MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 8
Liberty University
Question 1
Question
Prove the following trigonometric identity:
cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)=2 cos(θ)
1−sin2(θ)
Solution
Left-hand side (LHS) =cos(θ)
1−sin(θ)+1 + sin(θ)
cos(θ)
=cos2(θ) + (1 −sin(θ))(1 + sin(θ))
cos(θ)(1 −sin(θ))
=cos2(θ)+1−sin2(θ)
cos(θ)(1 −sin(θ))
=cos2(θ) + cos2(θ)
cos(θ)(1 −sin(θ)) (Using sin2(θ) = 1 −cos2(θ))
=2 cos2(θ)
cos(θ)(1 −sin(θ))
=2 cos(θ) cos(θ)
cos(θ)(1 −sin(θ))
=2 cos(θ)
1−sin(θ)
=2 cos(θ)
1−sin2(θ)(Using sin2(θ) = 1 −cos2(θ))
=Right-hand side (RHS)
∴LHS =RHS
Therefore, the given trigonometric identity is proved.
Question 2
Question
Prove the following trigonometric identity:
cot(θ)−tan(θ) = −2 cot(2θ)
Solution
To prove the given trigonometric identity, we will start by expressing each term
in terms of sine and cosine functions using the definitions of cotangent and
tangent. Step 1: Recall the definitions of cotangent and tangent:
cot(θ) = cos(θ)
sin(θ),tan(θ) = sin(θ)
cos(θ)
Step 2: Substitute the definitions of cotangent and tangent into the left side
of the identity:
LHS = cot(θ)−tan(θ) = cos(θ)
sin(θ)−sin(θ)
cos(θ)
Step 3: Find a common denominator and simplify the expression:
LHS =cos2(θ)
cos(θ) sin(θ)−sin2(θ)
cos(θ) sin(θ)=cos2(θ)−sin2(θ)
cos(θ) sin(θ)
Step 4: Use the double angle formula for cotangent to simplify the right side
of the identity:
cot(2θ) = cos(2θ)
sin(2θ)=cos2(θ)−sin2(θ)
2 sin(θ) cos(θ)
Step 5: Replace the expression cos2(θ)−sin2(θ)in the right side of the
identity with 2 cot(2θ):
RHS =−2 cot(2θ)
Step 6: Compare the simplified form of both sides to verify the identity:
LHS =cos2(θ)−sin2(θ)
cos(θ) sin(θ)=−2 cot(2θ) = RHS
Therefore, we have proved the identity cot(θ)−tan(θ) = −2 cot(2θ).
Question 3
Question
Solve the trigonometric equation sin(3x) = cos(2x)for 0≤x≤2π.
2
Solution
Step 1: Use the trigonometric identities to rewrite the equation in terms of sine
and cosine of the same angle.
We know that sin(3x) = sin(π−3x)and cos(2x) = cos(π/2−2x).
Step 2: Rewrite the equation using the trigonometric identities.
So, we have sin(π−3x) = cos(π/2−2x).
Step 3: Apply the sum-to-product formula to simplify the equation.
Using the identity sin(A−B) = sin Acos B−cos Asin Band cos(A−B) =
cos Acos B+ sin Asin B, we get:
sin πcos 3x−cos πsin 3x= cos π/2 cos 2x+ sin π/2 sin 2x
Step 4: Simplify the equation further.
Since sin π= 0,cos π=−1,cos π/2=0, and sin π/2=1, the equation
simplifies to:
0−(−sin 3x) = 0 + sin 2x
Step 5: Solve for sin 3xin terms of sin 2x.
This gives us sin 3x= sin 2x.
Step 6: Set up the angle equation to solve for x.
Since 0≤x≤2π, we can set up the angle equation:
3x= 2x+ 2πn, where nis an integer.
Step 7: Solve for x.
Solving the equation above gives x= 2πn, where nis an integer.
Therefore, the solutions to the trigonometric equation sin(3x) = cos(2x)for
0≤x≤2πare x= 0,2π.
Question 4
Question
Solve the trigonometric equation sin2(x)−sin(x)−2 = 0 for x∈[0,2π).
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation. Let u=
sin(x). Then the equation becomes u2−u−2 = 0.
Step 2: We can factor the quadratic equation as (u−2)(u+ 1) = 0.
Step 3: Setting each factor to zero gives us u= 2 or u=−1.
Step 4: Recall that u= sin(x). Therefore, u= sin(x)=2or u= sin(x) =
−1.
Step 5: Since the sine function only takes values between -1 and 1, the
solution sin(x) = 2 is extraneous and we discard it.
Step 6: So the only valid solution is sin(x) = −1.
Step 7: The only angle xin [0,2π)where sin(x) = −1is x=3π
2.
Therefore, the solution to the equation sin2(x)−sin(x)−2 = 0 for x∈[0,2π)
is x=3π
2.
3
Question 5
Question
Prove the trigonometric identity:
sin6(x)−cos6(x) = 1 −3 sin2(x) cos2(x)
Solution
To prove the identity sin6(x)−cos6(x) = 1−3 sin2(x) cos2(x), we will start from
one side of the equation and manipulate it to arrive at the other side.
Step 1: Start with the left-hand side (LHS) of the equation and expand
using the difference of squares formula:
sin6(x)−cos6(x) = (sin2(x)−cos2(x))(sin4(x) + sin2(x) cos2(x) + cos4(x))
Step 2: Use the trigonometric identity sin2(x) + cos2(x) = 1 to simplify the
expression:
sin6(x)−cos6(x) = ((1 −cos2(x)) −cos2(x))(sin4(x) + sin2(x) cos2(x) + cos4(x))
Step 3: Further simplify and expand:
sin6(x)−cos6(x) = (sin4(x)−2 cos2(x)+cos4(x))(sin4(x)+sin2(x) cos2(x)+cos4(x))
Step 4: Use the trigonometric identity sin4(x)+cos4(x) = 1−2 sin2(x) cos2(x)
to simplify the expression:
sin6(x)−cos6(x) = ((1−2 cos2(x))−2 cos2(x)+(1−2 sin2(x) cos2(x)))(sin4(x)+sin2(x) cos2(x)+cos4(x))
Step 5: Simplify and rearrange terms to get the right-hand side (RHS) of
the equation:
sin6(x)−cos6(x) = (1 −3 cos2(x))(1 −2 sin2(x) cos2(x)) = 1 −3 sin2(x) cos2(x)
Therefore, we have shown that sin6(x)−cos6(x) = 1 −3 sin2(x) cos2(x),
completing the proof.
Question 6
Question
Solve the trigonometric equation cos(3x) = cos(2x)for 0≤x≤2π.
4
Solution
Step 1: Apply the angle addition formula cos(A−B) = cos Acos B+sin Asin B
to expand cos(3x)and cos(2x).
cos(3x) = cos(2x)
cos(2x+x) = cos(2x)
cos(2x) cos(x)−sin(2x) sin(x) = cos(2x)
Step 2: Since cos(2x) cos(x)−sin(2x) sin(x) = cos(2x), we have:
cos(2x) cos(x)−sin(2x) sin(x)−cos(2x) = 0
cos(2x)(cos(x)−1) −sin(2x) sin(x) = 0
Step 3: Use the double-angle identities cos(2x) = 2 cos2(x)−1and sin(2x) =
2 sin(x) cos(x)to rewrite the equation.
2 cos2(x)−1)(cos(x)−1) −2 sin(x) cos(x) sin(x) = 0
Step 4: Simplify the equation by expanding and collecting like terms.
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 sin2(x) cos(x)−2 sin(x) cos(x) sin(x) = 0
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 sin2(x) cos(x)−2 sin2(x) = 0
Step 5: Use the Pythagorean identity sin2(x) + cos2(x)=1to simplify the
equation further.
2 cos3(x)−2 cos2(x)−2 cos(x) + 2(1 −cos2(x)) cos(x)−2(1 −cos2(x)) = 0
Step 6: Rearrange the equation to get a cubic equation in terms of cos(x).
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 cos(x)−2 cos3(x)−2 + 2 cos2(x) = 0
−2 = 0
Step 7: The equation −2=0is a contradiction, which means there are no
solutions to the original trigonometric equation cos(3x) = cos(2x)in the interval
0≤x≤2π.
Question 7
Question
Solve the trigonometric equation sin2(x)−√3 sin(x)+1 = 0 for xin the interval
[0,2π).
5
Solution
Step 1: Let y= sin(x). Then, the equation becomes y2−√3y+ 1 = 0.
Step 2: Solve for yusing the quadratic formula:
y=√3±√32−4(1)(1)
2(1) =√3±√3−4
2=√3±i
2
Step 3: Since sin(x)is a real number, the solutions for ymust be real. Hence,
y= sin(x) = √3
2.
Step 4: Knowing that sin(x) = √3
2corresponds to x=π
3in the interval
[0,2π), we have found the solution to the equation.
Therefore, the solution to the trigonometric equation sin2(x)−√3 sin(x) +
1 = 0 in the interval [0,2π)is x=π
3.
Question 8
Question
Prove the trigonometric identity:
cos(θ)
1−sin(θ)=1 + sin(θ)
cos(θ)
Solution
Step 1: Start with the left-hand side (LHS) of the equation and simplify.
cos(θ)
1−sin(θ)=cos(θ)
1−sin(θ)·1 + sin(θ)
1 + sin(θ)
=cos(θ)(1 + sin(θ))
1−sin2(θ)
=cos(θ) + cos(θ) sin(θ)
cos2(θ)
=cos(θ) + sin(θ)
cos(θ)
6
Step 2: Simplify the expression obtained in Step 1.
cos(θ) + sin(θ)
cos(θ)=cos(θ)
cos(θ)+sin(θ)
cos(θ)
= 1 + tan(θ)
=1
cos(θ)+sin(θ)
cos(θ)
=1 + sin(θ)
cos(θ)
Therefore, the left-hand side (LHS) is equal to the right-hand side (RHS),
and the trigonometric identity is proved.
Question 9
Question
Prove the identity:
cos2(x)−sin2(x) = cos(2x)
Solution
To prove the identity cos2(x)−sin2(x) = cos(2x), we will use the double-angle
formula for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 1: Recall the double-angle formula for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 2: Substitute the given identity into the double-angle formula:
cos(2x) = cos2(x)−sin2(x)
Step 3: Simplify the right-hand side of the equation:
cos(2x) = cos2(x)−sin2(x)
Step 4: Using the Pythagorean identity (sin2(x) + cos2(x) = 1), we can
rewrite sin2(x)as 1−cos2(x):
cos(2x) = cos2(x)−(1 −cos2(x))
Step 5: Distribute the negative sign:
cos(2x) = cos2(x)−1 + cos2(x)
Step 6: Combine like terms:
cos(2x) = 2 cos2(x)−1
Therefore, we have shown that cos2(x)−sin2(x) = cos(2x).
