MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 10
Liberty University
Question 1
Question
Prove the trigonometric identity:
sec(x)−cos(x) = sin(x)
cos(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it until we arrive at the right-hand side.
Step 1: Begin with the left-hand side of the identity.
sec(x)−cos(x)
Step 2: Express sec(x)in terms of cos(x).
sec(x) = 1
cos(x)
Step 3: Substitute 1
cos(x)for sec(x)in the expression.
1
cos(x)−cos(x)
Step 4: Find a common denominator for the terms.
1−cos2(x)
cos(x)
Step 5: Recall the Pythagorean identity: sin2(x) + cos2(x) = 1.
1−cos2(x) = sin2(x)
Step 6: Substitute sin2(x)for 1−cos2(x)in the expression.
sin2(x)
cos(x)
Step 7: Simplify the expression.
sin(x)
cos(x)
Therefore, we have shown that sec(x)−cos(x) = sin(x)
cos(x), completing the proof
of the trigonometric identity.
Question 2
Question
Solve the equation tan2(x) + sec(x) tan(x) = 0 for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation in terms of sine and cosine:
tan2(x) + sec(x) tan(x) = 0
sin2(x)
cos2(x)+1
cos(x)·sin(x)
cos(x)= 0
Step 2: Find a common denominator and combine the fractions:
sin2(x) + sin(x)
cos2(x) cos(x)= 0
Step 3: Since the numerator must equal 0, we have:
sin2(x) + sin(x) = 0
Step 4: Factor out sin(x)from the equation:
sin(x)(sin(x) + 1) = 0
Step 5: Set each factor equal to 0 and solve for x: For sin(x) = 0:
x= 0, π
2
For sin(x) + 1 = 0:
sin(x) = −1
x=3π
2
Step 6: Check the solutions in the original equation to ensure they are valid
in the given interval: Checking x= 0:
tan2(0) + sec(0) tan(0) = 0
0 + (1)(0) = 0
The solution x= 0 is valid.
Checking x=π:
tan2(π) + sec(π) tan(π) = 0
0+(−1)(0) = 0
The solution x=πis valid.
Checking x=3π
2:
tan2(3π
2)+ sec (3π
2)tan (3π
2)= 0
1 + (0)(−1) = 1
The solution x=3π
2is not valid in the given interval.
Therefore, the solutions to the equation are x= 0 and x=π.
Question 3
Question
Prove the following trigonometric identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Solution
To prove the trigonometric identity sin(3θ) = 3 sin(θ)−4 sin3(θ), we will use
the angle addition formula for sine:
sin(A+B) = sin Acos B+ cos Asin B
Step 1: Start by expanding sin(3θ)using the angle addition formula twice.
sin(3θ) = sin(2θ+θ)
Since sin(2θ) = 2 sin(θ) cos(θ), we have:
sin(3θ) = sin(2θ) cos(θ) + cos(2θ) sin(θ)
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Step 2: To simplify the expression, we need to compute cos(2θ). Using the
double-angle formula cos(2θ) = 2 cos2(θ)−1, we get:
sin(3θ) = (2 sin(θ) cos(θ)) cos(θ) + (2 cos2(θ)−1) sin(θ)
Step 3: Simplify the expression further by substituting sin2(θ) = 1−cos2(θ).
sin(3θ) = 2 sin(θ) cos2(θ) + 2 cos2(θ) sin(θ)−sin(θ)
sin(3θ) = 2 sin(θ)(1 −sin2(θ)) + 2(1 −sin2(θ)) sin(θ)−sin(θ)
Step 4: Simplify the expression and factor out sin(θ).
sin(3θ) = 2 sin(θ)−2 sin3(θ) + 2 sin(θ)−2 sin3(θ)−sin(θ)
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Therefore, we have successfully proved the trigonometric identity sin(3θ) =
3 sin(θ)−4 sin3(θ).
Question 4
Question
Prove the trigonometric identity: sin4(x)−cos4(x) = 1 −2 cos2(2x).
Solution
Step 1: Start with the left-hand side of the given identity and simplify.
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
= (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
Step 2: Use the double angle formula for cosine (cos(2θ) = 2 cos2(θ)−1) to
simplify sin2(x)−cos2(x).
sin2(x)−cos2(x) = sin2(x)−(1 −sin2(x))
= 2 sin2(x)−1
=−2 cos2(−x)
Step 3: Rewrite the expression −2 cos2(−x)in terms of 2x.
−2 cos2(−x) = −2 cos2(−(2x)) = −2 cos2(2x)
Step 4: Therefore, we have shown that sin4(x)−cos4(x) = −2 cos2(2x).
But the given identity is 1−2 cos2(2x). To make them match, simply multiply
by −1on both sides:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
So, the identity is proved.
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Question 5
Question
Solve the following trigonometric equation for xin the interval [0,2π]:sin(x) +
√3 cos(x) = 1.
Solution
Step 1: Rewrite the equation in terms of a single trigonometric function. Step
2: Square both sides of the equation. Step 3: Use a trigonometric identity
to express the squared trigonometric function in terms of one trigonometric
function. Step 4: Solve the resulting trigonometric equation. Step 5: Check the
solutions in the original equation for validity.
Question 6
Question
Solve the equation √3 cos(x) + sin(x) = 0 for xin the interval [0,2π).
Solution
Step 1: We can rewrite the given equation as sin (x+π
6)= 0, where we used
the trigonometric identity sin(a) cos(b) + cos(a) sin(b) = sin(a+b).
Step 2: We can conclude that x+π
6must be equal to kπ, where kis an
integer.
Step 3: Solving for xin the interval [0,2π), we have
x+π
6= 0
or
x+π
6=π
or
x+π
6= 2π.
Step 4: Solving these equations, we get
x=−π
6
or
x=5π
6
or
x=11π
6.
Thus, the solutions to the equation √3 cos(x) + sin(x)=0in the interval
[0,2π)are −π
6,5π
6, and 11π
6.
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Question 7
Question
Prove the trigonometric identity sin4(x)−cos4(x) = 2 sin2(x) cos2(x).
Solution
Step 1: Start with the left side of the equation and use the Pythagorean identity
sin2(x) + cos2(x) = 1.
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (1)(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x))
Step 2: Expand the expression (sin(x) + cos(x))(sin(x)−cos(x)) using the
distributive property.
(sin(x) + cos(x))(sin(x)−cos(x)) = sin2(x)−cos2(x) + sin(x) cos(x)−sin(x) cos(x)
= sin2(x)−cos2(x)
= 2 sin2(x) cos2(x)
Therefore, sin4(x)−cos4(x) = 2 sin2(x) cos2(x), as required.
Question 8
Question
Prove the following trigonometric identity:
cos2(x)
sin(x) cos(x)=1
sin(x)−1
cos(x)
6
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
until it is equivalent to the right-hand side.
LHS =cos2(x)
sin(x) cos(x)
=cos(x) cos(x)
sin(x) cos(x)
=cos(x)
sin(x)
=1
sin(x)·1
cos(x)
=1
sin(x)·1
1
cos(x)
=1
sin(x)·cos(x)
=cos(x)
sin(x)
=1
sin(x)·1
cos(x)
=1
sin(x)−1
cos(x)
=RHS
Therefore, the given trigonometric identity is true.
Question 9
Question
Prove the trigonometric identity
tan(θ) + sin(θ)
cos(θ)= sec(θ) + tan(θ)
Solution
Step 1: Recall that sec(θ) = 1
cos(θ)and tan(θ) = sin(θ)
cos(θ).
Step 2: Start by rewriting the left-hand side of the equation using the
definitions of sec(θ)and tan(θ):
tan(θ) + sin(θ)
cos(θ)=
sin(θ)
cos(θ)+ sin(θ)
cos(θ)
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Step 3: Combine the terms in the numerator:
sin(θ)
cos(θ)+ sin(θ)
cos(θ)=
sin(θ)+sin(θ) cos(θ)
cos(θ)
cos(θ)
Step 4: Factor out a sin(θ)from the numerator:
sin(θ)(1+cos(θ))
cos(θ)
cos(θ)
Step 5: Simplify the expression by canceling out cos(θ)in the numerator
and denominator: sin(θ)(1 + cos(θ))
cos2(θ)
Step 6: Recall that 1 + cos2(θ) = sec2(θ). Rewrite the simplified expression
in terms of sec(θ):
sin(θ) sec(θ)
cos(θ)
Step 7: Simplify further to obtain the right-hand side of the equation:
sin(θ)
cos(θ)·sec(θ) = tan(θ)·sec(θ) = sec(θ) + tan(θ)
Therefore, the trigonometric identity is proved.
Question 10
Question
Solve the trigonometric equation sin4(x)−sin2(x) = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the equation sin4(x)−sin2(x) = 0 as sin2(x)(sin2(x)−1) =
0.
Step 2: Therefore, we have two cases:
• Case 1: sin2(x) = 0
In this case, sin(x) = 0. The solutions to sin(x) = 0 in the interval [0,2π)
are x= 0, π.
• Case 2: sin2(x)−1 = 0
Solving sin2(x)−1 = 0, we get sin2(x) = 1, which implies sin(x) = ±1.
The solutions to sin(x) = 1 in the interval [0,2π)are x=π
2,5π
2, and the
solutions to sin(x) = −1in the interval [0,2π)is x=3π
2.
Step 3: Therefore, the solutions to the trigonometric equation sin4(x)−
sin2(x) = 0 in the interval [0,2π)are x= 0,π
2,3π
2, π, 5π
2.
