 1−2 cos θsin θ
cos2θ−sin2θ
Step 6: Use the Pythagorean identity again to simplify the denominator.
1−2 cos θsin θ
cos2θ−(1 −cos2θ)
1−2 cos θsin θ
2 cos2θ−1
Step 7: Recognize that 2 cos2θ−1 = cos 2θ(double angle formula for
cosine).
1−2 cos θsin θ
cos 2θ
Step 8: Use the double angle formula for sine, sin 2θ= 2 sin θcos θ.
1−2 sin θcos θ= 1 −sin 2θ= cos 2θ
Hence, the left-hand side is equal to the right-hand side, thus proving the
trigonometric identity.
Question 23
Question
Prove the trigonometric identity: tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y).
18
Solution
To prove the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y), we can start
by expressing tan(x)and tan(y)in terms of sine and cosine functions and then
simplify the left-hand side to match the right-hand side of the equation.
Step 1: Recall that tan(x) = sin(x)
cos(x)and tan(y) = sin(y)
cos(y).
Step 2: Substitute the expressions for tan(x)and tan(y)into the left-hand
side of the equation:
tan(x) tan(y) = (sin(x)
cos(x))(sin(y)
cos(y))
Step 3: Simplify the expression:
tan(x) tan(y) = sin(x) sin(y)
cos(x) cos(y)
Step 4: Notice that cos(x) cos(y)−sin(x) sin(y) = cos(x+y)using the
angle addition identity.
Step 5: Substitute the above formula into the original expression to obtain:
tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)
Therefore, the trigonometric identity tan(x) tan(y) = sin(x) sin(y)
1−cos(x) cos(y)is proven.
Question 24
Question
Prove the following trigonometric identity:
cos3(x)
sin(x)+1
cos2(x)=1
sin(x)
Solution
Step 1: Recall that sin2(x) + cos2(x) = 1. Step 2: Start with the left side of the
given identity:
cos3(x)
sin(x)+1
cos2(x)
Step 3: Rewrite 1
cos2(x)as sin2(x)
sin(x) cos2(x). Step 4: So, the left side becomes
cos3(x)
sin(x)+sin2(x)
sin(x) cos2(x)
19
Step 5: Factor out cos(x)from the first term and simplify:
=cos(x)·cos2(x)
sin(x)+sin2(x)
sin(x) cos2(x)
=cos(x) cos2(x) + sin2(x)
sin(x) cos2(x)
Step 6: Now, use the Pythagorean identity sin2(x) = 1 −cos2(x)to simplify the
numerator:
=cos(x) cos2(x)+1−cos2(x)
sin(x) cos2(x)
Step 7: Combine like terms in the numerator:
=1
sin(x) cos2(x)
Step 8: Now, notice that 1 = sin2(x) + cos2(x)can be written as cos2(x) =
1
sin2(x). Step 9: Substitute cos2(x) = 1
sin2(x)into the expression:
=1
sin(x)·1
sin2(x)
=1
1
sin(x)
=1
sin(x)
Step 10: Therefore, the left side equals the right side, proving the given trigono-
metric identity.
Question 25
Question
Prove the following trigonometric identity:
1 + sin x
cos x+1 + cos x
sin x= 2 csc xcot x
20
Solution
We start by manipulating the left-hand side of the equation:
Step 1: 1 + sin x
cos x+1 + cos x
sin x
=1
cos x+sin x
cos x+1
sin x+cos x
sin x
=1
cos x+sin x
cos x+1
sin x+cos x
sin x×cos x
cos x(rationalizing)
=1
cos x+sin x
cos x+cos x
sin x+cos2x
sin xcos x
=1 + sin x+ cos x+ cos2x
sin xcos x(combining terms)
=(sin x+ 1)(cos x+ 1)
sin xcos x
=sin xcos x+ sin x+ cos x+ 1
sin xcos x
=sin xcos x+ sin x+ cos x
sin xcos x+1
sin xcos x
=sin x(cos x+ 1) + cos x(sin x+ 1)
sin xcos x+1
sin xcos x
=sin x+ cos x
sin xcos x+ csc xcot x(using trigonometric identities)
=1
sin x+1
cos x+ csc xcot x
= csc x+ sec x+ csc xcot x
= csc x(1 + cot x) + sec x
= 2 csc xcot x+ sec x
= 2 csc xcot x(since sec x = 1/cos x)
Therefore, we have proved the given trigonometric identity.