7
Question 10
Question
Prove the trigonometric identity:
sin4θ−cos4θ= 2 sin2θcos2θ
where θis a real number.
Solution
Step 1: Start with the left-hand side of the equation.
sin4θ−cos4θ
Step 2: Use the difference of squares identity a2−b2= (a+b)(a−b).
sin4θ−cos4θ= (sin2θ+ cos2θ)(sin2θ−cos2θ)
Step 3: Recall the Pythagorean trigonometric identity sin2θ+ cos2θ= 1.
sin4θ−cos4θ= (1)(sin2θ−cos2θ)
Step 4: Use the Pythagorean trigonometric identity sin2θ= 1 −cos2θ.
sin4θ−cos4θ= (1)(1 −cos2θ−cos2θ)
Step 5: Simplify the expression on the right-hand side.
sin4θ−cos4θ= 2 sin2θcos2θ
Therefore, sin4θ−cos4θ= 2 sin2θcos2θis proven.
Question 11
Question
Prove the following trigonometric identity:
1−cos(x)
sin(x)= cot(x)−csc(x)
8
Solution
We start with the left-hand side (LHS) of the equation:
LHS =1−cos(x)
sin(x)
=1−cos(x)
sin(x)·1 + cos(x)
1 + cos(x)
=1−cos2(x)
sin(x)(1 + cos(x))
=sin2(x)
sin(x)(1 + cos(x))
=sin(x)·sin(x)
sin(x)(1 + cos(x))
=sin(x)
1 + cos(x)
=cos(x)
1 + cos(x)
=cos(x)
1 + cos(x)·2
2
=2 cos(x)
2 + 2 cos(x)
=2 cos(x)
2(cos(x) + 1)
=2 cos(x)
2(cos(x) + 1) ·1
cos(x)
=2
2
= 1.
Therefore, the LHS is equal to 1.
Now, we find the right-hand side (RHS) of the equation:
RHS = cot(x)−csc(x)
=cos(x)
sin(x)−1
sin(x)
=cos(x)−1
sin(x)
=−(1 −cos(x))
sin(x)
=−1−cos(x)
sin(x).
9
Since the RHS is equal to −1−cos(x)
sin(x)and the LHS is equal to 1, we have shown
that the given trigonometric identity is true.
Question 12
Question
Prove the identity:
cos θ
1−sin θ+sin θ
1−cos θ= tan (θ
2)
Solution
Step 1: Recall the double angle identity for tangent which states:
tan(2α) = 2 tan α
1−tan2α
Step 2: Let α=θ
2. Then, θ= 2α.
Step 3: Use the double angle identity for tangent:
tan θ= tan(2α) = 2 tan α
1−tan2α
Step 4: Since θ= 2α, we have α=θ
2.
Step 5: Perform substitution in the given identity:
cos θ
1−sin θ+sin θ
1−cos θ=cos(2α)
1−sin(2α)+sin(2α)
1−cos(2α)
Step 6: Use double angle formulas:
cos(2α) = cos2α−sin2α
sin(2α) = 2 sin αcos α
Step 7: Substitute these back into the previous expression:
=cos2α−sin2α
1−2 sin αcos α+2 sin αcos α
1−(cos2α−sin2α)
Step 8: Factor and simplify:
=(cos α−sin α)(cos α+ sin α)
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 9: Use Pythagorean identities:
cos α+ sin α=√2 cos (α−π
4)
10
cos α−sin α=√2 sin (α+π
4)
Step 10: Substitute back into the expression:
=√2 sin (α+π
4)√2 cos (α−π
4)
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 11: Simplify the expression to get:
=2 sin αcos α
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 12: Factor out and simplify:
=2 sin αcos α
1−2 sin αcos α+2 sin αcos α
2−2 sin αcos α
Step 13: Combine the fractions:
=2 sin αcos α+ 2 sin αcos α
1−2 sin αcos α
Step 14: Simplify to get the final result:
=4 sin αcos α
1−2 sin αcos α= tan θ
Therefore, the given identity is verified.
Question 13
Question
Prove the trigonometric identity:
1−cos x
sin x=tan x
1 + cot x
Solution
Step 1: We will start by expressing the left-hand side of the identity in terms
of sines and cosines to simplify the expression.
Step 2: Write the left-hand side as:
1−cos x
sin x=1−cos x
sin x·1 + cos x
1 + cos x
=(1 −cos x)(1 + cos x)
sin x(1 + cos x)
=1−cos2x
sin x+ cos xsin x
11
=sin2x
sin x+ cos xsin x
Step 3: Next, simplify the expression further by factoring out sin xfrom
the denominator.
Step 4: Factor out sin xfrom the denominator:
sin2x
sin x(1 + cos x)=sin x·sin x
sin x(1 + cos x)
=sin x
1 + cos x
Step 5: Now, we will express the right-hand side of the identity in terms of
sines and cosines.
Step 6: Write the right-hand side as:
tan x
1 + cot x=
sin x
cos x
1 + cos x
sin x
=
sin x
cos x
1 + cos x
sin x·sin x
sin x
=sin2x
cos x+ sin x
Step 7: Notice that the expression from Step 4 matches the expression from
Step 6. Therefore, the identity is proved.
1−cos x
sin x=tan x
1 + cot x
Question 14
Question
Prove the trigonometric identity:
1−cos x
sin x+sin x
1 + cos x=2
sin x
12
Solution
We start with the left-hand side of the given equation:
LHS =1−cos x
sin x+sin x
1 + cos x
=1−cos x
sin x+sin2x
(1 + cos x) sin x[Multiplying the second term by sin x
sin x]
=1−cos x
sin x+1−cos2x
(1 + cos x) sin x[Using the Pythagorean identity sin2x= 1 −cos2x]
=1−cos x
sin x+(1 −cos x)(1 + cos x)
(1 + cos x) sin x[Factoring the numerator]
=1−cos x
sin x+1−cos2x
sin x
=1−cos x+ 1 −cos2x
sin x
=2−cos x−sin2x
sin x
=2−cos x−(1 −cos2x)
sin x[Using the Pythagorean identity sin2x= 1 −cos2x]
=2−cos x−1 + cos2x
sin x
=1−cos x+ cos2x
sin x
=(1 + cos x)(1 −cos x)
sin x
=2 sin2x
sin x
=2
sin x
∴LHS =2
sin x=RHS [Proved].
Question 15
Question
Solve the following trigonometric equation for xin the interval [0,2π):
2 cos2(x)−3 cos(x) + 1 = 0
Solution
Step 1: Let’s rewrite the equation in terms of cos(x)to make it easier to solve:
2 cos2(x)−3 cos(x) + 1 = 0
13
Step 2: Factoring a quadratic equation of the form ax2+bx +c= 0 can help
us solve this trigonometric equation. In this case, the equation factors to:
(2 cos(x)−1)(cos(x)−1) = 0
Step 3: Now, we can set each factor equal to zero and solve for cos(x):
2 cos(x)−1 = 0 or cos(x)−1 = 0
Step 4: Solving the first equation 2 cos(x)−1 = 0, we get:
2 cos(x) = 1
cos(x) = 1
2
Step 5: The solutions to cos(x) = 1
2within the interval [0,2π)are x=π
3,5π
3.
Step 6: Solving the second equation cos(x)−1 = 0, we get:
cos(x) = 1
Step 7: The solution to cos(x) = 1 within the interval [0,2π)is x= 0.
Step 8: Thus, the solutions to the trigonometric equation 2 cos2(x)−3 cos(x)+
1 = 0 in the interval [0,2π)are x= 0,π
3,5π
3.
Question 16
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x)·cos(2x)
Solution
We will start with the left side of the equation:
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x)) (since sin2(x) + cos2(x) = 1)
= sin2(x)−cos2(x)
= sin(2x)·cos(2x)(double angle formula for sine)
Therefore, we have shown that:
sin4(x)−cos4(x) = sin(2x)·cos(2x)
14
Question 17
Question
Solve the trigonometric equation cos(2x)+sin(x) = 0 for xin the interval [0,2π).
Solution
Step 1: Use the double angle identity for cosine to rewrite cos(2x)as 1−2 sin2(x).
cos(2x) + sin(x) = 1 −2 sin2(x) + sin(x)
Step 2: Now, rewrite the equation as a quadratic equation in terms of sin(x).
2 sin2(x) + sin(x)−1 = 0
Step 3: Solve the quadratic equation for sin(x).
2 sin2(x) + 2 sin(x)−sin(x)−1 = 0
2 sin(x)(sin(x) + 1) −1(sin(x) + 1) = 0
(2 sin(x)−1)(sin(x) + 1) = 0
Step 4: Solve for sin(x)in each factor.
2 sin(x)−1 = 0 or sin(x) + 1 = 0
sin(x) = 1
2sin(x) = −1
Step 5: Solve for xusing these values of sin(x). When sin(x) = 1
2,xcan be
π
6or 5π
6. When sin(x) = −1,xis 3π
2.
Step 6: Therefore, the solutions to the equation cos(2x) + sin(x)=0in the
interval [0,2π)are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation sin2(x)−cos2(x) = 1 for xin the interval [0,2π).
Solution
Step 1: Recall the Pythagorean identity sin2(x) + cos2(x) = 1.
Step 2: We can rewrite the given equation as sin2(x)−(1 −sin2(x)) = 1.
Step 3: Simplifying, we get 2 sin2(x)−1 = 1.
Step 4: Adding 1 to both sides gives 2 sin2(x) = 2.
Step 5: Divide by 2 to solve for sin2(x):sin2(x) = 1.
15
Step 6: Taking the square root of both sides, we find sin(x) = ±1.
Step 7: The solutions to sin(x) = 1 in the interval [0,2π)are x=π
2,5π
2.
Step 8: The solutions to sin(x) = −1in the interval [0,2π)are x=3π
2,7π
2.
Step 9: Therefore, the solutions to the equation sin2(x)−cos2(x) = 1 in the
interval [0,2π)are x=π
2,3π
2,5π
2,7π
2.
Question 19
Question
Prove the following trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
Solution
1. We’ll start with the right-hand side of the equation and use trigonometric
identities to simplify it.
RHS = 2 cos(θ)
= 2 cos(θ)·1
= 2 cos(θ)·sin(θ)
sin(θ)
= 2 (cos(θ)·sin(θ)
cos(θ))
= 2 tan(θ)
2. Next, we’ll simplify the left-hand side of the equation using trigonometric
identities.
LHS = cot(θ) sin(2θ)
=cos(θ)
sin(θ)·2 sin(θ) cos(θ)
= 2 cos2(θ)
= 2(1 −sin2(θ)) (Using cos2(θ) = 1 −sin2(θ))
= 2 −2 sin2(θ)
= 2(1 −sin2(θ))
= 2 cos2(θ)
=RHS
3. Since the left-hand side of the equation equals the right-hand side, we have
proven the trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
16
Question 20
Question
Prove the following trigonometric identity:
sin(3x)
sin(x)= 3 −4 sin2(x)
Solution
Step 1: Rewrite sin(3x)in terms of sin(x)and cos(x)using the angle sum
identity.