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Question 11
Question
Prove the following trigonometric identity:
sin3x+ cos3x
sin x+ cos x= 1 −sin xcos x
Solution
Step 1: We will start by simplifying the left-hand side of the equation. Step 2:
Note that sin3x= sin x(sin2x)and cos3x= cos x(cos2x). Step 3: Substituting
these expressions into the numerator, we get:
sin x(sin2x) + cos x(cos2x)
Step 4: Now, we’ll factor out a sin xfrom the first term and a cos xfrom the
second term:
sin x(sin x2+ cos x2)
Step 5: Since sin2x+ cos2x= 1, the expression simplifies to sin x. Step 6:
Therefore, the numerator simplifies to sin x. Step 7: Simplifying the denomi-
nator sin x+ cos xstays as it is. Step 8: Substituting the simplified numerator
and denominator back into the original expression, we get:
sin x
sin x+ cos x
Step 9: Now, simplify the expression sin x
sin x+cos xto obtain 1−sin xcos x. Step 10:
The left-hand side is now simplified to the right-hand side, so the trigonometric
identity is proved.
Question 12
Question
Prove the trigonometric identity:
tan(x)−sin(x)
cos(x)= sec(x)−cos(x)
9
Solution
Step 1: Start with the left side of the equation and simplify:
tan(x)−sin(x)
cos(x)=
sin(x)
cos(x)−sin(x)
cos(x)
=sin(x)−sin(x) cos(x)
cos(x)
=sin(x)(1 −cos(x))
cos(x)
=sin(x) sin(x)
cos(x)
=sin2(x)
cos(x)
=1−cos2(x)
cos(x)[Using the Pythagorean identity: sin2(x) = 1 −cos2(x)]
=1
cos(x)−cos2(x)
cos(x)
= sec(x)−cos(x)
Therefore, the given trigonometric identity is proved.
Question 14
Question
Prove the following trigonometric identity:
cos4θ
1−sin θ+1
cos4θ=1 + sin2θ
cos2θ
Solution
We will start by simplifying each side of the equation separately.
10
Step 1: Simplify the left-hand side.
cos4θ
1−sin θ+1
cos4θ=cos4θ+ (1 −sin θ)
1−sin θ
=cos4θ+ 1 −sin θ
1−sin θ
=cos4θ+ cos2θ−cos2θ−sin θ+ 1
1−sin θ
=cos2θ(cos2θ+ 1) −cos2θ−sin θ+ 1
1−sin θ
=cos2θ(sin2θ+ 1) −cos2θ−sin θ+ 1
1−sin θ
=cos2θsin2θ+ cos2θ−cos2θ−sin θ+ 1
1−sin θ
=cos2θsin2θ−sin θ+ 1
1−sin θ
Step 2: Simplify the right-hand side.
1 + sin2θ
cos2θ=1 + sin2θ
cos2θ·1
1
=sin2θ+ 1
cos2θ
=cos2θsin2θ+ cos2θ
cos2θ
=cos2θ(sin2θ+ 1)
cos2θ
= sin2θ+ 1
= 1 + sin2θ
Step 3: Conclude. Since both sides simplify to 1 + sin2θ, we have shown
that cos4θ
1−sin θ+1
cos4θ=1 + sin2θ
cos2θ
and the trigonometric identity is proven.
Question 15
Question
Solve the trigonometric equation sin(3x) = cos(x)for xin the interval [0,2π).
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Solution
Step 1: Use the angle addition formula for sine and cosine to rewrite the equa-
tion.
sin(3x) = cos(x)
sin(2x+x) = cos(x)
Step 2: Apply the angle addition formulas:
sin(2x) cos(x) + cos(2x) sin(x) = cos(x)
Step 3: Replace sin(2x)and cos(2x)with the double angle formulas:
2 sin(x) cos(x) cos(x) + (2 cos2(x)−1) sin(x) = cos(x)
Step 4: Simplify the equation:
2 sin(x) cos2(x) + 2 cos2(x) sin(x)−sin(x) = cos(x)
2 sin(x) cos2(x)−sin(x) = cos(x)−2 cos2(x) sin(x)
sin(x)(2 cos2(x)−1) = cos(x)(1 −2 sin2(x))
Step 5: Work with the Pythagorean identities:
sin(x)(2 −sin2(x)) = cos(x)(1 −2 sin2(x))
(2 sin(x)−sin3(x)) = cos(x)−2 cos(x) sin2(x)
Step 6: Use the Pythagorean identity to substitute cos(x)and sin2(x):
2 sin(x)−sin3(x) = √1−sin2(x)−2√1−sin2(x) sin2(x)
Step 7: Let u= sin(x)and solve the resulting cubic equation:
2u−u3=√1−u2−2√1−u2u2
u3−2u2−2u+ 1 = 0
(u−1)(u2−u−1) = 0
Step 8: Solve for u: For u= 1,sin(x) = 1 =⇒x=π
2or 3π
2.
For u2−u−1 = 0,sin(x) = 1±√5
2. However, 1+√5
2is outside the interval [0,2π),
so we only consider sin(x) = 1−√5
2. This gives x= arcsin (1−√5
2).
Therefore, the solutions to the equation sin(3x) = cos(x)in the interval
[0,2π)are x=π
2,3π
2, and arcsin (1−√5
2).
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Question 16
Question
Prove the identity:
sin4(x)−cos4(x)
sin2(x) + cos2(x)= sin(2x)
Solution
Step 1: Simplify the left-hand side of the equation using the difference of squares:
sin4(x)−cos4(x)
sin2(x) + cos2(x)=(sin2(x)−cos2(x))(sin2(x) + cos2(x))
sin2(x) + cos2(x)
Step 2: Notice that sin2(x)−cos2(x) = −cos(2x)and simplify further:
(sin2(x)−cos2(x))(sin2(x) + cos2(x))
sin2(x) + cos2(x)=−cos(2x)·1
1=−cos(2x)
Step 3: Recall that the double angle formula for sin(2x)is:
sin(2x) = 2 sin(x) cos(x)
Step 4: Use the identity cos(2x) = 1−2 sin2(x)to rewrite −cos(2x)in terms
of sin(x)and cos(x):
−cos(2x) = −1 + 2 sin2(x)
Step 5: Recognize that 2 sin2(x)=1−cos(2x)and replace 2 sin2(x)with
1−cos(2x)in the previous expression:
−cos(2x) = −1+1−cos(2x) = −cos(2x)
Step 6: Hence, sin4(x)−cos4(x)
sin2(x)+cos2(x)= sin(2x)is proven.
Question 17
Question
Prove the identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
formula for sine: sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
13
Step 1: Express sin(3x)in terms of sin(x)Using the angle addition
formula for sine twice, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express sin(2x)in terms of sin(x)We can express sin(2x)using
the double angle formula: sin(2x) = 2 sin(x) cos(x). Substitute this into the
expression for sin(3x):
sin(3x) = (2 sin(x) cos(x)) cos(x) + cos(2x) sin(x)
= 2 sin(x) cos2(x) + cos(2x) sin(x)
Step 3: Express cos(2x)in terms of sin(x)and cos(x)Using the double
angle formula for cosine, cos(2x) = cos2(x)−sin2(x). Substitute this into the
expression for sin(3x):
sin(3x) = 2 sin(x) cos2(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x)(1 −sin2(x)) + (cos2(x)−sin2(x)) sin(x)
Step 4: Simplify the expression Now, simplify the expression:
sin(3x) = 2 sin(x)−2 sin3(x) + cos2(x) sin(x)−sin3(x) sin(x)
= 3 sin(x)−3 sin3(x) = 3 sin(x)−3 sin(x)3
= 3 sin(x)−4 sin3(x)
Therefore, we have proven the identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 18
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the angle addition formula for sine and some algebraic manipulation.
Step 1: Apply the angle addition formula for sine.
sin(3x) = sin(2x+x)
Step 2: Expand using the angle addition formula: sin(A+B) = sin(A) cos(B)+
cos(A) sin(B).
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
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Step 3: Express sin(2x)and cos(2x)in terms of sin(x)and cos(x).
sin(3x) = (2 sin(x) cos(x))(cos(x)) + (cos2(x)−sin2(x))(sin(x))
Step 4: Simplify the expression.
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
sin(3x) = 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, we have successfully proven the trigonometric identity sin(3x) =
3 sin(x)−4 sin3(x).
Question 19
Question
Prove the trigonometric identity:
sin(α+β) tan(α−β) = tan(α)−tan(β)
Solution
To prove the given identity, we will start with the left-hand side and manipulate
it to match the right-hand side.
Step 1: Expand sin(α+β)using the angle sum formula:
sin(α+β) = sin(α) cos(β) + cos(α) sin(β)
Step 2: Expand tan(α−β)using the tangent subtraction formula:
tan(α−β) = tan(α)−tan(β)
1 + tan(α) tan(β)
Step 3: Substitute the expansions from Step 1 and Step 2 into the left-hand
side of the identity:
sin(α+β) tan(α−β) = (sin(α) cos(β) + cos(α) sin(β)) ·(tan(α)−tan(β)
1 + tan(α) tan(β))
=sin(α) cos(β) tan(α)−sin(α) cos(β) tan(β) + cos(α) sin(β) tan(α)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
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Step 4: Simplify the expression by combining like terms:
sin(α) cos(β) tan(α)−sin(α) cos(β) tan(β) + cos(α) sin(β) tan(α)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
=sin(α) cos(β) tan(α) + cos(α) sin(β) tan(α)−sin(α) cos(β) tan(β)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
=sin(α) tan(α)(cos(β) + cos(β)) −sin(β) tan(β)(cos(α) + cos(α))
1 + tan(α) tan(β)
=sin(α) tan(α)−sin(β) tan(β)
1 + tan(α) tan(β)
= tan(α)−tan(β)
Therefore, we have shown that sin(α+β) tan(α−β) = tan(α)−tan(β), and
the trigonometric identity is proven.