Question 26
Question
Prove the following trigonometric identity:
sin6θ−cos6θ= sin 2θsin 4θ
21
Solution
To prove the given trigonometric identity, we will first express sin6θand cos6θ
in terms of sin 2θand cos 2θusing double angle formulas.
sin 2θ= 2 sin θcos θ
cos 2θ= cos2θ−sin2θ
Step 1: Express sin6θin terms of sin 2θand cos 2θ.
sin6θ= (sin2θ)3
= (sin2θ)2·sin2θ
= (1 −cos2θ)2·sin2θ
= (1 −cos2θ)2·(1 −cos2θ)
= (1 −2 cos2θ+ cos4θ)·(1 −cos2θ)
= 1 −3 cos2θ+ 2 cos4θ
Step 2: Express cos6θin terms of sin 2θand cos 2θ.
cos6θ= (cos2θ)3
= (cos2θ)2·cos2θ
= (1 −sin2θ)2·cos2θ
= (1 −sin2θ)2·(1 −sin2θ)
= (1 −2 sin2θ+ sin4θ)·(1 −sin2θ)
= 1 −3 sin2θ+ 2 sin4θ
Step 3: Substitute the expressions for sin6θand cos6θinto the original
identity and simplify.
sin6θ−cos6θ= (1 −3 cos2θ+ 2 cos4θ)−(1 −3 sin2θ+ 2 sin4θ)
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2 sin4θ
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2(1 −cos2θ)2
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2(1 −2 cos2θ+ cos4θ)
=−3 cos2θ+ 2 cos4θ+ 3 sin2θ−2 + 4 cos2θ−2 cos4θ
= 4 cos2θ−3 cos2θ+ 3 sin2θ−2
= cos2θ+ 3 sin2θ−2
= (1 −sin2θ) + 3 sin2θ−2
= 1 + 2 sin2θ−2
= 2 sin2θ
Finally, using the double angle formula for sine, sin 2θ= 2
22
Question 27
Question
Solve the trigonometric equation sin4(x)−cos4(x) = 1
2for 0◦≤x≤360◦.
Solution
Step 1: Start by applying the Pythagorean identity sin2(x) + cos2(x) = 1 to
obtain an expression involving only one trigonometric function.
sin4(x)−(1 −sin2(x))2=1
2
Step 2: Expand and simplify the equation.
sin4(x)−(1 −2 sin2(x) + sin4(x)) = 1
2
2 sin2(x)−1 = 1
2
Step 3: Simplify the equation by moving all terms to one side.
2 sin2(x)−3
2= 0
Step 4: Factor the quadratic equation.
2 sin2(x)−3
2= 0
4 sin2(x)−3 = 0
(2 sin(x)−√3)(2 sin(x) + √3) = 0
Step 5: Solve for sin(x).
sin(x) = √3
2or sin(x) = −√3
2
Step 6: Find the corresponding angles within the given interval. For sin(x) =
√3
2, the solutions are x= 60◦and x= 120◦. For sin(x) = −√3
2, there are no
solutions within the given interval.
Therefore, the solutions to the trigonometric equation sin4(x)−cos4(x) = 1
2
for 0◦≤x≤360◦are x= 60◦and x= 120◦.
Question 28
Question
Prove the following trigonometric identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
23
Solution
Step 1: We will start by using the angle addition formula for sine: sin(A+B) =
sin Acos B+ cos Asin B.