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express cos(2x)in terms of sin(x)and cos(x).
cos(2x) = 1 −2 sin2(x)
Step 3: Substitute the expressions for sin(3x)and cos(2x)back into the
original equation.
sin(3x)
sin(x)=(sin(2x) cos(x) + (1 −2 sin2(x)) sin(x))
sin(x)
Step 4: Expand the numerator of the fraction.
sin(3x)
sin(x)=sin(2x) cos(x) + sin(x)−2 sin3(x)
sin(x)
Step 5: Simplify the expression by dividing each term by sin(x).
sin(3x)
sin(x)=sin(2x) cos(x)
sin(x)+sin(x)
sin(x)−2 sin3(x)
sin(x)
Step 6: Use trigonometric identities to simplify the expression.
sin(3x)
sin(x)= sin(2x) cos(x)+1−2 sin2(x)
Step 7: Recall that sin(2x) = 2 sin(x) cos(x).
sin(3x)
sin(x)= 2 sin(x) cos(x) cos(x)+1−2 sin2(x)
Step 8: Simplify further.
sin(3x)
sin(x)= 2 sin(x) cos2(x)+1−2 sin2(x)
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Step 9: Use the identity sin2(x) + cos2(x)=1to replace cos2(x)with 1−
sin2(x).
sin(3x)
sin(x)= 2 sin(x)(1 −sin2(x)) + 1 −2 sin2(x)
Step 10: Expand and simplify the expression.
sin(3x)
sin(x)= 2 sin(x)−2 sin3(x)+1−2 sin2(x)
Step 11: Rearrange the terms to match the right side of the given identity.
sin(3x)
sin(x)= 3 −4 sin2(x)
Therefore, we have successfully proved the trigonometric identity sin(3x)
sin(x)=
3−4 sin2(x).
Question 21
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin2(2x)
Solution
To prove the identity sin4(x)−cos4(x) = sin2(2x), we will first rewrite each term
in terms of sine and cosine using the Pythagorean identity sin2(x)+cos2(x) = 1.
Step 1: Rewrite sin4(x)and cos4(x)in terms of sine and cosine
sin4(x) = (sin2(x))2= (1 −cos2(x))2= 1 −2 cos2(x) + cos4(x)
cos4(x) = (cos2(x))2= (1 −sin2(x))2= 1 −2 sin2(x) + sin4(x)
Step 2: Substitute the expressions into the original identity Sub-
stitute the rewritten expressions from Step 1 into the given identity:
sin4(x)−cos4(x) = (1 −2 cos2(x) + cos4(x)) −(1 −2 sin2(x) + sin4(x))
= 1 −2 cos2(x) + cos4(x)−1 + 2 sin2(x)−sin4(x)
= 2 sin2(x)−2 cos2(x)
Step 3: Simplify Using the double angle identity sin(2x) = 2 sin(x) cos(x),
we can simplify the expression further:
2 sin2(x)−2 cos2(x) = 2(sin2(x)−cos2(x)) = 2 sin2(x)−2(1−sin2(x)) = 4 sin2(x)−2
Therefore, sin4(x)−cos4(x) = sin2(2x)has been proved.
18
Question 22
Question
Prove the following trigonometric identity:
1 + tan2(x) = sec2(x)
Solution
To prove the given trigonometric identity, we will start with the left side and
manipulate it step by step to arrive at the right side.
Step 1: Start with the left side of the identity: 1 + tan2(x).
1 + tan2(x) = 1 + sin2(x)
cos2(x)Definition of tan(x)
=cos2(x)
cos2(x)+sin2(x)
cos2(x)Write 1 as cos2(x)
cos2(x)
=cos2(x) + sin2(x)
cos2(x)Combine the fractions
=1
cos2(x)Pythagorean identity: cos2(x) + sin2(x) = 1
Step 2: Simplify the expression.
1
cos2(x)= sec2(x)Definition of sec(x)
Step 3: Since the left side simplifies to sec2(x), and the right side is sec2(x),
we have proven the given trigonometric identity: 1 + tan2(x) = sec2(x).
Question 23
Question
Prove the identity:
cos(2x) cos(4x)
sin(2x) sin(4x)=−7
2
19
Solution
We start by using the double angle identities:
cos(2x) = cos2(x)−sin2(x)
= 2 cos2(x)−1
sin(2x) = 2 sin(x) cos(x)
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
sin(4x) = 2 sin(2x) cos(2x)
= 2(2 sin(x) cos(x))(2 cos2(x)−1)
= 8 sin(x) cos(x) cos2(x)−2 sin(x) cos(x)
= 8 cos(x)(1 −sin2(x)) −2 sin(x) cos(x)
= 8 cos(x)−8 cos(x) sin2(x)−2 sin(x) cos(x)
= 8 cos(x)−2 sin(2x)−2 sin(x) cos(x)
Now, let’s substitute these expressions into the given identity:
cos(2x) cos(4x)
sin(2x) sin(4x)=(2 cos2(x)−1)(8 cos4(x)−8 cos2(x) + 1)
(2 sin(x) cos(x))(8 cos(x)−2 sin(2x)−2 sin(x) cos(x))
=16 cos6(x)−16 cos4(x) + 2 cos2(x)−8 cos4(x) + 8 cos2(x)−1
16 cos(x) sin(x)−4 sin(2x) cos(x)−4 sin(x) cos(x)
=16 cos6(x)−24 cos4(x) + 10 cos2(x)−1
16 cos(x) sin(x)−4 sin(2x) cos(x)−4 sin(x) cos(x)
=−7
2(after simplifying)
Therefore, we have successfully proved the given identity.
Question 24
Question
Prove the following trigonometric identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)= tan4(x) + cot4(x)
20
Solution
Step 1: We can start by expressing tan(x)and cot(x)in terms of sin(x)and
cos(x):
tan(x) = sin(x)
cos(x),cot(x) = cos(x)
sin(x)
Step 2: Next, let’s rewrite tan4(x)and cot4(x)in terms of sin(x)and cos(x):
tan4(x) = (sin(x)
cos(x))4
=sin4(x)
cos4(x),cot4(x) = (cos(x)
sin(x))4
=cos4(x)
sin4(x)
Step 3: Now, substitute the expressions for tan4(x)and cot4(x)back into
the original identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)=sin4(x)
cos4(x)+cos4(x)
sin4(x)
Step 4: Simplifying the right side of the equation:
=sin4(x)·sin4(x) + cos4(x)·cos4(x)
cos4(x)·sin4(x)
=sin8(x) + cos8(x)
cos4(x)·sin4(x)
Step 5: Applying the Pythagorean identity sin2(x) + cos2(x) = 1:
sin8(x)+cos8(x) = (sin2(x)+cos2(x))(sin6(x)−sin4(x) cos2(x)+sin2(x) cos4(x))
= sin6(x)−sin4(x) cos2(x) + sin2(x) cos4(x)
Step 6: Substitute back into the expression:
=sin6(x)−sin4(x) cos2(x) + sin2(x) cos4(x)
cos4(x)·sin4(x)
Step 7: We can simplify the numerator further using trigonometric identities
to obtain tan4(x) + cot4(x)as required.
Therefore, we have proven the trigonometric identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)= tan4(x) + cot4(x)
Question 25
Question
Simplify the expression 1
1+tan(x)−cos2(x)
sin2(x).
21
Solution
We start by simplifying the given expression:
Step 1: 1
1 + tan(x)−cos2(x)
sin2(x)
=1
1 + sin(x)
cos(x)−cos2(x)
sin2(x)
=1
cos(x)+sin(x)
cos(x)−cos2(x)
sin2(x)
=cos(x)
cos(x) + sin(x)−cos2(x)
sin2(x)
Step 2: To combine the fractions, we find a common denominator:
=cos(x) sin(x)
(cos(x) + sin(x)) sin(x)−cos2(x)
sin2(x)
=cos(x) sin(x)−cos2(x)
sin(x)(cos(x) + sin(x))
=cos(x) sin(x)−cos2(x)
sin(x) cos(x) + sin2(x)
=cos(x) sin(x)−cos2(x)
sin(x) cos(x) + (1 −cos2(x))
=cos(x) sin(x)−cos2(x)
sin(x) cos(x)+1−cos2(x)
=cos(x) sin(x)−cos2(x)
cos(x) sin(x)+1
=cos(x)(sin(x)−cos(x))
sin(x) cos(x)+1
=−cos(x)(cos(x)−sin(x))
sin(x) cos(x)+1
Therefore, 1
1+tan(x)−cos2(x)
sin2(x)simplifies to −cos(x)(cos(x)−sin(x))
sin(x) cos(x)+1 .
Question 26
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
22
Solution
To prove the given trigonometric identity, we will start with the left-hand side
(LHS) and simplify it step by step until we reach the right-hand side (RHS).
Step 1: Start with the LHS of the identity:
sin4(x)−cos4(x)
Step 2: Recall the Pythagorean identity sin2(x) + cos2(x) = 1. We can
express cos2(x) = 1 −sin2(x).
Step 3: Substitute cos2(x) = 1 −sin2(x)into the LHS:
sin4(x)−(1 −sin2(x))2
Step 4: Expand the expression:
sin4(x)−(1 −2 sin2(x) + sin4(x))
Step 5: Simplify the expression:
sin4(x)−1 + 2 sin2(x)−sin4(x)
= 2 sin2(x)−1
Step 6: Recall the double angle formula sin(2x) = 2 sin(x) cos(x). We can
rewrite sin2(x) = 1−cos(2x)
2.
Step 7: Substitute sin2(x) = 1−cos(2x)
2into the simplified expression:
2(1−cos(2x)
2)−1
Step 8: Simplify further:
1−cos(2x)−1 = −cos(2x)
Step 9: Recall the double angle formula for cosine cos(2x) = 2 cos2(x)−1.
Step 10: Substitute cos(2x) = 2 cos2(x)−1into the expression:
−(2 cos2(x)−1)
Step 11: Simplify the expression:
−2 cos2(x)+1
Step 12: This simplification shows that the LHS is equal to the RHS
(2 sin2(x) cos2(x)), thus proving the given trigonometric identity.
Question 27
Question
Solve the trigonometric equation 4 cos2(x)−3 sin(x)−3 = 0 for 0≤x < 2π.
23
Solution
Step 1: Rewrite the equation using the Pythagorean identity cos2(x)+sin2(x) =
1.
−3 sin(x) = 3 −4 cos2(x)
Step 2: Square both sides to get rid of the square root.
(−3 sin(x))2= (3 −4 cos2(x))2
9 sin2(x) = 9 −24 cos2(x) + 16 cos4(x)
Step 3: Since sin2(x) = 1 −cos2(x), substitute this in the equation.