Question 20
Question
Prove the trigonometric identity:
1 + cot2(θ)
1 + tan2(θ)= sec2(θ)
16
Solution
Step 1: Start with the left-hand side of the identity and simplify using the
definitions of cotangent, tangent, and secant:
1 + cot2(θ)
1 + tan2(θ)=1 + cos2(θ)
sin2(θ)
1 + sin2(θ)
cos2(θ)
=
sin2(θ)+cos2(θ)
sin2(θ)
cos2(θ)+sin2(θ)
cos2(θ)
=
1
sin2(θ)
1
cos2(θ)
=cos2(θ)
sin2(θ)
=cos2(θ)
1−cos2(θ)[using sin2(θ) = 1 −cos2(θ)]
=cos2(θ)
sin2(θ)·1
1−cos2(θ)
=cos2(θ)
sin2(θ)·1
sin2(θ)
=cos2(θ)
sin3(θ)
=1
sin(θ)
= sec(θ)
= sec2(θ)
Therefore, we have shown that 1+cot2(θ)
1+tan2(θ)= sec2(θ), completing the proof of
the identity.
Question 21
Question
Prove the following trigonometric identity:
sin4(x) + cos4(x) = 3
2−1
2cos(4x)
Solution
To prove the given trigonometric identity, we will first manipulate both sides
using trigonometric identities and then simplify the expressions until they are
17
equivalent.
Step 1: Start with the left side of the identity.
sin4(x) + cos4(x)
Step 2: Use the Pythagorean identity: sin2(x) + cos2(x) = 1, to express
sin4(x)and cos4(x)in terms of sin2(x)and cos2(x).
= (sin2(x))2+ (cos2(x))2
Step 3: Expand the expressions.
= sin4(x) + 2 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x)
Step 4: Factor the middle terms as (sin2(x) + cos2(x))2.
= (sin2(x) + cos2(x))2−2 sin2(x) cos2(x)
Step 5: Use the Pythagorean identity and simplify.
= 12−2 sin2(x) cos2(x) = 1 −2 sin2(x) cos2(x)
Step 6: Use the double angle identity: sin(2θ) = 2 sin(θ) cos(θ)to express
sin2(x) cos2(x)in terms of cos(2x).
= 1 −cos(2x)
2
Step 7: Multiply by 2 to get rid of the fraction.
= 2 −cos(2x)
Step 8: Substitute cos(2x)with the identity cos(2x) = 2 cos2(x)−1.
= 2 −(2 cos2(x)−1) = 3 −2 cos2(x)
Step 9: Use the Pythagorean identity to express cos2(x)in terms of sin2(x).
= 3 −2(1 −sin2(x)) = 3 −2 + 2 sin2(x) = 1 + 2 sin2(x)
Step 10: Finally, substitute sin2(x)with 1
2−1
2cos(2x).
= 1 + 2 (1
2−1
2cos(2x))= 1 + 1 −cos(2x) = 2 −cos(2x)
Step 11: This matches the right side of the given identity.
=3
2−1
2cos(4x)
Therefore, we have shown that sin4(x)+cos4(x) = 3
2−1
2cos(4x), as required.
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Question 22
Question
Prove the trigonometric identity:
tan4(x)−sin4(x) = 2 tan2(x) sin2(x)
Solution
tan4(x)−sin4(x) = tan4(x)−(sin2(x))2
= tan4(x)−(sin2(x) + cos2(x))2
= tan4(x)−(sin2(x)
cos2(x)+ 1)2
= tan4(x)−sin4(x)
cos4(x)−2sin2(x)
cos2(x)−1
= tan4(x)−sin4(x)
cos4(x)−2 tan2(x)−1
=sin4(x)
cos4(x)−sin4(x)
cos4(x)−2 tan2(x)−1
= 2 tan2(x)−1
= 2 tan2(x)(1 −sin2(x))
= 2 tan2(x) sin2(x)
Therefore, the trigonometric identity tan4(x)−sin4(x) = 2 tan2(x) sin2(x)is
proven.
Question 23
Question
Solve the trigonometric equation: sin2(x)−cos(x)−1 = 0, for 0≤x≤2π.
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1:
sin2(x)−(1 −sin2(x)) −1 = 0
Step 2: Simplify the equation:
2 sin2(x)−2 = 0
19
Step 3: Divide by 2 to isolate sin2(x):
sin2(x)−1 = 0
Step 4: Solve for sin(x)by taking the square root of both sides:
sin(x) = ±1
Step 5: Find the solutions for xwithin the interval 0≤x≤2π.
When sin(x) = 1, we have x=π
2since sin(π
2)= 1.
When sin(x) = −1, we have x=3π
2since sin(3π
2)=−1.
Therefore, the solutions to the equation sin2(x)−cos(x)−1 = 0 for 0≤x≤
2πare x=π
2and x=3π
2.
Question 24
Question
Solve the trigonometric equation 2 sin2(x)−3 cos(x) = 2 for 0◦≤x≤360◦.
Solution
Step 1: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to express sin2(x)
in terms of cos(x). Step 2: Substitute the expression for sin2(x)into the given
equation to obtain a quadratic equation in cos(x). Step 3: Solve the quadratic
equation for cos(x)using factoring or the quadratic formula. Step 4: Find the
corresponding values of xby using the connection between cos(x)and sin(x)in
the first quadrant. Step 5: Check the solutions obtained by substituting them
back into the original equation to ensure they satisfy the equation.
Let’s follow these steps:
Step 1: We know that sin2(x) = 1−cos2(x), so 2(1−cos2(x))−3 cos(x) = 2.
Step 2: Simplifying the equation gives 2−2 cos2(x)−3 cos(x) = 2.
Step 3: Rearranging terms, we get −2 cos2(x)−3 cos(x) = 0.
Factoring out −cos(x), we have −cos(x)(2 cos(x) + 3) = 0.
So, cos(x) = 0 or cos(x) = −3
2.
Since −3
2is not in the range of [−1,1], we discard it.
Hence, cos(x) = 0 which implies x= 90◦or x= 270◦.
Step 4: In terms of sin(x),x= 90◦corresponds to sin(x) = 1 and x= 270◦
corresponds to sin(x) = −1.
Step 5: Checking the solutions:
For x= 90◦:2(sin2(90◦)) −3 cos(90◦) = 2(1) −3(0) = 2. This is true.
For x= 270◦:2(sin2(270◦)) −3 cos(270◦) = 2(1) −3(0) = 2. This is also
true.
Therefore, the solutions to the equation are x= 90◦and x= 270◦for
0◦≤x≤360◦.
20
Question 25
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the trigonometric identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will
first rewrite the left-hand side using the Pythagorean identity:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
Step 1: Apply the Pythagorean identity (sin2(x) + cos2(x) = 1) to simplify
the expression:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))
Step 2: Now, use the double-angle formula sin(2x) = 2 sin(x) cos(x)to
expand the right-hand side of the identity:
sin(2x) sin(4x) = 2 sin(x) cos(x)·sin(4x)
Step 3: Using the double-angle formula for sine sin(2θ) = 2 sin(θ) cos(θ),
the expression becomes:
2 sin(x) cos(x)·sin(4x) = 2 sin(x) cos(x)·(2 sin(2x) cos(2x))
Step 4: Further simplify the expression by expanding sin(2x)and cos(2x):
2 sin(x) cos(x)·(2 sin(2x) cos(2x)) = 4 sin(x) cos(x)·(sin(2x) cos(2x))
Step 5: Apply the double-angle formulas for sine and cosine to rewrite the
expression:
4 sin(x) cos(x)·(sin(2x) cos(2x)) = 4 sin(x) cos(x)·(1
2sin(4x))
= 2 sin(x) cos(x) sin(4x)
Step 6: Finally, simplify the right-hand side:
2 sin(x) cos(x) sin(4x) = sin(2x) sin(4x)
Therefore, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x), and the
trigonometric identity is proved.
Question 26
Question
Solve the equation sin(2x) + √3 cos(x) = 0 for xin the interval [0,2π).
21
Solution
Step 1: Use the double angle identity for sine to rewrite sin(2x):
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the original equation:
2 sin(x) cos(x) + √3 cos(x) = 0
Step 3: Factor out cos(x)from the equation:
cos(x)(2 sin(x) + √3) = 0
Step 4: Set each factor equal to zero and solve for x:
• For cos(x) = 0, we have x=π
2and x=3π
2.
• For 2 sin(x)+√3 = 0, we have sin(x) = −√3
2. This occurs in the third and
fourth quadrants where the reference angle is π
3. So, x=2π
3and x=5π
3.
Step 5: The solutions in the interval [0,2π)are x=π
2,2π
3, and 5π
3.
Question 27
Question
Prove the following trigonometric identity:
sin4(x)
1 + cos(x)+cos4(x)
1 + sin(x)= 1 −sin(x) cos(x)
Solution
Step 1: Begin by expressing sin4(x)and cos4(x)in terms of sin(2x)and cos(2x).
sin4(x) = (1−cos(2x)
2)2
=1−2 cos(2x) + cos2(2x)
4
cos4(x) = (1 + cos(2x)
2)2
=1 + 2 cos(2x) + cos2(2x)
4
Step 2: Substitute these expressions back into the given identity.
1−2 cos(2x)+cos2(2x)
4
1 + cos(x)+
1+2 cos(2x)+cos2(2x)
4
1 + sin(x)= 1 −sin(x) cos(x)
Step 3: Expand and simplify each term in the identity.
1−2 cos(2x) + cos2(2x)
4(1 + cos(x)) +1 + 2 cos(2x) + cos2(2x)
4(1 + sin(x)) = 1 −sin(x) cos(x)
22
Step 4: Clear the denominators by finding a common denominator.
(1 −2 cos(2x) + cos2(2x))(1 + sin(x)) + (1 + 2 cos(2x) + cos2(2x))(1 + cos(x))
4(1 + cos(x))(1 + sin(x)) = 1−sin(x) cos(x)
Step 5: Expand the numerator and simplify the expression.