sin(3θ) = sin(2θ+θ)
Step 2: Now, expand sin(2θ+θ)using the angle addition formula:
sin(3θ) = sin(2θ) cos(θ) + cos(2θ) sin(θ)
Step 3: Use the double angle formulas for sine and cosine:
sin(2θ) = 2 sin(θ) cos(θ)and cos(2θ) = 2 cos2(θ)−1
Step 4: Substitute the double angle formulas into the expression for sin(3θ):
sin(3θ) = 2(2 sin(θ) cos(θ)) cos(θ) + (2 cos2(θ)−1) sin(θ)
Step 5: Simplify the expression:
sin(3θ) = 4 sin(θ) cos2(θ) + 2 cos2(θ) sin(θ)−sin(θ)
Step 6: Recall that cos2(θ) = 1 −sin2(θ):
sin(3θ) = 4 sin(θ)(1 −sin2(θ)) + 2(1 −sin2(θ)) sin(θ)−sin(θ)
Step 7: Distribute and simplify the expression:
sin(3θ) = 4 sin(θ)−4 sin3(θ) + 2 sin(θ)−2 sin3(θ)−sin(θ)
Step 8: Combine like terms to obtain the desired identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Therefore, the trigonometric identity sin(3θ) = 3 sin(θ)−4 sin3(θ)is proven.
Question 29
Question
Prove the trigonometric identity: cot2(θ)−1 = csc2(θ).
Solution
To prove the trigonometric identity cot2(θ)−1 = csc2(θ), we will start with the
left-hand side of the equation and manipulate it step by step until we reach the
right-hand side of the equation.
24
Step 1: Start with the left-hand side of the equation.
cot2(θ)−1
Step 2: Rewrite cot2(θ)in terms of sine and cosine.
cos2(θ)
sin2(θ)−1
Step 3: Find a common denominator to combine the fractions.
cos2(θ)−sin2(θ)
sin2(θ)
Step 4: Recall the Pythagorean identity: cos2(θ)−sin2(θ) = 1.
1
sin2(θ)
Step 5: Rewrite 1
sin2(θ)in terms of csc2(θ).
csc2(θ)
Therefore, we have shown that cot2(θ)−1 = csc2(θ), completing the proof
of the trigonometric identity.
Question 30
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: Start with the left side of the identity and find a common denominator.
Step 2: Use trigonometric identities to simplify the expression. Step 3: Simplify
the expression further to show that it equals 1.
Step 1: Start with the left side of the identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)
Finding a common denominator, we have:
sin2(x)(1 + sin(x))
(1 + cos(x))(1 + sin(x)) +cos2(x)(1 + cos(x))
(1 + cos(x))(1 + sin(x))
25
Step 2: Now, use trigonometric identities to simplify the expression:
sin2(x) + sin3(x) + cos2(x) + cos3(x)
1 + sin(x) + cos(x) + sin(x) cos(x)
Since sin2(x) + cos2(x) = 1, the expression becomes:
1 + (sin3(x) + cos3(x))
1 + sin(x) + cos(x) + sin(x) cos(x)
Step 3: Now, simplify the expression further:
1 + (sin(x) + cos(x))(1 −sin(x) cos(x))
1 + sin(x) + cos(x) + sin(x) cos(x)
=1 + sin(x) + cos(x)−sin(x) cos(x)
1 + sin(x) + cos(x) + sin(x) cos(x)
Since 1 + sin(x) + cos(x) + sin(x) cos(x) = 1 + (sin(x) + cos(x))(1 + sin(x)) =
1 + sin(x) + cos(x), the expression simplifies to:
1 + sin(x) + cos(x)−sin(x) cos(x)
1 + sin(x) + cos(x)= 1
Therefore, we have proved the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Question 31
Question
Solve the trigonometric equation cos(θ) + cos(2θ) = 1 for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity cos(2θ) = 2 cos2(θ)−1to substitute
cos(2θ)in the equation, giving us:
cos(θ) + 2 cos2(θ)−1 = 1
Step 2: Rearrange the equation to get all terms on one side of the equation:
2 cos2(θ) + cos(θ)−2 = 0
Step 3: Now, let x= cos(θ). Substituting xinto the quadratic equation, we
get:
2x2+x−2 = 0
26
Step 4: Solve for xusing the quadratic formula x=−b±√b2−4ac
2awhere a= 2,
b= 1, and c=−2.
Step 5: Plugging in the values and simplifying, we get:
x=−1±√1 + 16
4=−1±√17
4
Step 6: Since −1≤cos(θ)≤1, we have to discard the extraneous solution
where x > 1. Thus, the two valid solutions for cos(θ)are −1−√17
4and −1+√17
4.
Step 7: For each valid solution of cos(θ), find the corresponding values of θby
taking the inverse cosine. Hence, the solutions for θare θ= cos−1(−1−√17
4andθ =
cos−1(−1+√17
4for0≤θ≤2π.