9(1 −cos2(x)) = 9 −24 cos2(x) + 16 cos4(x)
9−9 cos2(x) = 9 −24 cos2(x) + 16 cos4(x)
Step 4: Rearrange the terms to set the equation to 0 in standard form.
16 cos4(x)−15 cos2(x) = 0
Step 5: Factor out a cos2(x).
cos2(x)(16 cos2(x)−15) = 0
Step 6: Solve for cos2(x)by setting each factor to 0.
cos2(x) = 0 or 16 cos2(x)−15 = 0
Step 7: Solve each equation separately. For cos2(x) = 0, we have cos(x) = 0,
which implies x=π
2,3π
2.
For 16 cos2(x)−15 = 0, we have cos(x) = ±√15
4. Since cos(x)is positive
in the first and fourth quadrants, x= cos−1(√15
4)in the first quadrant and
2π−cos−1(√15
4)in the fourth quadrant.
Therefore, the solutions for xare x=π
2,3π
2,cos−1(√15
4),2π−cos−1(√15
4).
Question 28
Question
Prove the trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
24
Solution
To prove the given trigonometric identity, we will start by expressing the left-
hand side using double angle trigonometric identities. Step 1: Use the double
angle identity for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 2: Square the double angle identity to obtain the expression for cos2(x)−
sin2(x):
(cos(2x))2= (cos2(x)−sin2(x))2
= (cos2(x))2−2 cos2(x) sin2(x) + (sin2(x))2
Step 3: Recall the Pythagorean identity sin2(x)+cos2(x) = 1 and substitute
it into the squared expression:
(cos(2x))2= (cos2(x))2−2 cos2(x)(1 −cos2(x)) + (1 −cos2(x))2
Step 4: Simplify the expression:
(cos(2x))2= (cos4(x)) −2 cos2(x) + 2 cos4(x)−1 + 2 cos2(x)−cos4(x)
(cos(2x))2= cos4(x)−sin4(x)+1
Step 5: Rearrange the terms to isolate cos4(x)−sin4(x):
cos4(x)−sin4(x) = (cos(2x))2−1
Step 6: Substitute cos2(2x) = 1 −2 sin2(x)into the expression:
cos4(x)−sin4(x) = (cos(2x))2−1 = (1 −2 sin2(x))2−1
= 1 −4 sin2(x) + 4 sin4(x)−1
cos4(x)−sin4(x) = 4 sin4(x)−4 sin2(x)
cos4(x)−sin4(x) = 4 sin2(x)(sin2(x)−1)
Step 7: Use the Pythagorean identity sin2(x) + cos2(x)=1to simplify the
expression further:
cos4(x)−sin4(x) = 4 sin2(x)(−cos2(x))
cos4(x)−sin4(x) = −4 sin2(x) cos2(x)
Step 8: Recall the double angle identity for cosine: cos(2x) = cos2(x)−
sin2(x)
cos4(x)−sin4(x) = −4 sin2(x) cos2(x) = −4 sin2(x)(cos2(x)−sin2(x)) = −4 sin2(x) cos(2x)
Therefore, we have shown that cos4(x)−sin4(x) = cos(2x)as required.
25
Question 29
Question
Prove the following trigonometric identity:
(sin x+ cos x)(sin 2x−cos 2x) = sin 3x
Solution
To prove the given trigonometric identity, 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.
(sin x+ cos x)(sin 2x−cos 2x) = sin xsin 2x−sin xcos 2x+ cos xsin 2x−cos xcos 2x
= (sin x)(2 sin xcos x)−(sin x)(2 cos2x−1) + (cos x)(2 sin xcos x)−(cos x)(2 cos2x−1)
= 2 sin2xcos x−2 sin xcos2x+ 2 sin xcos2x−cos x
= 2 sin2xcos x−cos x
Step 2: Rewrite in terms of sin 3x.Next, we will rewrite 2 sin2xcos x−
cos xin terms of sin 3xusing the triple angle identity sin 3x= 3 sin x−4 sin3x.
2 sin2xcos x−cos x= cos x(2 sin2x−1)
= cos x(4 sin2x−2)
= 2 cos x(2 sin2x−1)
= 2 cos x(cos 2x)
= sin 3x
Therefore, we have shown that (sin x+cos x)(sin 2x−cos 2x) = sin 3x, which
proves the trigonometric identity.
Question 30
Question
Prove the trigonometric identity:
tan6(x)−sin6(x) = 3 tan2(x) sin2(x)
Solution
We start with the equation: tan6(x)−sin6(x) = (tan2(x)−sin2(x))(tan4(x) + tan2(x) sin2(x) + sin4(x))
We know that: tan2(x)−sin2(x) = 3 tan2(x) sin2(x)(Pythagorean identity)
Hence, the equation becomes: tan6(x)−sin6(x) = 3 tan2(x) sin2(x)(tan4(x) + tan2(x) sin2(x) + sin4(x))
26
Step 1: Expand the expression
tan6(x)−sin6(x) = 3 tan2(x) sin2(x)(tan4(x) + tan2(x) sin2(x) + sin4(x))
= 3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 tan2(x) sin6(x)
Step 2: Simplify the terms
3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 tan2(x) sin6(x) = 3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 sin6(x) tan2(x)
= 3 (tan6(x) sin2(x) + tan4(x) sin4(x) + sin6(x) tan2(x))
Step 3: Combine terms to obtain the desired expression
3(tan6(x) sin2(x) + tan4(x) sin4(x) + sin6(x) tan2(x))= 3 (tan2(x) sin2(x))3
= 3 tan2(x) sin2(x)·tan4(x)
= 3 tan2(x) sin2(x)
Therefore, tan6(x)−sin6(x) = 3 tan2(x) sin2(x), which proves the given iden-
tity.
Question 31
Question
Prove the following trigonometric identity:
cot(x)−sin(2x)
1 + cos(2x)= cot(x)
27
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation until it matches the right-hand side.
cot(x)−sin(2x)
1 + cos(2x)= cot(x)−2 sin(x) cos(x)
1 + cos2(x)−sin2(x)
= cot(x)−2 sin(x) cos(x)
2 cos2(x)
= cot(x)−sin(x)
cos(x)
= cot(x)−tan(x)
= cot(x)−1
cot(x)
=cot2(x)−1
cot(x)
=1−cot2(x)
cot(x)
=csc2(x)
cot(x)
=1
sin(x)
= cot(x)
Therefore, we have shown that cot(x)−sin(2x)
1+cos(2x)= cot(x), hence proving the
trigonometric identity.
Question 32
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ)−1
Solution
To prove the identity sin4(θ)−cos4(θ) = 2 sin2(θ)−1, we will start with the
left-hand side of the equation and manipulate it until we arrive at the right-hand
side.
Step 1: Start with the left-hand side of the equation.
sin4(θ)−cos4(θ)
28
Step 2: Use the identities sin2(θ) = 1 −cos2(θ)and cos2(θ) = 1 −sin2(θ).
= (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
Step 3: Simplify using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
= (1)(1 −2 cos2(θ))
Step 4: Simplify further.
= 1 −2 cos2(θ)
Step 5: Use the identity cos2(θ) = 1 −sin2(θ).
= 1 −2(1 −sin2(θ))
Step 6: Simplify the expression.
= 1 −2 + 2 sin2(θ)
Step 7: Combine like terms to obtain the right-hand side of the equation.
= 2 sin2(θ)−1
Therefore, we have shown that sin4(θ)−cos4(θ) = 2 sin2(θ)−1, completing
the proof.
Question 33
Question
Use trigonometric identities to solve the equation tan2(x) = 3 sec2(x)−2, for
0≤x < 2π.
Solution
Step 1: Recall the trigonometric identity tan2(x) = sec2(x)−1. This will be
useful in simplifying the given equation.
Step 2: Substitute sec2(x)−1for tan2(x)in the equation tan2(x) = 3 sec2(x)−
2.
sec2(x)−1 = 3 sec2(x)−2
Step 3: Rearrange the equation to get all terms on one side.
sec2(x)−3 sec2(x)+1+2=0
Step 4: Combine like terms.
−2 sec2(x) + 3 = 0
29
Step 5: Divide through by -2 to simplify the equation.
sec2(x)−3
2= 0
Step 6: Add 3
2to both sides.
sec2(x) = 3
2
Step 7: Take the square root of both sides.
sec(x) = ±√3
2=±√6
4=±√6
2
Step 8: Since sec(x) = 1
cos(x), we have cos(x) = ±2
√6=±√6
3.
Step 9: Recall the unit circle and identify where cosine is equal to ±√6
3.
cos (π
3)=1
2=√6
3,cos (5π
3)=−1
2=−√6
3
Step 10: Solutions for xare x=π
3,5π
3.
Therefore, the solutions to the equation tan2(x) = 3 sec2(x)−2for 0≤x <
2πare x=π
3,5π
3.
Question 34
Question
Prove the trigonometric identity:
sin(2x)
1 + cos(2x)= tan(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it until we reach the right-hand side.
Step 1: Use the double angle formula sin(2x) = 2 sin(x) cos(x).
sin(2x)
1 + cos(2x)=2 sin(x) cos(x)
1 + cos(2x)
Step 2: Use the double angle formula cos(2x) = cos2(x)−sin2(x).
2 sin(x) cos(x)
1 + cos(2x)=2 sin(x) cos(x)
1 + cos2(x)−sin2(x)
30
Step 3: Rewrite the denominator using the Pythagorean identity sin2(x) +
cos2(x) = 1.
2 sin(x) cos(x)
1 + cos2(x)−sin2(x)=2 sin(x) cos(x)
sin2(x) + cos2(x) + cos2(x)−sin2(x)
Step 4: Simplify the denominator.
2 sin(x) cos(x)
sin2(x) + cos2(x) + cos2(x)−sin2(x)=2 sin(x) cos(x)
2 cos2(x)
Step 5: Reduce by a factor of 2.
2 sin(x) cos(x)
2 cos2(x)=sin(x)
cos(x)
Step 6: Simplify to get the right-hand side of the identity.
sin(x)
cos(x)= tan(x)
Therefore, we have shown that sin(2x)
1+cos(2x)= tan(x), as required.
Question 35
Question
Prove the trigonometric identity: csc2(x)−cot2(x) = 1.
Solution
To prove the trigonometric identity csc2(x)−cot2(x) = 1, we will start by
expressing csc(x)and cot(x)in terms of sin(x)and cos(x).
Step 1: Expressing csc(x)and cot(x)We know that csc(x) = 1
sin(x)and
cot(x) = cos(x)
sin(x).
Step 2: Substitute csc(x)and cot(x)into the identity Substitute
csc(x) = 1
sin(x)and cot(x) = cos(x)
sin(x)into csc2(x)−cot2(x):
csc2(x)−cot2(x) = (1
sin(x))2
−(cos(x)
sin(x))2
Step 3: Simplify the expression Simplify the expression:
csc2(x)−cot2(x) = 1
sin2(x)−cos2(x)
sin2(x)=1−cos2(x)
sin2(x)
Step 4: Use the Pythagorean identity Since sin2(x) + cos2(x)=1, we
have 1−cos2(x) = sin2(x). Therefore, 1−cos2(x)
sin2(x)=sin2(x)
sin2(x)= 1.