1−cos2(2x)−2 cos(2x)−2 sin(x) cos(2x) + cos2(2x) + 2 cos2(2x) + 2 cos(2x) + 2 sin(x) cos(2x)
4(1 + cos(x))(1 + sin(x)) = 1−sin(x) cos(x)
1−cos2(2x) + cos2(2x)
4(1 + cos(x))(1 + sin(x)) = 1 −sin(x) cos(x)
1
4(1 + cos(x))(1 + sin(x)) = 1 −sin(x) cos(x)
1
4(1 + cos(x))(1 + sin(x)) =4 sin(x) cos(x)
2
1
(1 + cos(x))(1 + sin(x)) = 2 sin(x) cos(x)
Step 6: We know that the Pythagorean identity sin2(x) + cos2(x)=1.
Let’s express denumerator in terms of sin(x) and cos(x), sin2(x) + cos2(x) +
2 sin(x) cos(x) = 1 + 2 sin(x) cos(x).
1
(1 + cos(x))(1 + sin(x)) = 2 sin(x) cos(x)
Step 7: Finally, substitute back in the Pythagorean identity to get:
1
1 + 2 sin(x) cos(x)= 2 sin(x) cos(x)
1
1 + 2 sin(x) cos(x)= 2 sin(x) cos(x)
Therefore, the given trigonometric identity is proved.
Question 28
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)=1 + cos(x)
sin(x)
23
Solution
To prove the identity, we will start by simplifying each side separately using
trigonometric identities.
Step 1: Simplify the left side We have:
sin2(x)
1−cos(x)=sin(x) sin(x)
1−cos(x)=sin(x)(1 −cos2(x))
1−cos(x)
Using the Pythagorean identity sin2(x) = 1 −cos2(x), we get:
sin(x)(1 −cos2(x))
1−cos(x)=sin(x) sin2(x)
1−cos(x)
Further simplify:
sin(x) sin2(x)
1−cos(x)=sin(x)(1 −cos(x)) sin(x)
1−cos(x)= sin(x) sin(x) = sin2(x)
Step 2: Simplify the right side We have:
1 + cos(x)
sin(x)=1
sin(x)+cos(x)
sin(x)= csc(x) + cot(x)
Step 3: Compare the simplified forms Since sin2(x)= csc(x) + cot(x),
the given identity
sin2(x)
1−cos(x)=1 + cos(x)
sin(x)
is not true for all values of x. Thus, the identity does not hold.
Question 29
Question
Prove the identity sin2(x) cos2(x) = 1
4sin22x.
Solution
To prove the given identity, we will start with the right-hand side and simplify
it step by step to match the left-hand side.
Step 1: Start with the right-hand side of the identity:
1
4sin22x
Step 2: Recall the double angle identity: sin2(2x) = 1−cos(4x)
2.
24
Step 3: Substitute sin2(2x)using the double angle identity:
1
4(1 −cos(4x))
Step 4: Recall the double angle identity: cos(2θ) = 1 −2 sin2(θ).
Step 5: Substitute cos(4x)using the double angle identity:
1
4(1−(1 −2 sin2(2x)))
Step 6: Simplify the expression:
1
4(1−1 + 2 sin2(2x))
1
4(2 sin2(2x))
Step 7: Simplify further:
sin2(2x)
Step 8: Recall the double angle identity: sin(2θ) = 2 sin(θ) cos(θ).
Step 9: Substitute sin(2x)using the double angle identity:
(2 sin(x) cos(x))2
Step 10: Simplify the expression:
(2 sin(x) cos(x))2
(2 sin(x) cos(x))(2 sin(x) cos(x))
4 sin2(x) cos2(x)
Step 11: Therefore, the right-hand side equals the left-hand side of the
identity, so sin2(x) cos2(x) = 1
4sin22x.
Question 30
Question
Prove the following trigonometric identity:
1
sin(x)+1
cos(x)=1
sin(x) cos(x)
25
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until it simplifies to the right-hand side.
Step 1: Rewrite the left-hand side using a common denominator.
1
sin(x)+1
cos(x)=cos(x)
sin(x) cos(x)+sin(x)
sin(x) cos(x)
Step 2: Combine the fractions over a common denominator.
cos(x) + sin(x)
sin(x) cos(x)
Step 3: Use the trigonometric Pythagorean identity sin2(x) + cos2(x)=1
to simplify the numerator.
cos(x) + sin(x) = cos(x)(cos(x) + sin(x)) + sin(x)(cos(x) + sin(x))
= cos2(x) + cos(x) sin(x) + cos(x) sin(x) + sin2(x)
= cos2(x) + 2 cos(x) sin(x) + sin2(x)
= (cos(x) + sin(x))2
Step 4: Substitute the simplified numerator back into the fraction.
(cos(x) + sin(x))2
sin(x) cos(x)
Step 5: Expand the squared binomial in the numerator.
cos2(x) + 2 cos(x) sin(x) + sin2(x)
sin(x) cos(x)
Step 6: Use the Pythagorean identity again to simplify the numerator to 1.
1
sin(x) cos(x)
Since the left-hand side is equal to the right-hand side, the trigonometric
identity has been proven.
Question 31
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
26
Solution
Step 1: Rewrite cos(x)in terms of sin(x)using the Pythagorean identity sin2(x)+
cos2(x) = 1.
cos(x) = √1−sin2(x)
Step 2: Substitute cos(x)in the given equation to get:
sin2(x) + √1−sin2(x) = 1
Step 3: Rearrange the equation and isolate sin(x):
sin2(x) = 1 −√1−sin2(x)
Step 4: Square both sides of the equation to eliminate the square root:
sin4(x) = (1 −√1−sin2(x))2
Step 5: Expand the right side of the equation and simplify:
sin4(x) = 1 −2√1−sin2(x) + (1 −sin2(x))
Step 6: Combine like terms on the right side:
sin4(x) = 2 −2√1−sin2(x)−sin2(x)
Step 7: Let u= sin2(x)for simplicity. The equation becomes:
u2= 2 −2√1−u−u
Step 8: Solve the quadratic equation for u:
u2+u−2+2√1−u= 0
Step 9: Factor the equation:
(u+ 2)(u−1+2√1−u) = 0
Step 10: Solve for uusing each factor separately: 1) u+ 2 = 0 =⇒u=−2
(not in the domain [0,2π)) 2) u−1+2√1−u= 0
Step 11: Solve u−1+2√1−u= 0 for u:
2√1−u= 1 −u
4(1 −u) = (1 −u)2
Step 12: Expand and simplify:
4−4u= 1 −2u+u2
27
Step 13: Rearrange the equation as a quadratic:
u2−2u−3 = 0
Step 14: Factor the equation:
(u−3)(u+ 1) = 0
Step 15: Solve for uusing each factor separately: 1) u−3 = 0 =⇒u= 3
(not in the domain [0,2π)) 2) u+1 = 0 =⇒u=−1(not in the domain [0,2π))
Step 16: Since there are no solutions for uin the given domain, the original
equation sin2(x) + cos(x) = 1 has no solutions in the interval [0,2π).
Question 32
Question
Prove the trigonometric identity:
1
sin(θ) + cos(θ)=√2−1
2·1
sin (θ−π
4)
Solution
Step 1: Rewrite the denominator using a trigonometric identity. Let’s
start by rewriting the denominator using the angle difference formula for sine:
sin (θ−π
4)= sin(θ) cos (π
4)−cos(θ) sin (π
4)
=√2
2sin(θ)−√2
2cos(θ) = √2
2(sin(θ)−cos(θ))
Step 2: Substitute the rewritten denominator back into the equa-
tion. Substitute the rewritten denominator back into the equation:
1
sin(θ) + cos(θ)=√2−1
2·1
√2
2(sin(θ)−cos(θ))
Step 3: Simplify and manipulate the right-hand side. Simplify the
expression on the right-hand side by dividing the fraction and simplifying:
√2−1
2·1
√2
2(sin(θ)−cos(θ)) = (√2−1) ·2
√2(sin(θ)−cos(θ))
= (√2−1) ·2
√2 sin(θ)−√2 cos(θ)= (√2−1) ·2
√2(sin(θ)−cos(θ))
28
= (√2−1) ·2
√2
2(sin(θ)−cos(θ)) = (√2−1) ·4 = 4√2−4
Step 4: Verify the identity. Therefore, we have shown that:
1
sin(θ) + cos(θ)=√2−1
2·1
sin (θ−π
4)= 4√2−4
Hence, the trigonometric identity is proven.
Question 33
Question
Prove the following trigonometric identity:
cos4x−sin4x= cos 2x.
Solution
Step 1: Start with the left-hand side of the equation and use the difference of
squares formula:
cos4x−sin4x= (cos2x+ sin2x)(cos2x−sin2x).
Step 2: Recall the Pythagorean identity cos2x+ sin2x= 1 and rewrite the
expression in terms of sin 2x:
(cos2x+ sin2x)(cos2x−sin2x) = (1)(cos2x−sin2x) = cos2x−sin2x.
Step 3: Use the double angle formula for cosine: cos 2x= cos2x−sin2xto
substitute in the expression:
cos2x−sin2x= cos 2x.
Step 4: Therefore, we have shown that cos4x−sin4x= cos 2x, as required.
Question 34
Question
Prove the following trigonometric identity:
cos(2x)·sin(3x)
cos(x)= 2 sin(2x) sin(3x)
29
Solution
We start by simplifying the left side of the equation:
Step 1: cos(2x)·sin(3x)
cos(x)
Step 2: =cos(2x)·(sin(x) cos(2x) + cos(x) sin(2x))
cos(x)(Expand sin(3x)using the angle sum formula)
Step 3: = cos(2x) sin(x) + sin(2x)
Step 4: = 2 cos(x) sin(x)·sin(x) + 2 sin(x) cos(x)(Expand using cos(2x) = 2 cos(x) sin(x))
Step 5: = 2 sin(2x) sin(3x)(Combine terms)
Thus, we have shown that cos(2x)·sin(3x)
cos(x)= 2 sin(2x) sin(3x)
Question 35
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)= cot(θ)−csc(θ)
Solution
To prove the given trigonometric identity, we will first simplify the left-hand side
of the equation using trigonometric identities, and then show that it is equal to
the right-hand side of the equation.