Question 32
Question
Prove the trigonometric identity:
cot(x) + csc(x) = sin(x)
cos(x)
Solution
To prove the trigonometric identity, we will manipulate the left-hand side until
it matches the right-hand side.
Step 1: Rewrite cot(x)and csc(x)in terms of sine and cosine:
cot(x) = cos(x)
sin(x)
csc(x) = 1
sin(x)
Step 2: Add cot(x)and csc(x)together:
cot(x) + csc(x) = cos(x)
sin(x)+1
sin(x)
Step 3: Find a common denominator and combine the fractions:
cot(x) + csc(x) = cos(x)+1
sin(x)
Step 4: Multiply the numerator and denominator by cos(x)to rationalize
the expression:
cot(x) + csc(x) = (cos(x) + 1) ·cos(x)
sin(x)·cos(x)
27
Step 5: Simplify the expression:
cot(x) + csc(x) = cos2(x) + cos(x)
sin(x) cos(x)
Step 6: Recall the Pythagorean identity sin2(x) + cos2(x) = 1:
cos2(x) = 1 −sin2(x)
Step 7: Substitute cos2(x)in terms of sin(x):
cot(x) + csc(x) = 1−sin2(x) + cos(x)
sin(x) cos(x)
Step 8: Simplify the numerator:
cot(x) + csc(x) = 1 + cos(x)
sin(x) cos(x)
Step 9: Recall the double angle identity cos(2x) = cos2(x)−sin2(x):
cos(2x) = 2 cos2(x)−1
cos(x) = cos(2x)+1
2
Step 10: Substitute cos(x)in terms of cos(2x):
cot(x) + csc(x) = 1 + cos(2x)+1
2
sin(x) cos(x)
Step 11: Simplify further:
cot(x) + csc(x) =
cos(2x)+3
2
sin(x) cos(x)
Step 12: Simplify:
cot(x) + csc(x) = cos(2x)+3
2 sin(x) cos(x)
Step 13: Recall the double angle identity sin(2x) = 2 sin(x) cos(x):
cot(x) + csc(x) = cos(2x)+3
2 sin(x) cos(x)=sin(x)
cos(x)
Therefore, the trigonometric identity is proven:
cot(x) + csc(x) = sin(x)
cos(x)
28
Question 33
Question
Prove the following trigonometric identity:
1 + tan(θ)
1−tan(θ)= sec2(θ)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it using trigonometric identities until we arrive at the right-hand
side.
Step 1: Rewrite tan(θ)in terms of sin(θ)and cos(θ)using the definition of
tangent:
tan(θ) = sin(θ)
cos(θ)
Step 2: Substitute sin(θ)
cos(θ)for tan(θ)in 1+tan(θ)
1−tan(θ):
1 + sin(θ)
cos(θ)
1−sin(θ)
cos(θ)
Step 3: Simplify the expression by combining the fractions:
cos(θ) + sin(θ)
cos(θ)−sin(θ)
Step 4: Multiply the numerator and denominator by (cos(θ) + sin(θ)) to
rationalize the expression:
(cos(θ) + sin(θ))(cos(θ) + sin(θ))
(cos(θ)−sin(θ))(cos(θ) + sin(θ))
Step 5: Expand the numerator and the denominator:
cos2(θ) + 2 sin(θ) cos(θ) + sin2(θ)
cos2(θ)−sin2(θ)
Step 6: Recall the Pythagorean identities: cos2(θ) = 1 −sin2(θ)and
sin2(θ) = 1 −cos2(θ). Substitute them into the expression:
1−cos2(θ) + 2 sin(θ) cos(θ)+1−cos2(θ)
1−cos2(θ)−1 + cos2(θ)
Step 7: Simplify the expression further:
2−2 cos2(θ) + 2 sin(θ) cos(θ)
−2 sin(θ) cos(θ)
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Step 8: Simplify by factoring out a negative sign and canceling out common
factors:
−2(1 −cos2(θ)−sin(θ) cos(θ))
2 sin(θ) cos(θ)
Step 9: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to simplify
further:
−2(2 −1)
2 sin(θ) cos(θ)
Step 10: Simplify the expression to obtain the right-hand side:
−2
2 sin(θ) cos(θ)=−1
sin(θ) cos(θ)=−csc(θ) cot(θ) = −1
sin(θ)·1
cos(θ)=−sec(θ) csc(θ) = −sec(θ) cot(θ) = −sec(θ) cot(θ)
Therefore, we have proven the identity 1+tan(θ)
1−tan(θ)= sec2(θ).