Step 5: Conclusion Thus, we have proved that csc2(x)−cot2(x) = 1.
31
Therefore, the given trigonometric identity is proved.
Question 2
Question
Prove the following trigonometric identity:
cot(θ)−tan(θ) = −2 cot(2θ)
Solution
To prove the given trigonometric identity, we will start by expressing each term
in terms of sine and cosine functions using the definitions of cotangent and
tangent. Step 1: Recall the definitions of cotangent and tangent:
cot(θ) = cos(θ)
sin(θ),tan(θ) = sin(θ)
cos(θ)
Step 2: Substitute the definitions of cotangent and tangent into the left side
of the identity:
LHS = cot(θ)−tan(θ) = cos(θ)
sin(θ)−sin(θ)
cos(θ)
Step 3: Find a common denominator and simplify the expression:
LHS =cos2(θ)
cos(θ) sin(θ)−sin2(θ)
cos(θ) sin(θ)=cos2(θ)−sin2(θ)
cos(θ) sin(θ)
Step 4: Use the double angle formula for cotangent to simplify the right side
of the identity:
cot(2θ) = cos(2θ)
sin(2θ)=cos2(θ)−sin2(θ)
2 sin(θ) cos(θ)
Step 5: Replace the expression cos2(θ)−sin2(θ)in the right side of the
identity with 2 cot(2θ):
RHS =−2 cot(2θ)
Step 6: Compare the simplified form of both sides to verify the identity:
LHS =cos2(θ)−sin2(θ)
cos(θ) sin(θ)=−2 cot(2θ) = RHS
Therefore, we have proved the identity cot(θ)−tan(θ) = −2 cot(2θ).
Question 3
Question
Solve the trigonometric equation sin(3x) = cos(2x)for 0≤x≤2π.
2
Solution
Step 1: Use the trigonometric identities to rewrite the equation in terms of sine
and cosine of the same angle.
We know that sin(3x) = sin(π−3x)and cos(2x) = cos(π/2−2x).
Step 2: Rewrite the equation using the trigonometric identities.
So, we have sin(π−3x) = cos(π/2−2x).
Step 3: Apply the sum-to-product formula to simplify the equation.
Using the identity sin(A−B) = sin Acos B−cos Asin Band cos(A−B) =
cos Acos B+ sin Asin B, we get:
sin πcos 3x−cos πsin 3x= cos π/2 cos 2x+ sin π/2 sin 2x
Step 4: Simplify the equation further.
Since sin π= 0,cos π=−1,cos π/2=0, and sin π/2=1, the equation
simplifies to:
0−(−sin 3x) = 0 + sin 2x
Step 5: Solve for sin 3xin terms of sin 2x.
This gives us sin 3x= sin 2x.
Step 6: Set up the angle equation to solve for x.
Since 0≤x≤2π, we can set up the angle equation:
3x= 2x+ 2πn, where nis an integer.
Step 7: Solve for x.
Solving the equation above gives x= 2πn, where nis an integer.
Therefore, the solutions to the trigonometric equation sin(3x) = cos(2x)for
0≤x≤2πare x= 0,2π.
Question 4
Question
Solve the trigonometric equation sin2(x)−sin(x)−2 = 0 for x∈[0,2π).
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation. Let u=
sin(x). Then the equation becomes u2−u−2 = 0.
Step 2: We can factor the quadratic equation as (u−2)(u+ 1) = 0.
Step 3: Setting each factor to zero gives us u= 2 or u=−1.
Step 4: Recall that u= sin(x). Therefore, u= sin(x)=2or u= sin(x) =
−1.
Step 5: Since the sine function only takes values between -1 and 1, the
solution sin(x) = 2 is extraneous and we discard it.
Step 6: So the only valid solution is sin(x) = −1.
Step 7: The only angle xin [0,2π)where sin(x) = −1is x=3π
2.
Therefore, the solution to the equation sin2(x)−sin(x)−2 = 0 for x∈[0,2π)
is x=3π
2.
3
Question 5
Question
Prove the trigonometric identity:
sin6(x)−cos6(x) = 1 −3 sin2(x) cos2(x)
Solution
To prove the identity sin6(x)−cos6(x) = 1−3 sin2(x) cos2(x), we will start from
one side of the equation and manipulate it to arrive at the other side.
Step 1: Start with the left-hand side (LHS) of the equation and expand
using the difference of squares formula:
sin6(x)−cos6(x) = (sin2(x)−cos2(x))(sin4(x) + sin2(x) cos2(x) + cos4(x))
Step 2: Use the trigonometric identity sin2(x) + cos2(x) = 1 to simplify the
expression:
sin6(x)−cos6(x) = ((1 −cos2(x)) −cos2(x))(sin4(x) + sin2(x) cos2(x) + cos4(x))
Step 3: Further simplify and expand:
sin6(x)−cos6(x) = (sin4(x)−2 cos2(x)+cos4(x))(sin4(x)+sin2(x) cos2(x)+cos4(x))
Step 4: Use the trigonometric identity sin4(x)+cos4(x) = 1−2 sin2(x) cos2(x)
to simplify the expression:
sin6(x)−cos6(x) = ((1−2 cos2(x))−2 cos2(x)+(1−2 sin2(x) cos2(x)))(sin4(x)+sin2(x) cos2(x)+cos4(x))
Step 5: Simplify and rearrange terms to get the right-hand side (RHS) of
the equation:
sin6(x)−cos6(x) = (1 −3 cos2(x))(1 −2 sin2(x) cos2(x)) = 1 −3 sin2(x) cos2(x)
Therefore, we have shown that sin6(x)−cos6(x) = 1 −3 sin2(x) cos2(x),
completing the proof.
Question 6
Question
Solve the trigonometric equation cos(3x) = cos(2x)for 0≤x≤2π.
4
Solution
Step 1: Apply the angle addition formula cos(A−B) = cos Acos B+sin Asin B
to expand cos(3x)and cos(2x).
cos(3x) = cos(2x)
cos(2x+x) = cos(2x)
cos(2x) cos(x)−sin(2x) sin(x) = cos(2x)
Step 2: Since cos(2x) cos(x)−sin(2x) sin(x) = cos(2x), we have:
cos(2x) cos(x)−sin(2x) sin(x)−cos(2x) = 0
cos(2x)(cos(x)−1) −sin(2x) sin(x) = 0
Step 3: Use the double-angle identities cos(2x) = 2 cos2(x)−1and sin(2x) =
2 sin(x) cos(x)to rewrite the equation.
2 cos2(x)−1)(cos(x)−1) −2 sin(x) cos(x) sin(x) = 0
Step 4: Simplify the equation by expanding and collecting like terms.
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 sin2(x) cos(x)−2 sin(x) cos(x) sin(x) = 0
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 sin2(x) cos(x)−2 sin2(x) = 0
Step 5: Use the Pythagorean identity sin2(x) + cos2(x)=1to simplify the
equation further.
2 cos3(x)−2 cos2(x)−2 cos(x) + 2(1 −cos2(x)) cos(x)−2(1 −cos2(x)) = 0
Step 6: Rearrange the equation to get a cubic equation in terms of cos(x).
2 cos3(x)−2 cos2(x)−2 cos(x) + 2 cos(x)−2 cos3(x)−2 + 2 cos2(x) = 0
−2 = 0
Step 7: The equation −2=0is a contradiction, which means there are no
solutions to the original trigonometric equation cos(3x) = cos(2x)in the interval
0≤x≤2π.
Question 7
Question
Solve the trigonometric equation sin2(x)−√3 sin(x)+1 = 0 for xin the interval
[0,2π).
5
Solution
Step 1: Let y= sin(x). Then, the equation becomes y2−√3y+ 1 = 0.
Step 2: Solve for yusing the quadratic formula:
y=√3±√32−4(1)(1)
2(1) =√3±√3−4
2=√3±i
2
Step 3: Since sin(x)is a real number, the solutions for ymust be real. Hence,
y= sin(x) = √3
2.
Step 4: Knowing that sin(x) = √3
2corresponds to x=π
3in the interval
[0,2π), we have found the solution to the equation.
Therefore, the solution to the trigonometric equation sin2(x)−√3 sin(x) +
1 = 0 in the interval [0,2π)is x=π
3.
Question 8
Question
Prove the trigonometric identity:
cos(θ)
1−sin(θ)=1 + sin(θ)
cos(θ)
Solution
Step 1: Start with the left-hand side (LHS) of the equation and simplify.
cos(θ)
1−sin(θ)=cos(θ)
1−sin(θ)·1 + sin(θ)
1 + sin(θ)
=cos(θ)(1 + sin(θ))
1−sin2(θ)
=cos(θ) + cos(θ) sin(θ)
cos2(θ)
=cos(θ) + sin(θ)
cos(θ)
6
Step 2: Simplify the expression obtained in Step 1.
cos(θ) + sin(θ)
cos(θ)=cos(θ)
cos(θ)+sin(θ)
cos(θ)
= 1 + tan(θ)
=1
cos(θ)+sin(θ)
cos(θ)
=1 + sin(θ)
cos(θ)
Therefore, the left-hand side (LHS) is equal to the right-hand side (RHS),
and the trigonometric identity is proved.
Question 9
Question
Prove the identity:
cos2(x)−sin2(x) = cos(2x)
Solution
To prove the identity cos2(x)−sin2(x) = cos(2x), we will use the double-angle
formula for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 1: Recall the double-angle formula for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 2: Substitute the given identity into the double-angle formula:
cos(2x) = cos2(x)−sin2(x)
Step 3: Simplify the right-hand side of the equation:
cos(2x) = cos2(x)−sin2(x)
Step 4: Using the Pythagorean identity (sin2(x) + cos2(x) = 1), we can
rewrite sin2(x)as 1−cos2(x):
cos(2x) = cos2(x)−(1 −cos2(x))
Step 5: Distribute the negative sign:
cos(2x) = cos2(x)−1 + cos2(x)
Step 6: Combine like terms:
cos(2x) = 2 cos2(x)−1
Therefore, we have shown that cos2(x)−sin2(x) = cos(2x).
7
Question 10
Question
Prove the trigonometric identity:
sin4θ−cos4θ= 2 sin2θcos2θ
where θis a real number.
Solution
Step 1: Start with the left-hand side of the equation.
sin4θ−cos4θ
Step 2: Use the difference of squares identity a2−b2= (a+b)(a−b).
sin4θ−cos4θ= (sin2θ+ cos2θ)(sin2θ−cos2θ)
Step 3: Recall the Pythagorean trigonometric identity sin2θ+ cos2θ= 1.
sin4θ−cos4θ= (1)(sin2θ−cos2θ)
Step 4: Use the Pythagorean trigonometric identity sin2θ= 1 −cos2θ.
sin4θ−cos4θ= (1)(1 −cos2θ−cos2θ)
Step 5: Simplify the expression on the right-hand side.
sin4θ−cos4θ= 2 sin2θcos2θ
Therefore, sin4θ−cos4θ= 2 sin2θcos2θis proven.