Step 1: Simplify the left-hand side Let’s simplify the left-hand side of
the equation:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)=sin2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
=sin2(θ) + sin2(θ)
sin(θ)(1 −cos(θ))
=2 sin2(θ)
sin(θ)(1 −cos(θ))
=2 sin(θ)
1−cos(θ)
Step 2: Further simplify the left-hand side Using the double angle
identity sin(2θ) = 2 sin(θ) cos(θ), we can express the numerator in terms of
cos(θ):
2 sin(θ)
1−cos(θ)=2 sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)=2 sin(θ)(1 + cos(θ))
1−cos2(θ)=2 sin(θ)(1 + cos(θ))
sin2(θ)
30
Step 5: Recall the Pythagorean identity: sin2(x) + cos2(x) = 1.
1−cos2(x) = sin2(x)
Step 6: Substitute sin2(x)for 1−cos2(x)in the expression.
sin2(x)
cos(x)
Step 7: Simplify the expression.
sin(x)
cos(x)
Therefore, we have shown that sec(x)−cos(x) = sin(x)
cos(x), completing the proof
of the trigonometric identity.
Question 2
Question
Solve the equation tan2(x) + sec(x) tan(x) = 0 for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation in terms of sine and cosine:
tan2(x) + sec(x) tan(x) = 0
sin2(x)
cos2(x)+1
cos(x)·sin(x)
cos(x)= 0
Step 2: Find a common denominator and combine the fractions:
sin2(x) + sin(x)
cos2(x) cos(x)= 0
Step 3: Since the numerator must equal 0, we have:
sin2(x) + sin(x) = 0
Step 4: Factor out sin(x)from the equation:
sin(x)(sin(x) + 1) = 0
Step 5: Set each factor equal to 0 and solve for x: For sin(x) = 0:
x= 0, π
2
For sin(x) + 1 = 0:
sin(x) = −1
x=3π
2
Step 6: Check the solutions in the original equation to ensure they are valid
in the given interval: Checking x= 0:
tan2(0) + sec(0) tan(0) = 0
0 + (1)(0) = 0
The solution x= 0 is valid.
Checking x=π:
tan2(π) + sec(π) tan(π) = 0
0+(−1)(0) = 0
The solution x=πis valid.
Checking x=3π
2:
tan2(3π
2)+ sec (3π
2)tan (3π
2)= 0
1 + (0)(−1) = 1
The solution x=3π
2is not valid in the given interval.
Therefore, the solutions to the equation are x= 0 and x=π.
Question 3
Question
Prove the following trigonometric identity:
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Solution
To prove the trigonometric identity sin(3θ) = 3 sin(θ)−4 sin3(θ), we will use
the angle addition formula for sine:
sin(A+B) = sin Acos B+ cos Asin B
Step 1: Start by expanding sin(3θ)using the angle addition formula twice.
sin(3θ) = sin(2θ+θ)
Since sin(2θ) = 2 sin(θ) cos(θ), we have:
sin(3θ) = sin(2θ) cos(θ) + cos(2θ) sin(θ)
3
Step 2: To simplify the expression, we need to compute cos(2θ). Using the
double-angle formula cos(2θ) = 2 cos2(θ)−1, we get:
sin(3θ) = (2 sin(θ) cos(θ)) cos(θ) + (2 cos2(θ)−1) sin(θ)
Step 3: Simplify the expression further by substituting sin2(θ) = 1−cos2(θ).
sin(3θ) = 2 sin(θ) cos2(θ) + 2 cos2(θ) sin(θ)−sin(θ)
sin(3θ) = 2 sin(θ)(1 −sin2(θ)) + 2(1 −sin2(θ)) sin(θ)−sin(θ)
Step 4: Simplify the expression and factor out sin(θ).
sin(3θ) = 2 sin(θ)−2 sin3(θ) + 2 sin(θ)−2 sin3(θ)−sin(θ)
sin(3θ) = 3 sin(θ)−4 sin3(θ)
Therefore, we have successfully proved the trigonometric identity sin(3θ) =
3 sin(θ)−4 sin3(θ).
Question 4
Question
Prove the trigonometric identity: sin4(x)−cos4(x) = 1 −2 cos2(2x).
Solution
Step 1: Start with the left-hand side of the given identity and simplify.
sin4(x)−cos4(x) = (sin2(x))2−(cos2(x))2
= (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
Step 2: Use the double angle formula for cosine (cos(2θ) = 2 cos2(θ)−1) to
simplify sin2(x)−cos2(x).
sin2(x)−cos2(x) = sin2(x)−(1 −sin2(x))
= 2 sin2(x)−1
=−2 cos2(−x)
Step 3: Rewrite the expression −2 cos2(−x)in terms of 2x.
−2 cos2(−x) = −2 cos2(−(2x)) = −2 cos2(2x)
Step 4: Therefore, we have shown that sin4(x)−cos4(x) = −2 cos2(2x).
But the given identity is 1−2 cos2(2x). To make them match, simply multiply
by −1on both sides:
sin4(x)−cos4(x) = 1 −2 cos2(2x)
So, the identity is proved.
4
Question 5
Question
Solve the following trigonometric equation for xin the interval [0,2π]:sin(x) +
√3 cos(x) = 1.
Solution
Step 1: Rewrite the equation in terms of a single trigonometric function. Step
2: Square both sides of the equation. Step 3: Use a trigonometric identity
to express the squared trigonometric function in terms of one trigonometric
function. Step 4: Solve the resulting trigonometric equation. Step 5: Check the
solutions in the original equation for validity.
Question 6
Question
Solve the equation √3 cos(x) + sin(x) = 0 for xin the interval [0,2π).
Solution
Step 1: We can rewrite the given equation as sin (x+π
6)= 0, where we used
the trigonometric identity sin(a) cos(b) + cos(a) sin(b) = sin(a+b).
Step 2: We can conclude that x+π
6must be equal to kπ, where kis an
integer.
Step 3: Solving for xin the interval [0,2π), we have
x+π
6= 0
or
x+π
6=π
or
x+π
6= 2π.
Step 4: Solving these equations, we get
x=−π
6
or
x=5π
6
or
x=11π
6.
Thus, the solutions to the equation √3 cos(x) + sin(x)=0in the interval
[0,2π)are −π
6,5π
6, and 11π
6.
5
Question 7
Question
Prove the trigonometric identity sin4(x)−cos4(x) = 2 sin2(x) cos2(x).
Solution
Step 1: Start with the left side of the equation and use the Pythagorean identity
sin2(x) + cos2(x) = 1.
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (1)(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x))
Step 2: Expand the expression (sin(x) + cos(x))(sin(x)−cos(x)) using the
distributive property.
(sin(x) + cos(x))(sin(x)−cos(x)) = sin2(x)−cos2(x) + sin(x) cos(x)−sin(x) cos(x)
= sin2(x)−cos2(x)
= 2 sin2(x) cos2(x)
Therefore, sin4(x)−cos4(x) = 2 sin2(x) cos2(x), as required.
Question 8
Question
Prove the following trigonometric identity:
cos2(x)
sin(x) cos(x)=1
sin(x)−1
cos(x)
6
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
until it is equivalent to the right-hand side.
LHS =cos2(x)
sin(x) cos(x)
=cos(x) cos(x)
sin(x) cos(x)
=cos(x)
sin(x)
=1
sin(x)·1
cos(x)
=1
sin(x)·1
1
cos(x)
=1
sin(x)·cos(x)
=cos(x)
sin(x)
=1
sin(x)·1
cos(x)
=1
sin(x)−1
cos(x)
=RHS
Therefore, the given trigonometric identity is true.
Question 9
Question
Prove the trigonometric identity
tan(θ) + sin(θ)
cos(θ)= sec(θ) + tan(θ)
Solution
Step 1: Recall that sec(θ) = 1
cos(θ)and tan(θ) = sin(θ)
cos(θ).
Step 2: Start by rewriting the left-hand side of the equation using the
definitions of sec(θ)and tan(θ):
tan(θ) + sin(θ)
cos(θ)=
sin(θ)
cos(θ)+ sin(θ)
cos(θ)
7
Step 3: Combine the terms in the numerator:
sin(θ)
cos(θ)+ sin(θ)
cos(θ)=
sin(θ)+sin(θ) cos(θ)
cos(θ)
cos(θ)
Step 4: Factor out a sin(θ)from the numerator:
sin(θ)(1+cos(θ))
cos(θ)
cos(θ)
Step 5: Simplify the expression by canceling out cos(θ)in the numerator
and denominator: sin(θ)(1 + cos(θ))
cos2(θ)
Step 6: Recall that 1 + cos2(θ) = sec2(θ). Rewrite the simplified expression
in terms of sec(θ):
sin(θ) sec(θ)
cos(θ)
Step 7: Simplify further to obtain the right-hand side of the equation:
sin(θ)
cos(θ)·sec(θ) = tan(θ)·sec(θ) = sec(θ) + tan(θ)
Therefore, the trigonometric identity is proved.
Question 10
Question
Solve the trigonometric equation sin4(x)−sin2(x) = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s rewrite the equation sin4(x)−sin2(x) = 0 as sin2(x)(sin2(x)−1) =
0.
Step 2: Therefore, we have two cases:
• Case 1: sin2(x) = 0
In this case, sin(x) = 0. The solutions to sin(x) = 0 in the interval [0,2π)
are x= 0, π.
• Case 2: sin2(x)−1 = 0
Solving sin2(x)−1 = 0, we get sin2(x) = 1, which implies sin(x) = ±1.