Question 34
Question
Solve the equation sin4(x) + cos4(x) = 5
8for 0≤x≤2π.
Solution
Step 1: Use the Pythagorean identity sin2(x)+cos2(x) = 1 to simplify the given
equation:
sin4(x)+cos4(x) = (sin2(x))2+(cos2(x))2= sin2(x)+cos2(x)−2 sin2(x) cos2(x)
Step 2: Substitute sin2(x) + cos2(x) = 1 into the simplified equation:
sin4(x) + cos4(x) = 1 −2 sin2(x) cos2(x)
Step 3: Rearrange the equation:
2 sin2(x) cos2(x) = 1 −sin4(x)−cos4(x)
Step 4: Using the given equation sin4(x) + cos4(x) = 5
8, substitute into the
rearranged equation:
2 sin2(x) cos2(x) = 1 −5
8=3
8
Step 5: Simplify the equation further:
sin2(x) cos2(x) = 3
16
30
Step 6: Use the double angle identity sin(2x) = 2 sin(x) cos(x)to rewrite the
equation: 1
4sin(2x) = 3
16
Step 7: Solve for sin(2x):
sin(2x) = 3
4
Step 8: Find the solutions for 2x: Since sin(2x) = 3
4is positive in both the
first and second quadrants, we have:
2x=π
3,2π
3
Therefore, the solutions for xare:
x=π
6,π
3
Question 35
Question
Prove the trigonometric identity:
sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
for all x∈R.
Solution
To prove the trigonometric identity sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x),
we will use the trigonometric identity sin(A+B) = sin(A) cos(B)+cos(A) sin(B)
repeatedly.
Step 1: Apply the angle addition formula:
sin(5x) = sin(4x+x) = sin(4x) cos(x) + cos(4x) sin(x)
Step 2: Express sin(4x)using the double angle identity:
sin(4x) = 2 sin(2x) cos(2x) = 2(2 sin(x) cos(x))(2 cos2(x)−1)
= 8 sin(x) cos(x)(2 cos2(x)−1) = 8 sin(x) cos(x)(2 −2 sin2(x)−1)
= 8 sin(x) cos(x)(1 −2 sin2(x)) = 8 sin(x) cos(x)−16 sin3(x) cos(x)
Step 3: Substitute sin(4x)back into the expression for sin(5x):
31
sin(5x) = (8 sin(x) cos(x)−16 sin3(x) cos(x)) cos(x) + cos(4x) sin(x)
= 8 sin(x) cos2(x)−16 sin3(x) cos2(x) + cos(4x) sin(x)
= 8 sin(x)(1 −sin2(x)) −16 sin3(x)(1 −sin2(x)) + cos(4x) sin(x)
= 8 sin(x)−8 sin3(x)−16 sin3(x) + 16 sin5(x) + cos(4x) sin(x)
= 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + cos(4x) sin(x)
Step 4: Use the double angle identity to express cos(4x):
cos(4x) = 2 cos2(2x)−1 = 2(2 cos2(x)−1)2−1
= 2(4 cos4(x)−4 cos2(x) + 1) −1 = 8 cos4(x)−8 cos2(x)+1−1
= 8 cos4(x)−8 cos2(x)
= 8(1 −sin2(x))2−8(1 −sin2(x)) = 8(1 −2 sin2(x) + sin4(x)) −8 + 8 sin2(x)
= 8 −16 sin2(x) + 8 sin4(x)−8 + 8 sin2(x) = 8 sin4(x)−8 sin2(x)
Step 5: Substitute cos(4x)back into the expression for sin(5x):
sin(5x) = 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + cos(4x) sin(x)
= 16 sin5(x)−20 sin3(x) + 8 sin(x)−8 sin3(x) + (8 sin4(x)−8 sin2(x)) sin(x)
=
32