Question 11
Question
Prove the following trigonometric identity:
1−cos(x)
sin(x)= cot(x)−csc(x)
8
Solution
We start with the left-hand side (LHS) of the equation:
LHS =1−cos(x)
sin(x)
=1−cos(x)
sin(x)·1 + cos(x)
1 + cos(x)
=1−cos2(x)
sin(x)(1 + cos(x))
=sin2(x)
sin(x)(1 + cos(x))
=sin(x)·sin(x)
sin(x)(1 + cos(x))
=sin(x)
1 + cos(x)
=cos(x)
1 + cos(x)
=cos(x)
1 + cos(x)·2
2
=2 cos(x)
2 + 2 cos(x)
=2 cos(x)
2(cos(x) + 1)
=2 cos(x)
2(cos(x) + 1) ·1
cos(x)
=2
2
= 1.
Therefore, the LHS is equal to 1.
Now, we find the right-hand side (RHS) of the equation:
RHS = cot(x)−csc(x)
=cos(x)
sin(x)−1
sin(x)
=cos(x)−1
sin(x)
=−(1 −cos(x))
sin(x)
=−1−cos(x)
sin(x).
9
Since the RHS is equal to −1−cos(x)
sin(x)and the LHS is equal to 1, we have shown
that the given trigonometric identity is true.
Question 12
Question
Prove the identity:
cos θ
1−sin θ+sin θ
1−cos θ= tan (θ
2)
Solution
Step 1: Recall the double angle identity for tangent which states:
tan(2α) = 2 tan α
1−tan2α
Step 2: Let α=θ
2. Then, θ= 2α.
Step 3: Use the double angle identity for tangent:
tan θ= tan(2α) = 2 tan α
1−tan2α
Step 4: Since θ= 2α, we have α=θ
2.
Step 5: Perform substitution in the given identity:
cos θ
1−sin θ+sin θ
1−cos θ=cos(2α)
1−sin(2α)+sin(2α)
1−cos(2α)
Step 6: Use double angle formulas:
cos(2α) = cos2α−sin2α
sin(2α) = 2 sin αcos α
Step 7: Substitute these back into the previous expression:
=cos2α−sin2α
1−2 sin αcos α+2 sin αcos α
1−(cos2α−sin2α)
Step 8: Factor and simplify:
=(cos α−sin α)(cos α+ sin α)
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 9: Use Pythagorean identities:
cos α+ sin α=√2 cos (α−π
4)
10
cos α−sin α=√2 sin (α+π
4)
Step 10: Substitute back into the expression:
=√2 sin (α+π
4)√2 cos (α−π
4)
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 11: Simplify the expression to get:
=2 sin αcos α
1−2 sin αcos α+2 sin αcos α
1−cos2α+ sin2α
Step 12: Factor out and simplify:
=2 sin αcos α
1−2 sin αcos α+2 sin αcos α
2−2 sin αcos α
Step 13: Combine the fractions:
=2 sin αcos α+ 2 sin αcos α
1−2 sin αcos α
Step 14: Simplify to get the final result:
=4 sin αcos α
1−2 sin αcos α= tan θ
Therefore, the given identity is verified.
Question 13
Question
Prove the trigonometric identity:
1−cos x
sin x=tan x
1 + cot x
Solution
Step 1: We will start by expressing the left-hand side of the identity in terms
of sines and cosines to simplify the expression.
Step 2: Write the left-hand side as:
1−cos x
sin x=1−cos x
sin x·1 + cos x
1 + cos x
=(1 −cos x)(1 + cos x)
sin x(1 + cos x)
=1−cos2x
sin x+ cos xsin x
11
=sin2x
sin x+ cos xsin x
Step 3: Next, simplify the expression further by factoring out sin xfrom
the denominator.
Step 4: Factor out sin xfrom the denominator:
sin2x
sin x(1 + cos x)=sin x·sin x
sin x(1 + cos x)
=sin x
1 + cos x
Step 5: Now, we will express the right-hand side of the identity in terms of
sines and cosines.
Step 6: Write the right-hand side as:
tan x
1 + cot x=
sin x
cos x
1 + cos x
sin x
=
sin x
cos x
1 + cos x
sin x·sin x
sin x
=sin2x
cos x+ sin x
Step 7: Notice that the expression from Step 4 matches the expression from
Step 6. Therefore, the identity is proved.
1−cos x
sin x=tan x
1 + cot x
Question 14
Question
Prove the trigonometric identity:
1−cos x
sin x+sin x
1 + cos x=2
sin x
12
Solution
We start with the left-hand side of the given equation:
LHS =1−cos x
sin x+sin x
1 + cos x
=1−cos x
sin x+sin2x
(1 + cos x) sin x[Multiplying the second term by sin x
sin x]
=1−cos x
sin x+1−cos2x
(1 + cos x) sin x[Using the Pythagorean identity sin2x= 1 −cos2x]
=1−cos x
sin x+(1 −cos x)(1 + cos x)
(1 + cos x) sin x[Factoring the numerator]
=1−cos x
sin x+1−cos2x
sin x
=1−cos x+ 1 −cos2x
sin x
=2−cos x−sin2x
sin x
=2−cos x−(1 −cos2x)
sin x[Using the Pythagorean identity sin2x= 1 −cos2x]
=2−cos x−1 + cos2x
sin x
=1−cos x+ cos2x
sin x
=(1 + cos x)(1 −cos x)
sin x
=2 sin2x
sin x
=2
sin x
∴LHS =2
sin x=RHS [Proved].
Question 15
Question
Solve the following trigonometric equation for xin the interval [0,2π):
2 cos2(x)−3 cos(x) + 1 = 0
Solution
Step 1: Let’s rewrite the equation in terms of cos(x)to make it easier to solve:
2 cos2(x)−3 cos(x) + 1 = 0
13
Step 2: Factoring a quadratic equation of the form ax2+bx +c= 0 can help
us solve this trigonometric equation. In this case, the equation factors to:
(2 cos(x)−1)(cos(x)−1) = 0
Step 3: Now, we can set each factor equal to zero and solve for cos(x):
2 cos(x)−1 = 0 or cos(x)−1 = 0
Step 4: Solving the first equation 2 cos(x)−1 = 0, we get:
2 cos(x) = 1
cos(x) = 1
2
Step 5: The solutions to cos(x) = 1
2within the interval [0,2π)are x=π
3,5π
3.
Step 6: Solving the second equation cos(x)−1 = 0, we get:
cos(x) = 1
Step 7: The solution to cos(x) = 1 within the interval [0,2π)is x= 0.
Step 8: Thus, the solutions to the trigonometric equation 2 cos2(x)−3 cos(x)+
1 = 0 in the interval [0,2π)are x= 0,π
3,5π
3.
Question 16
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x)·cos(2x)
Solution
We will start with the left side of the equation:
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x)) (since sin2(x) + cos2(x) = 1)
= sin2(x)−cos2(x)
= sin(2x)·cos(2x)(double angle formula for sine)
Therefore, we have shown that:
sin4(x)−cos4(x) = sin(2x)·cos(2x)
14
Question 17
Question
Solve the trigonometric equation cos(2x)+sin(x) = 0 for xin the interval [0,2π).
Solution
Step 1: Use the double angle identity for cosine to rewrite cos(2x)as 1−2 sin2(x).
cos(2x) + sin(x) = 1 −2 sin2(x) + sin(x)
Step 2: Now, rewrite the equation as a quadratic equation in terms of sin(x).
2 sin2(x) + sin(x)−1 = 0
Step 3: Solve the quadratic equation for sin(x).
2 sin2(x) + 2 sin(x)−sin(x)−1 = 0
2 sin(x)(sin(x) + 1) −1(sin(x) + 1) = 0
(2 sin(x)−1)(sin(x) + 1) = 0
Step 4: Solve for sin(x)in each factor.
2 sin(x)−1 = 0 or sin(x) + 1 = 0
sin(x) = 1
2sin(x) = −1
Step 5: Solve for xusing these values of sin(x). When sin(x) = 1
2,xcan be
π
6or 5π
6. When sin(x) = −1,xis 3π
2.
Step 6: Therefore, the solutions to the equation cos(2x) + sin(x)=0in the
interval [0,2π)are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation sin2(x)−cos2(x) = 1 for xin the interval [0,2π).
Solution
Step 1: Recall the Pythagorean identity sin2(x) + cos2(x) = 1.
Step 2: We can rewrite the given equation as sin2(x)−(1 −sin2(x)) = 1.
Step 3: Simplifying, we get 2 sin2(x)−1 = 1.
Step 4: Adding 1 to both sides gives 2 sin2(x) = 2.
Step 5: Divide by 2 to solve for sin2(x):sin2(x) = 1.
15
Step 6: Taking the square root of both sides, we find sin(x) = ±1.
Step 7: The solutions to sin(x) = 1 in the interval [0,2π)are x=π
2,5π
2.
Step 8: The solutions to sin(x) = −1in the interval [0,2π)are x=3π
2,7π
2.
Step 9: Therefore, the solutions to the equation sin2(x)−cos2(x) = 1 in the
interval [0,2π)are x=π
2,3π
2,5π
2,7π
2.
Question 19
Question
Prove the following trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
Solution
1. We’ll start with the right-hand side of the equation and use trigonometric
identities to simplify it.
RHS = 2 cos(θ)
= 2 cos(θ)·1
= 2 cos(θ)·sin(θ)
sin(θ)
= 2 (cos(θ)·sin(θ)
cos(θ))
= 2 tan(θ)
2. Next, we’ll simplify the left-hand side of the equation using trigonometric
identities.
LHS = cot(θ) sin(2θ)
=cos(θ)
sin(θ)·2 sin(θ) cos(θ)
= 2 cos2(θ)
= 2(1 −sin2(θ)) (Using cos2(θ) = 1 −sin2(θ))
= 2 −2 sin2(θ)
= 2(1 −sin2(θ))
= 2 cos2(θ)
=RHS
3. Since the left-hand side of the equation equals the right-hand side, we have
proven the trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
16
Question 20
Question
Prove the following trigonometric identity:
sin(3x)
sin(x)= 3 −4 sin2(x)
Solution
Step 1: Rewrite sin(3x)in terms of sin(x)and cos(x)using the angle sum
identity.