The solutions to sin(x) = 1 in the interval [0,2π)are x=π
2,5π
2, and the
solutions to sin(x) = −1in the interval [0,2π)is x=3π
2.
Step 3: Therefore, the solutions to the trigonometric equation sin4(x)−
sin2(x) = 0 in the interval [0,2π)are x= 0,π
2,3π
2, π, 5π
2.
8
Question 11
Question
Prove the following trigonometric identity:
sin3x+ cos3x
sin x+ cos x= 1 −sin xcos x
Solution
Step 1: We will start by simplifying the left-hand side of the equation. Step 2:
Note that sin3x= sin x(sin2x)and cos3x= cos x(cos2x). Step 3: Substituting
these expressions into the numerator, we get:
sin x(sin2x) + cos x(cos2x)
Step 4: Now, we’ll factor out a sin xfrom the first term and a cos xfrom the
second term:
sin x(sin x2+ cos x2)
Step 5: Since sin2x+ cos2x= 1, the expression simplifies to sin x. Step 6:
Therefore, the numerator simplifies to sin x. Step 7: Simplifying the denomi-
nator sin x+ cos xstays as it is. Step 8: Substituting the simplified numerator
and denominator back into the original expression, we get:
sin x
sin x+ cos x
Step 9: Now, simplify the expression sin x
sin x+cos xto obtain 1−sin xcos x. Step 10:
The left-hand side is now simplified to the right-hand side, so the trigonometric
identity is proved.
Question 12
Question
Prove the trigonometric identity:
tan(x)−sin(x)
cos(x)= sec(x)−cos(x)
9
Solution
Step 1: Start with the left side of the equation and simplify:
tan(x)−sin(x)
cos(x)=
sin(x)
cos(x)−sin(x)
cos(x)
=sin(x)−sin(x) cos(x)
cos(x)
=sin(x)(1 −cos(x))
cos(x)
=sin(x) sin(x)
cos(x)
=sin2(x)
cos(x)
=1−cos2(x)
cos(x)[Using the Pythagorean identity: sin2(x) = 1 −cos2(x)]
=1
cos(x)−cos2(x)
cos(x)
= sec(x)−cos(x)
Therefore, the given trigonometric identity is proved.
Question 14
Question
Prove the following trigonometric identity:
cos4θ
1−sin θ+1
cos4θ=1 + sin2θ
cos2θ
Solution
We will start by simplifying each side of the equation separately.
10
Step 1: Simplify the left-hand side.
cos4θ
1−sin θ+1
cos4θ=cos4θ+ (1 −sin θ)
1−sin θ
=cos4θ+ 1 −sin θ
1−sin θ
=cos4θ+ cos2θ−cos2θ−sin θ+ 1
1−sin θ
=cos2θ(cos2θ+ 1) −cos2θ−sin θ+ 1
1−sin θ
=cos2θ(sin2θ+ 1) −cos2θ−sin θ+ 1
1−sin θ
=cos2θsin2θ+ cos2θ−cos2θ−sin θ+ 1
1−sin θ
=cos2θsin2θ−sin θ+ 1
1−sin θ
Step 2: Simplify the right-hand side.
1 + sin2θ
cos2θ=1 + sin2θ
cos2θ·1
1
=sin2θ+ 1
cos2θ
=cos2θsin2θ+ cos2θ
cos2θ
=cos2θ(sin2θ+ 1)
cos2θ
= sin2θ+ 1
= 1 + sin2θ
Step 3: Conclude. Since both sides simplify to 1 + sin2θ, we have shown
that cos4θ
1−sin θ+1
cos4θ=1 + sin2θ
cos2θ
and the trigonometric identity is proven.
Question 15
Question
Solve the trigonometric equation sin(3x) = cos(x)for xin the interval [0,2π).
11
Solution
Step 1: Use the angle addition formula for sine and cosine to rewrite the equa-
tion.
sin(3x) = cos(x)
sin(2x+x) = cos(x)
Step 2: Apply the angle addition formulas:
sin(2x) cos(x) + cos(2x) sin(x) = cos(x)
Step 3: Replace sin(2x)and cos(2x)with the double angle formulas:
2 sin(x) cos(x) cos(x) + (2 cos2(x)−1) sin(x) = cos(x)
Step 4: Simplify the equation:
2 sin(x) cos2(x) + 2 cos2(x) sin(x)−sin(x) = cos(x)
2 sin(x) cos2(x)−sin(x) = cos(x)−2 cos2(x) sin(x)
sin(x)(2 cos2(x)−1) = cos(x)(1 −2 sin2(x))
Step 5: Work with the Pythagorean identities:
sin(x)(2 −sin2(x)) = cos(x)(1 −2 sin2(x))
(2 sin(x)−sin3(x)) = cos(x)−2 cos(x) sin2(x)
Step 6: Use the Pythagorean identity to substitute cos(x)and sin2(x):
2 sin(x)−sin3(x) = √1−sin2(x)−2√1−sin2(x) sin2(x)
Step 7: Let u= sin(x)and solve the resulting cubic equation:
2u−u3=√1−u2−2√1−u2u2
u3−2u2−2u+ 1 = 0
(u−1)(u2−u−1) = 0
Step 8: Solve for u: For u= 1,sin(x) = 1 =⇒x=π
2or 3π
2.
For u2−u−1 = 0,sin(x) = 1±√5
2. However, 1+√5
2is outside the interval [0,2π),
so we only consider sin(x) = 1−√5
2. This gives x= arcsin (1−√5
2).
Therefore, the solutions to the equation sin(3x) = cos(x)in the interval
[0,2π)are x=π
2,3π
2, and arcsin (1−√5
2).
12
Question 16
Question
Prove the identity:
sin4(x)−cos4(x)
sin2(x) + cos2(x)= sin(2x)
Solution
Step 1: Simplify the left-hand side of the equation using the difference of squares:
sin4(x)−cos4(x)
sin2(x) + cos2(x)=(sin2(x)−cos2(x))(sin2(x) + cos2(x))
sin2(x) + cos2(x)
Step 2: Notice that sin2(x)−cos2(x) = −cos(2x)and simplify further:
(sin2(x)−cos2(x))(sin2(x) + cos2(x))
sin2(x) + cos2(x)=−cos(2x)·1
1=−cos(2x)
Step 3: Recall that the double angle formula for sin(2x)is:
sin(2x) = 2 sin(x) cos(x)
Step 4: Use the identity cos(2x) = 1−2 sin2(x)to rewrite −cos(2x)in terms
of sin(x)and cos(x):
−cos(2x) = −1 + 2 sin2(x)
Step 5: Recognize that 2 sin2(x)=1−cos(2x)and replace 2 sin2(x)with
1−cos(2x)in the previous expression:
−cos(2x) = −1+1−cos(2x) = −cos(2x)
Step 6: Hence, sin4(x)−cos4(x)
sin2(x)+cos2(x)= sin(2x)is proven.
Question 17
Question
Prove the identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the identity sin(3x) = 3 sin(x)−4 sin3(x), we will use the angle addition
formula for sine: sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
13
Step 1: Express sin(3x)in terms of sin(x)Using the angle addition
formula for sine twice, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Express sin(2x)in terms of sin(x)We can express sin(2x)using
the double angle formula: sin(2x) = 2 sin(x) cos(x). Substitute this into the
expression for sin(3x):
sin(3x) = (2 sin(x) cos(x)) cos(x) + cos(2x) sin(x)
= 2 sin(x) cos2(x) + cos(2x) sin(x)
Step 3: Express cos(2x)in terms of sin(x)and cos(x)Using the double
angle formula for cosine, cos(2x) = cos2(x)−sin2(x). Substitute this into the
expression for sin(3x):
sin(3x) = 2 sin(x) cos2(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x)(1 −sin2(x)) + (cos2(x)−sin2(x)) sin(x)
Step 4: Simplify the expression Now, simplify the expression:
sin(3x) = 2 sin(x)−2 sin3(x) + cos2(x) sin(x)−sin3(x) sin(x)
= 3 sin(x)−3 sin3(x) = 3 sin(x)−3 sin(x)3
= 3 sin(x)−4 sin3(x)
Therefore, we have proven the identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 18
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the angle addition formula for sine and some algebraic manipulation.
Step 1: Apply the angle addition formula for sine.
sin(3x) = sin(2x+x)
Step 2: Expand using the angle addition formula: sin(A+B) = sin(A) cos(B)+
cos(A) sin(B).
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
14
Step 3: Express sin(2x)and cos(2x)in terms of sin(x)and cos(x).
sin(3x) = (2 sin(x) cos(x))(cos(x)) + (cos2(x)−sin2(x))(sin(x))
Step 4: Simplify the expression.
sin(3x) = 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
sin(3x) = 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, we have successfully proven the trigonometric identity sin(3x) =
3 sin(x)−4 sin3(x).
Question 19
Question
Prove the trigonometric identity:
sin(α+β) tan(α−β) = tan(α)−tan(β)
Solution
To prove the given identity, we will start with the left-hand side and manipulate
it to match the right-hand side.
Step 1: Expand sin(α+β)using the angle sum formula:
sin(α+β) = sin(α) cos(β) + cos(α) sin(β)
Step 2: Expand tan(α−β)using the tangent subtraction formula:
tan(α−β) = tan(α)−tan(β)
1 + tan(α) tan(β)
Step 3: Substitute the expansions from Step 1 and Step 2 into the left-hand
side of the identity:
sin(α+β) tan(α−β) = (sin(α) cos(β) + cos(α) sin(β)) ·(tan(α)−tan(β)
1 + tan(α) tan(β))
=sin(α) cos(β) tan(α)−sin(α) cos(β) tan(β) + cos(α) sin(β) tan(α)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
15
Step 4: Simplify the expression by combining like terms:
sin(α) cos(β) tan(α)−sin(α) cos(β) tan(β) + cos(α) sin(β) tan(α)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
=sin(α) cos(β) tan(α) + cos(α) sin(β) tan(α)−sin(α) cos(β) tan(β)−cos(α) sin(β) tan(β)
1 + tan(α) tan(β)
=sin(α) tan(α)(cos(β) + cos(β)) −sin(β) tan(β)(cos(α) + cos(α))
1 + tan(α) tan(β)
=sin(α) tan(α)−sin(β) tan(β)
1 + tan(α) tan(β)
= tan(α)−tan(β)
Therefore, we have shown that sin(α+β) tan(α−β) = tan(α)−tan(β), and
the trigonometric identity is proven.