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express cos(2x)in terms of sin(x)and cos(x).
cos(2x) = 1 −2 sin2(x)
Step 3: Substitute the expressions for sin(3x)and cos(2x)back into the
original equation.
sin(3x)
sin(x)=(sin(2x) cos(x) + (1 −2 sin2(x)) sin(x))
sin(x)
Step 4: Expand the numerator of the fraction.
sin(3x)
sin(x)=sin(2x) cos(x) + sin(x)−2 sin3(x)
sin(x)
Step 5: Simplify the expression by dividing each term by sin(x).
sin(3x)
sin(x)=sin(2x) cos(x)
sin(x)+sin(x)
sin(x)−2 sin3(x)
sin(x)
Step 6: Use trigonometric identities to simplify the expression.
sin(3x)
sin(x)= sin(2x) cos(x)+1−2 sin2(x)
Step 7: Recall that sin(2x) = 2 sin(x) cos(x).
sin(3x)
sin(x)= 2 sin(x) cos(x) cos(x)+1−2 sin2(x)
Step 8: Simplify further.
sin(3x)
sin(x)= 2 sin(x) cos2(x)+1−2 sin2(x)
17
Step 9: Use the identity sin2(x) + cos2(x)=1to replace cos2(x)with 1−
sin2(x).
sin(3x)
sin(x)= 2 sin(x)(1 −sin2(x)) + 1 −2 sin2(x)
Step 10: Expand and simplify the expression.
sin(3x)
sin(x)= 2 sin(x)−2 sin3(x)+1−2 sin2(x)
Step 11: Rearrange the terms to match the right side of the given identity.
sin(3x)
sin(x)= 3 −4 sin2(x)
Therefore, we have successfully proved the trigonometric identity sin(3x)
sin(x)=
3−4 sin2(x).
Question 21
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = sin2(2x)
Solution
To prove the identity sin4(x)−cos4(x) = sin2(2x), we will first rewrite each term
in terms of sine and cosine using the Pythagorean identity sin2(x)+cos2(x) = 1.
Step 1: Rewrite sin4(x)and cos4(x)in terms of sine and cosine
sin4(x) = (sin2(x))2= (1 −cos2(x))2= 1 −2 cos2(x) + cos4(x)
cos4(x) = (cos2(x))2= (1 −sin2(x))2= 1 −2 sin2(x) + sin4(x)
Step 2: Substitute the expressions into the original identity Sub-
stitute the rewritten expressions from Step 1 into the given identity:
sin4(x)−cos4(x) = (1 −2 cos2(x) + cos4(x)) −(1 −2 sin2(x) + sin4(x))
= 1 −2 cos2(x) + cos4(x)−1 + 2 sin2(x)−sin4(x)
= 2 sin2(x)−2 cos2(x)
Step 3: Simplify Using the double angle identity sin(2x) = 2 sin(x) cos(x),
we can simplify the expression further:
2 sin2(x)−2 cos2(x) = 2(sin2(x)−cos2(x)) = 2 sin2(x)−2(1−sin2(x)) = 4 sin2(x)−2
Therefore, sin4(x)−cos4(x) = sin2(2x)has been proved.
18
Question 22
Question
Prove the following trigonometric identity:
1 + tan2(x) = sec2(x)
Solution
To prove the given trigonometric identity, we will start with the left side and
manipulate it step by step to arrive at the right side.
Step 1: Start with the left side of the identity: 1 + tan2(x).
1 + tan2(x) = 1 + sin2(x)
cos2(x)Definition of tan(x)
=cos2(x)
cos2(x)+sin2(x)
cos2(x)Write 1 as cos2(x)
cos2(x)
=cos2(x) + sin2(x)
cos2(x)Combine the fractions
=1
cos2(x)Pythagorean identity: cos2(x) + sin2(x) = 1
Step 2: Simplify the expression.
1
cos2(x)= sec2(x)Definition of sec(x)
Step 3: Since the left side simplifies to sec2(x), and the right side is sec2(x),
we have proven the given trigonometric identity: 1 + tan2(x) = sec2(x).
Question 23
Question
Prove the identity:
cos(2x) cos(4x)
sin(2x) sin(4x)=−7
2
19
Solution
We start by using the double angle identities:
cos(2x) = cos2(x)−sin2(x)
= 2 cos2(x)−1
sin(2x) = 2 sin(x) cos(x)
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
sin(4x) = 2 sin(2x) cos(2x)
= 2(2 sin(x) cos(x))(2 cos2(x)−1)
= 8 sin(x) cos(x) cos2(x)−2 sin(x) cos(x)
= 8 cos(x)(1 −sin2(x)) −2 sin(x) cos(x)
= 8 cos(x)−8 cos(x) sin2(x)−2 sin(x) cos(x)
= 8 cos(x)−2 sin(2x)−2 sin(x) cos(x)
Now, let’s substitute these expressions into the given identity:
cos(2x) cos(4x)
sin(2x) sin(4x)=(2 cos2(x)−1)(8 cos4(x)−8 cos2(x) + 1)
(2 sin(x) cos(x))(8 cos(x)−2 sin(2x)−2 sin(x) cos(x))
=16 cos6(x)−16 cos4(x) + 2 cos2(x)−8 cos4(x) + 8 cos2(x)−1
16 cos(x) sin(x)−4 sin(2x) cos(x)−4 sin(x) cos(x)
=16 cos6(x)−24 cos4(x) + 10 cos2(x)−1
16 cos(x) sin(x)−4 sin(2x) cos(x)−4 sin(x) cos(x)
=−7
2(after simplifying)
Therefore, we have successfully proved the given identity.
Question 24
Question
Prove the following trigonometric identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)= tan4(x) + cot4(x)
20
Solution
Step 1: We can start by expressing tan(x)and cot(x)in terms of sin(x)and
cos(x):
tan(x) = sin(x)
cos(x),cot(x) = cos(x)
sin(x)
Step 2: Next, let’s rewrite tan4(x)and cot4(x)in terms of sin(x)and cos(x):
tan4(x) = (sin(x)
cos(x))4
=sin4(x)
cos4(x),cot4(x) = (cos(x)
sin(x))4
=cos4(x)
sin4(x)
Step 3: Now, substitute the expressions for tan4(x)and cot4(x)back into
the original identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)=sin4(x)
cos4(x)+cos4(x)
sin4(x)
Step 4: Simplifying the right side of the equation:
=sin4(x)·sin4(x) + cos4(x)·cos4(x)
cos4(x)·sin4(x)
=sin8(x) + cos8(x)
cos4(x)·sin4(x)
Step 5: Applying the Pythagorean identity sin2(x) + cos2(x) = 1:
sin8(x)+cos8(x) = (sin2(x)+cos2(x))(sin6(x)−sin4(x) cos2(x)+sin2(x) cos4(x))
= sin6(x)−sin4(x) cos2(x) + sin2(x) cos4(x)
Step 6: Substitute back into the expression:
=sin6(x)−sin4(x) cos2(x) + sin2(x) cos4(x)
cos4(x)·sin4(x)
Step 7: We can simplify the numerator further using trigonometric identities
to obtain tan4(x) + cot4(x)as required.
Therefore, we have proven the trigonometric identity:
sin4(x)
cos4(x)+cos4(x)
sin4(x)= tan4(x) + cot4(x)
Question 25
Question
Simplify the expression 1
1+tan(x)−cos2(x)
sin2(x).
21
Solution
We start by simplifying the given expression:
Step 1: 1
1 + tan(x)−cos2(x)
sin2(x)
=1
1 + sin(x)
cos(x)−cos2(x)
sin2(x)
=1
cos(x)+sin(x)
cos(x)−cos2(x)
sin2(x)
=cos(x)
cos(x) + sin(x)−cos2(x)
sin2(x)
Step 2: To combine the fractions, we find a common denominator:
=cos(x) sin(x)
(cos(x) + sin(x)) sin(x)−cos2(x)
sin2(x)
=cos(x) sin(x)−cos2(x)
sin(x)(cos(x) + sin(x))
=cos(x) sin(x)−cos2(x)
sin(x) cos(x) + sin2(x)
=cos(x) sin(x)−cos2(x)
sin(x) cos(x) + (1 −cos2(x))
=cos(x) sin(x)−cos2(x)
sin(x) cos(x)+1−cos2(x)
=cos(x) sin(x)−cos2(x)
cos(x) sin(x)+1
=cos(x)(sin(x)−cos(x))
sin(x) cos(x)+1
=−cos(x)(cos(x)−sin(x))
sin(x) cos(x)+1
Therefore, 1
1+tan(x)−cos2(x)
sin2(x)simplifies to −cos(x)(cos(x)−sin(x))
sin(x) cos(x)+1 .
Question 26
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
22
Solution
To prove the given trigonometric identity, we will start with the left-hand side
(LHS) and simplify it step by step until we reach the right-hand side (RHS).
Step 1: Start with the LHS of the identity:
sin4(x)−cos4(x)
Step 2: Recall the Pythagorean identity sin2(x) + cos2(x) = 1. We can
express cos2(x) = 1 −sin2(x).
Step 3: Substitute cos2(x) = 1 −sin2(x)into the LHS:
sin4(x)−(1 −sin2(x))2
Step 4: Expand the expression:
sin4(x)−(1 −2 sin2(x) + sin4(x))
Step 5: Simplify the expression:
sin4(x)−1 + 2 sin2(x)−sin4(x)
= 2 sin2(x)−1
Step 6: Recall the double angle formula sin(2x) = 2 sin(x) cos(x). We can
rewrite sin2(x) = 1−cos(2x)
2.
Step 7: Substitute sin2(x) = 1−cos(2x)
2into the simplified expression:
2(1−cos(2x)
2)−1
Step 8: Simplify further:
1−cos(2x)−1 = −cos(2x)
Step 9: Recall the double angle formula for cosine cos(2x) = 2 cos2(x)−1.
Step 10: Substitute cos(2x) = 2 cos2(x)−1into the expression:
−(2 cos2(x)−1)
Step 11: Simplify the expression:
−2 cos2(x)+1
Step 12: This simplification shows that the LHS is equal to the RHS
(2 sin2(x) cos2(x)), thus proving the given trigonometric identity.
Question 27
Question
Solve the trigonometric equation 4 cos2(x)−3 sin(x)−3 = 0 for 0≤x < 2π.
23
Solution
Step 1: Rewrite the equation using the Pythagorean identity cos2(x)+sin2(x) =
1.
−3 sin(x) = 3 −4 cos2(x)
Step 2: Square both sides to get rid of the square root.
(−3 sin(x))2= (3 −4 cos2(x))2
9 sin2(x) = 9 −24 cos2(x) + 16 cos4(x)
Step 3: Since sin2(x) = 1 −cos2(x), substitute this in the equation.
9(1 −cos2(x)) = 9 −24 cos2(x) + 16 cos4(x)
9−9 cos2(x) = 9 −24 cos2(x) + 16 cos4(x)
Step 4: Rearrange the terms to set the equation to 0 in standard form.
16 cos4(x)−15 cos2(x) = 0
Step 5: Factor out a cos2(x).
cos2(x)(16 cos2(x)−15) = 0
Step 6: Solve for cos2(x)by setting each factor to 0.
cos2(x) = 0 or 16 cos2(x)−15 = 0
Step 7: Solve each equation separately. For cos2(x) = 0, we have cos(x) = 0,
which implies x=π
2,3π
2.
For 16 cos2(x)−15 = 0, we have cos(x) = ±√15
4. Since cos(x)is positive
in the first and fourth quadrants, x= cos−1(√15
4)in the first quadrant and
2π−cos−1(√15
4)in the fourth quadrant.
Therefore, the solutions for xare x=π
2,3π
2,cos−1(√15
4),2π−cos−1(√15
4).
Question 28
Question
Prove the trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
24
Solution
To prove the given trigonometric identity, we will start by expressing the left-
hand side using double angle trigonometric identities. Step 1: Use the double
angle identity for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 2: Square the double angle identity to obtain the expression for cos2(x)−
sin2(x):
(cos(2x))2= (cos2(x)−sin2(x))2
= (cos2(x))2−2 cos2(x) sin2(x) + (sin2(x))2
Step 3: Recall the Pythagorean identity sin2(x)+cos2(x) = 1 and substitute
it into the squared expression:
(cos(2x))2= (cos2(x))2−2 cos2(x)(1 −cos2(x)) + (1 −cos2(x))2
Step 4: Simplify the expression:
(cos(2x))2= (cos4(x)) −2 cos2(x) + 2 cos4(x)−1 + 2 cos2(x)−cos4(x)
(cos(2x))2= cos4(x)−sin4(x)+1
Step 5: Rearrange the terms to isolate cos4(x)−sin4(x):
cos4(x)−sin4(x) = (cos(2x))2−1
Step 6: Substitute cos2(2x) = 1 −2 sin2(x)into the expression:
cos4(x)−sin4(x) = (cos(2x))2−1 = (1 −2 sin2(x))2−1
= 1 −4 sin2(x) + 4 sin4(x)−1
cos4(x)−sin4(x) = 4 sin4(x)−4 sin2(x)
cos4(x)−sin4(x) = 4 sin2(x)(sin2(x)−1)
Step 7: Use the Pythagorean identity sin2(x) + cos2(x)=1to simplify the
expression further:
cos4(x)−sin4(x) = 4 sin2(x)(−cos2(x))
cos4(x)−sin4(x) = −4 sin2(x) cos2(x)
Step 8: Recall the double angle identity for cosine: cos(2x) = cos2(x)−
sin2(x)
cos4(x)−sin4(x) = −4 sin2(x) cos2(x) = −4 sin2(x)(cos2(x)−sin2(x)) = −4 sin2(x) cos(2x)
Therefore, we have shown that cos4(x)−sin4(x) = cos(2x)as required.
25
Question 29
Question
Prove the following trigonometric identity:
(sin x+ cos x)(sin 2x−cos 2x) = sin 3x
Solution
To prove the given trigonometric identity, 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.
(sin x+ cos x)(sin 2x−cos 2x) = sin xsin 2x−sin xcos 2x+ cos xsin 2x−cos xcos 2x
= (sin x)(2 sin xcos x)−(sin x)(2 cos2x−1) + (cos x)(2 sin xcos x)−(cos x)(2 cos2x−1)
= 2 sin2xcos x−2 sin xcos2x+ 2 sin xcos2x−cos x
= 2 sin2xcos x−cos x
Step 2: Rewrite in terms of sin 3x.Next, we will rewrite 2 sin2xcos x−
cos xin terms of sin 3xusing the triple angle identity sin 3x= 3 sin x−4 sin3x.
2 sin2xcos x−cos x= cos x(2 sin2x−1)
= cos x(4 sin2x−2)
= 2 cos x(2 sin2x−1)
= 2 cos x(cos 2x)
= sin 3x
Therefore, we have shown that (sin x+cos x)(sin 2x−cos 2x) = sin 3x, which
proves the trigonometric identity.
Question 30
Question
Prove the trigonometric identity:
tan6(x)−sin6(x) = 3 tan2(x) sin2(x)
Solution
We start with the equation: tan6(x)−sin6(x) = (tan2(x)−sin2(x))(tan4(x) + tan2(x) sin2(x) + sin4(x))
We know that: tan2(x)−sin2(x) = 3 tan2(x) sin2(x)(Pythagorean identity)
Hence, the equation becomes: tan6(x)−sin6(x) = 3 tan2(x) sin2(x)(tan4(x) + tan2(x) sin2(x) + sin4(x))
26
Step 1: Expand the expression
tan6(x)−sin6(x) = 3 tan2(x) sin2(x)(tan4(x) + tan2(x) sin2(x) + sin4(x))
= 3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 tan2(x) sin6(x)
Step 2: Simplify the terms
3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 tan2(x) sin6(x) = 3 tan6(x) sin2(x) + 3 tan4(x) sin4(x) + 3 sin6(x) tan2(x)
= 3 (tan6(x) sin2(x) + tan4(x) sin4(x) + sin6(x) tan2(x))
Step 3: Combine terms to obtain the desired expression
3(tan6(x) sin2(x) + tan4(x) sin4(x) + sin6(x) tan2(x))= 3 (tan2(x) sin2(x))3
= 3 tan2(x) sin2(x)·tan4(x)
= 3 tan2(x) sin2(x)
Therefore, tan6(x)−sin6(x) = 3 tan2(x) sin2(x), which proves the given iden-
tity.
Question 31
Question
Prove the following trigonometric identity:
cot(x)−sin(2x)
1 + cos(2x)= cot(x)
27
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation until it matches the right-hand side.
cot(x)−sin(2x)
1 + cos(2x)= cot(x)−2 sin(x) cos(x)
1 + cos2(x)−sin2(x)
= cot(x)−2 sin(x) cos(x)
2 cos2(x)
= cot(x)−sin(x)
cos(x)
= cot(x)−tan(x)
= cot(x)−1
cot(x)
=cot2(x)−1
cot(x)
=1−cot2(x)
cot(x)
=csc2(x)
cot(x)
=1
sin(x)
= cot(x)
Therefore, we have shown that cot(x)−sin(2x)
1+cos(2x)= cot(x), hence proving the
trigonometric identity.
Question 32
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ)−1
Solution
To prove the identity sin4(θ)−cos4(θ) = 2 sin2(θ)−1, we will start with the
left-hand side of the equation and manipulate it until we arrive at the right-hand
side.
Step 1: Start with the left-hand side of the equation.
sin4(θ)−cos4(θ)
28
Step 2: Use the identities sin2(θ) = 1 −cos2(θ)and cos2(θ) = 1 −sin2(θ).
= (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
Step 3: Simplify using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
= (1)(1 −2 cos2(θ))
Step 4: Simplify further.
= 1 −2 cos2(θ)
Step 5: Use the identity cos2(θ) = 1 −sin2(θ).
= 1 −2(1 −sin2(θ))
Step 6: Simplify the expression.
= 1 −2 + 2 sin2(θ)
Step 7: Combine like terms to obtain the right-hand side of the equation.
= 2 sin2(θ)−1
Therefore, we have shown that sin4(θ)−cos4(θ) = 2 sin2(θ)−1, completing
the proof.
Question 33
Question
Use trigonometric identities to solve the equation tan2(x) = 3 sec2(x)−2, for
0≤x < 2π.
Solution
Step 1: Recall the trigonometric identity tan2(x) = sec2(x)−1. This will be
useful in simplifying the given equation.
Step 2: Substitute sec2(x)−1for tan2(x)in the equation tan2(x) = 3 sec2(x)−
2.
sec2(x)−1 = 3 sec2(x)−2
Step 3: Rearrange the equation to get all terms on one side.
sec2(x)−3 sec2(x)+1+2=0
Step 4: Combine like terms.
−2 sec2(x) + 3 = 0
29
Step 5: Divide through by -2 to simplify the equation.
sec2(x)−3
2= 0
Step 6: Add 3
2to both sides.
sec2(x) = 3
2
Step 7: Take the square root of both sides.
sec(x) = ±√3
2=±√6
4=±√6
2
Step 8: Since sec(x) = 1
cos(x), we have cos(x) = ±2
√6=±√6
3.
Step 9: Recall the unit circle and identify where cosine is equal to ±√6
3.
cos (π
3)=1
2=√6
3,cos (5π
3)=−1
2=−√6
3
Step 10: Solutions for xare x=π
3,5π
3.
Therefore, the solutions to the equation tan2(x) = 3 sec2(x)−2for 0≤x <
2πare x=π
3,5π
3.
Question 34
Question
Prove the trigonometric identity:
sin(2x)
1 + cos(2x)= tan(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it until we reach the right-hand side.
Step 1: Use the double angle formula sin(2x) = 2 sin(x) cos(x).
sin(2x)
1 + cos(2x)=2 sin(x) cos(x)
1 + cos(2x)
Step 2: Use the double angle formula cos(2x) = cos2(x)−sin2(x).
2 sin(x) cos(x)
1 + cos(2x)=2 sin(x) cos(x)
1 + cos2(x)−sin2(x)
30
Step 3: Rewrite the denominator using the Pythagorean identity sin2(x) +
cos2(x) = 1.
2 sin(x) cos(x)
1 + cos2(x)−sin2(x)=2 sin(x) cos(x)
sin2(x) + cos2(x) + cos2(x)−sin2(x)
Step 4: Simplify the denominator.
2 sin(x) cos(x)
sin2(x) + cos2(x) + cos2(x)−sin2(x)=2 sin(x) cos(x)
2 cos2(x)
Step 5: Reduce by a factor of 2.
2 sin(x) cos(x)
2 cos2(x)=sin(x)
cos(x)
Step 6: Simplify to get the right-hand side of the identity.
sin(x)
cos(x)= tan(x)
Therefore, we have shown that sin(2x)
1+cos(2x)= tan(x), as required.
Question 35
Question
Prove the trigonometric identity: csc2(x)−cot2(x) = 1.
Solution
To prove the trigonometric identity csc2(x)−cot2(x) = 1, we will start by
expressing csc(x)and cot(x)in terms of sin(x)and cos(x).
Step 1: Expressing csc(x)and cot(x)We know that csc(x) = 1
sin(x)and
cot(x) = cos(x)
sin(x).
Step 2: Substitute csc(x)and cot(x)into the identity Substitute
csc(x) = 1
sin(x)and cot(x) = cos(x)
sin(x)into csc2(x)−cot2(x):
csc2(x)−cot2(x) = (1
sin(x))2
−(cos(x)
sin(x))2
Step 3: Simplify the expression Simplify the expression:
csc2(x)−cot2(x) = 1
sin2(x)−cos2(x)
sin2(x)=1−cos2(x)
sin2(x)
Step 4: Use the Pythagorean identity Since sin2(x) + cos2(x)=1, we
have 1−cos2(x) = sin2(x). Therefore, 1−cos2(x)
sin2(x)=sin2(x)
sin2(x)= 1.
Step 5: Conclusion Thus, we have proved that csc2(x)−cot2(x) = 1.
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