Question 20
Question
Prove the trigonometric identity:
1 + cot2(θ)
1 + tan2(θ)= sec2(θ)
16
Solution
Step 1: Start with the left-hand side of the identity and simplify using the
definitions of cotangent, tangent, and secant:
1 + cot2(θ)
1 + tan2(θ)=1 + cos2(θ)
sin2(θ)
1 + sin2(θ)
cos2(θ)
=
sin2(θ)+cos2(θ)
sin2(θ)
cos2(θ)+sin2(θ)
cos2(θ)
=
1
sin2(θ)
1
cos2(θ)
=cos2(θ)
sin2(θ)
=cos2(θ)
1−cos2(θ)[using sin2(θ) = 1 −cos2(θ)]
=cos2(θ)
sin2(θ)·1
1−cos2(θ)
=cos2(θ)
sin2(θ)·1
sin2(θ)
=cos2(θ)
sin3(θ)
=1
sin(θ)
= sec(θ)
= sec2(θ)
Therefore, we have shown that 1+cot2(θ)
1+tan2(θ)= sec2(θ), completing the proof of
the identity.
Question 21
Question
Prove the following trigonometric identity:
sin4(x) + cos4(x) = 3
2−1
2cos(4x)
Solution
To prove the given trigonometric identity, we will first manipulate both sides
using trigonometric identities and then simplify the expressions until they are
17
equivalent.
Step 1: Start with the left side of the identity.
sin4(x) + cos4(x)
Step 2: Use the Pythagorean identity: sin2(x) + cos2(x) = 1, to express
sin4(x)and cos4(x)in terms of sin2(x)and cos2(x).
= (sin2(x))2+ (cos2(x))2
Step 3: Expand the expressions.
= sin4(x) + 2 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x)
Step 4: Factor the middle terms as (sin2(x) + cos2(x))2.
= (sin2(x) + cos2(x))2−2 sin2(x) cos2(x)
Step 5: Use the Pythagorean identity and simplify.
= 12−2 sin2(x) cos2(x) = 1 −2 sin2(x) cos2(x)
Step 6: Use the double angle identity: sin(2θ) = 2 sin(θ) cos(θ)to express
sin2(x) cos2(x)in terms of cos(2x).
= 1 −cos(2x)
2
Step 7: Multiply by 2 to get rid of the fraction.
= 2 −cos(2x)
Step 8: Substitute cos(2x)with the identity cos(2x) = 2 cos2(x)−1.
= 2 −(2 cos2(x)−1) = 3 −2 cos2(x)
Step 9: Use the Pythagorean identity to express cos2(x)in terms of sin2(x).
= 3 −2(1 −sin2(x)) = 3 −2 + 2 sin2(x) = 1 + 2 sin2(x)
Step 10: Finally, substitute sin2(x)with 1
2−1
2cos(2x).
= 1 + 2 (1
2−1
2cos(2x))= 1 + 1 −cos(2x) = 2 −cos(2x)
Step 11: This matches the right side of the given identity.
=3
2−1
2cos(4x)
Therefore, we have shown that sin4(x)+cos4(x) = 3
2−1
2cos(4x), as required.
18
Question 22
Question
Prove the trigonometric identity:
tan4(x)−sin4(x) = 2 tan2(x) sin2(x)
Solution
tan4(x)−sin4(x) = tan4(x)−(sin2(x))2
= tan4(x)−(sin2(x) + cos2(x))2
= tan4(x)−(sin2(x)
cos2(x)+ 1)2
= tan4(x)−sin4(x)
cos4(x)−2sin2(x)
cos2(x)−1
= tan4(x)−sin4(x)
cos4(x)−2 tan2(x)−1
=sin4(x)
cos4(x)−sin4(x)
cos4(x)−2 tan2(x)−1
= 2 tan2(x)−1
= 2 tan2(x)(1 −sin2(x))
= 2 tan2(x) sin2(x)
Therefore, the trigonometric identity tan4(x)−sin4(x) = 2 tan2(x) sin2(x)is
proven.
Question 23
Question
Solve the trigonometric equation: sin2(x)−cos(x)−1 = 0, for 0≤x≤2π.
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1:
sin2(x)−(1 −sin2(x)) −1 = 0
Step 2: Simplify the equation:
2 sin2(x)−2 = 0
19
Step 3: Divide by 2 to isolate sin2(x):
sin2(x)−1 = 0
Step 4: Solve for sin(x)by taking the square root of both sides:
sin(x) = ±1
Step 5: Find the solutions for xwithin the interval 0≤x≤2π.
When sin(x) = 1, we have x=π
2since sin(π
2)= 1.
When sin(x) = −1, we have x=3π
2since sin(3π
2)=−1.
Therefore, the solutions to the equation sin2(x)−cos(x)−1 = 0 for 0≤x≤
2πare x=π
2and x=3π
2.
Question 24
Question
Solve the trigonometric equation 2 sin2(x)−3 cos(x) = 2 for 0◦≤x≤360◦.
Solution
Step 1: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to express sin2(x)
in terms of cos(x). Step 2: Substitute the expression for sin2(x)into the given
equation to obtain a quadratic equation in cos(x). Step 3: Solve the quadratic
equation for cos(x)using factoring or the quadratic formula. Step 4: Find the
corresponding values of xby using the connection between cos(x)and sin(x)in
the first quadrant. Step 5: Check the solutions obtained by substituting them
back into the original equation to ensure they satisfy the equation.
Let’s follow these steps:
Step 1: We know that sin2(x) = 1−cos2(x), so 2(1−cos2(x))−3 cos(x) = 2.
Step 2: Simplifying the equation gives 2−2 cos2(x)−3 cos(x) = 2.
Step 3: Rearranging terms, we get −2 cos2(x)−3 cos(x) = 0.
Factoring out −cos(x), we have −cos(x)(2 cos(x) + 3) = 0.
So, cos(x) = 0 or cos(x) = −3
2.
Since −3
2is not in the range of [−1,1], we discard it.
Hence, cos(x) = 0 which implies x= 90◦or x= 270◦.
Step 4: In terms of sin(x),x= 90◦corresponds to sin(x) = 1 and x= 270◦
corresponds to sin(x) = −1.
Step 5: Checking the solutions:
For x= 90◦:2(sin2(90◦)) −3 cos(90◦) = 2(1) −3(0) = 2. This is true.
For x= 270◦:2(sin2(270◦)) −3 cos(270◦) = 2(1) −3(0) = 2. This is also
true.
Therefore, the solutions to the equation are x= 90◦and x= 270◦for
0◦≤x≤360◦.
20
Question 25
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the trigonometric identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will
first rewrite the left-hand side using the Pythagorean identity:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
Step 1: Apply the Pythagorean identity (sin2(x) + cos2(x) = 1) to simplify
the expression:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))
Step 2: Now, use the double-angle formula sin(2x) = 2 sin(x) cos(x)to
expand the right-hand side of the identity:
sin(2x) sin(4x) = 2 sin(x) cos(x)·sin(4x)
Step 3: Using the double-angle formula for sine sin(2θ) = 2 sin(θ) cos(θ),
the expression becomes:
2 sin(x) cos(x)·sin(4x) = 2 sin(x) cos(x)·(2 sin(2x) cos(2x))
Step 4: Further simplify the expression by expanding sin(2x)and cos(2x):
2 sin(x) cos(x)·(2 sin(2x) cos(2x)) = 4 sin(x) cos(x)·(sin(2x) cos(2x))
Step 5: Apply the double-angle formulas for sine and cosine to rewrite the
expression:
4 sin(x) cos(x)·(sin(2x) cos(2x)) = 4 sin(x) cos(x)·(1
2sin(4x))
= 2 sin(x) cos(x) sin(4x)
Step 6: Finally, simplify the right-hand side:
2 sin(x) cos(x) sin(4x) = sin(2x) sin(4x)
Therefore, we have shown that sin4(x)−cos4(x) = sin(2x) sin(4x), and the
trigonometric identity is proved.
Question 26
Question
Solve the equation sin(2x) + √3 cos(x) = 0 for xin the interval [0,2π).
21
Solution
Step 1: Use the double angle identity for sine to rewrite sin(2x):
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the original equation:
2 sin(x) cos(x) + √3 cos(x) = 0
Step 3: Factor out cos(x)from the equation:
cos(x)(2 sin(x) + √3) = 0
Step 4: Set each factor equal to zero and solve for x:
• For cos(x) = 0, we have x=π
2and x=3π
2.
• For 2 sin(x)+√3 = 0, we have sin(x) = −√3
2. This occurs in the third and
fourth quadrants where the reference angle is π
3. So, x=2π
3and x=5π
3.
Step 5: The solutions in the interval [0,2π)are x=π
2,2π
3, and 5π
3.
Question 27
Question
Prove the following trigonometric identity:
sin4(x)
1 + cos(x)+cos4(x)
1 + sin(x)= 1 −sin(x) cos(x)
Solution
Step 1: Begin by expressing sin4(x)and cos4(x)in terms of sin(2x)and cos(2x).
sin4(x) = (1−cos(2x)
2)2
=1−2 cos(2x) + cos2(2x)
4
cos4(x) = (1 + cos(2x)
2)2
=1 + 2 cos(2x) + cos2(2x)
4
Step 2: Substitute these expressions back into the given identity.
1−2 cos(2x)+cos2(2x)
4
1 + cos(x)+
1+2 cos(2x)+cos2(2x)
4
1 + sin(x)= 1 −sin(x) cos(x)
Step 3: Expand and simplify each term in the identity.
1−2 cos(2x) + cos2(2x)
4(1 + cos(x)) +1 + 2 cos(2x) + cos2(2x)
4(1 + sin(x)) = 1 −sin(x) cos(x)
22
Step 4: Clear the denominators by finding a common denominator.
(1 −2 cos(2x) + cos2(2x))(1 + sin(x)) + (1 + 2 cos(2x) + cos2(2x))(1 + cos(x))
4(1 + cos(x))(1 + sin(x)) = 1−sin(x) cos(x)
Step 5: Expand the numerator and simplify the expression.
1−cos2(2x)−2 cos(2x)−2 sin(x) cos(2x) + cos2(2x) + 2 cos2(2x) + 2 cos(2x) + 2 sin(x) cos(2x)
4(1 + cos(x))(1 + sin(x)) = 1−sin(x) cos(x)
1−cos2(2x) + cos2(2x)
4(1 + cos(x))(1 + sin(x)) = 1 −sin(x) cos(x)
1
4(1 + cos(x))(1 + sin(x)) = 1 −sin(x) cos(x)
1
4(1 + cos(x))(1 + sin(x)) =4 sin(x) cos(x)
2
1
(1 + cos(x))(1 + sin(x)) = 2 sin(x) cos(x)
Step 6: We know that the Pythagorean identity sin2(x) + cos2(x)=1.
Let’s express denumerator in terms of sin(x) and cos(x), sin2(x) + cos2(x) +
2 sin(x) cos(x) = 1 + 2 sin(x) cos(x).
1
(1 + cos(x))(1 + sin(x)) = 2 sin(x) cos(x)
Step 7: Finally, substitute back in the Pythagorean identity to get:
1
1 + 2 sin(x) cos(x)= 2 sin(x) cos(x)
1
1 + 2 sin(x) cos(x)= 2 sin(x) cos(x)
Therefore, the given trigonometric identity is proved.
Question 28
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)=1 + cos(x)
sin(x)
23
Solution
To prove the identity, we will start by simplifying each side separately using
trigonometric identities.
Step 1: Simplify the left side We have:
sin2(x)
1−cos(x)=sin(x) sin(x)
1−cos(x)=sin(x)(1 −cos2(x))
1−cos(x)
Using the Pythagorean identity sin2(x) = 1 −cos2(x), we get:
sin(x)(1 −cos2(x))
1−cos(x)=sin(x) sin2(x)
1−cos(x)
Further simplify:
sin(x) sin2(x)
1−cos(x)=sin(x)(1 −cos(x)) sin(x)
1−cos(x)= sin(x) sin(x) = sin2(x)
Step 2: Simplify the right side We have:
1 + cos(x)
sin(x)=1
sin(x)+cos(x)
sin(x)= csc(x) + cot(x)
Step 3: Compare the simplified forms Since sin2(x)= csc(x) + cot(x),
the given identity
sin2(x)
1−cos(x)=1 + cos(x)
sin(x)
is not true for all values of x. Thus, the identity does not hold.
Question 29
Question
Prove the identity sin2(x) cos2(x) = 1
4sin22x.
Solution
To prove the given identity, we will start with the right-hand side and simplify
it step by step to match the left-hand side.
Step 1: Start with the right-hand side of the identity:
1
4sin22x
Step 2: Recall the double angle identity: sin2(2x) = 1−cos(4x)
2.
24
Step 3: Substitute sin2(2x)using the double angle identity:
1
4(1 −cos(4x))
Step 4: Recall the double angle identity: cos(2θ) = 1 −2 sin2(θ).
Step 5: Substitute cos(4x)using the double angle identity:
1
4(1−(1 −2 sin2(2x)))
Step 6: Simplify the expression:
1
4(1−1 + 2 sin2(2x))
1
4(2 sin2(2x))
Step 7: Simplify further:
sin2(2x)
Step 8: Recall the double angle identity: sin(2θ) = 2 sin(θ) cos(θ).
Step 9: Substitute sin(2x)using the double angle identity:
(2 sin(x) cos(x))2
Step 10: Simplify the expression:
(2 sin(x) cos(x))2
(2 sin(x) cos(x))(2 sin(x) cos(x))
4 sin2(x) cos2(x)
Step 11: Therefore, the right-hand side equals the left-hand side of the
identity, so sin2(x) cos2(x) = 1
4sin22x.
Question 30
Question
Prove the following trigonometric identity:
1
sin(x)+1
cos(x)=1
sin(x) cos(x)
25
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until it simplifies to the right-hand side.
Step 1: Rewrite the left-hand side using a common denominator.
1
sin(x)+1
cos(x)=cos(x)
sin(x) cos(x)+sin(x)
sin(x) cos(x)
Step 2: Combine the fractions over a common denominator.
cos(x) + sin(x)
sin(x) cos(x)
Step 3: Use the trigonometric Pythagorean identity sin2(x) + cos2(x)=1
to simplify the numerator.
cos(x) + sin(x) = cos(x)(cos(x) + sin(x)) + sin(x)(cos(x) + sin(x))
= cos2(x) + cos(x) sin(x) + cos(x) sin(x) + sin2(x)
= cos2(x) + 2 cos(x) sin(x) + sin2(x)
= (cos(x) + sin(x))2
Step 4: Substitute the simplified numerator back into the fraction.
(cos(x) + sin(x))2
sin(x) cos(x)
Step 5: Expand the squared binomial in the numerator.
cos2(x) + 2 cos(x) sin(x) + sin2(x)
sin(x) cos(x)
Step 6: Use the Pythagorean identity again to simplify the numerator to 1.
1
sin(x) cos(x)
Since the left-hand side is equal to the right-hand side, the trigonometric
identity has been proven.
Question 31
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
26
Solution
Step 1: Rewrite cos(x)in terms of sin(x)using the Pythagorean identity sin2(x)+
cos2(x) = 1.
cos(x) = √1−sin2(x)
Step 2: Substitute cos(x)in the given equation to get:
sin2(x) + √1−sin2(x) = 1
Step 3: Rearrange the equation and isolate sin(x):
sin2(x) = 1 −√1−sin2(x)
Step 4: Square both sides of the equation to eliminate the square root:
sin4(x) = (1 −√1−sin2(x))2
Step 5: Expand the right side of the equation and simplify:
sin4(x) = 1 −2√1−sin2(x) + (1 −sin2(x))
Step 6: Combine like terms on the right side:
sin4(x) = 2 −2√1−sin2(x)−sin2(x)
Step 7: Let u= sin2(x)for simplicity. The equation becomes:
u2= 2 −2√1−u−u
Step 8: Solve the quadratic equation for u:
u2+u−2+2√1−u= 0
Step 9: Factor the equation:
(u+ 2)(u−1+2√1−u) = 0
Step 10: Solve for uusing each factor separately: 1) u+ 2 = 0 =⇒u=−2
(not in the domain [0,2π)) 2) u−1+2√1−u= 0
Step 11: Solve u−1+2√1−u= 0 for u:
2√1−u= 1 −u
4(1 −u) = (1 −u)2
Step 12: Expand and simplify:
4−4u= 1 −2u+u2
27
Step 13: Rearrange the equation as a quadratic:
u2−2u−3 = 0
Step 14: Factor the equation:
(u−3)(u+ 1) = 0
Step 15: Solve for uusing each factor separately: 1) u−3 = 0 =⇒u= 3
(not in the domain [0,2π)) 2) u+1 = 0 =⇒u=−1(not in the domain [0,2π))
Step 16: Since there are no solutions for uin the given domain, the original
equation sin2(x) + cos(x) = 1 has no solutions in the interval [0,2π).
Question 32
Question
Prove the trigonometric identity:
1
sin(θ) + cos(θ)=√2−1
2·1
sin (θ−π
4)
Solution
Step 1: Rewrite the denominator using a trigonometric identity. Let’s
start by rewriting the denominator using the angle difference formula for sine:
sin (θ−π
4)= sin(θ) cos (π
4)−cos(θ) sin (π
4)
=√2
2sin(θ)−√2
2cos(θ) = √2
2(sin(θ)−cos(θ))
Step 2: Substitute the rewritten denominator back into the equa-
tion. Substitute the rewritten denominator back into the equation:
1
sin(θ) + cos(θ)=√2−1
2·1
√2
2(sin(θ)−cos(θ))
Step 3: Simplify and manipulate the right-hand side. Simplify the
expression on the right-hand side by dividing the fraction and simplifying:
√2−1
2·1
√2
2(sin(θ)−cos(θ)) = (√2−1) ·2
√2(sin(θ)−cos(θ))
= (√2−1) ·2
√2 sin(θ)−√2 cos(θ)= (√2−1) ·2
√2(sin(θ)−cos(θ))
28
= (√2−1) ·2
√2
2(sin(θ)−cos(θ)) = (√2−1) ·4 = 4√2−4
Step 4: Verify the identity. Therefore, we have shown that:
1
sin(θ) + cos(θ)=√2−1
2·1
sin (θ−π
4)= 4√2−4
Hence, the trigonometric identity is proven.
Question 33
Question
Prove the following trigonometric identity:
cos4x−sin4x= cos 2x.
Solution
Step 1: Start with the left-hand side of the equation and use the difference of
squares formula:
cos4x−sin4x= (cos2x+ sin2x)(cos2x−sin2x).
Step 2: Recall the Pythagorean identity cos2x+ sin2x= 1 and rewrite the
expression in terms of sin 2x:
(cos2x+ sin2x)(cos2x−sin2x) = (1)(cos2x−sin2x) = cos2x−sin2x.
Step 3: Use the double angle formula for cosine: cos 2x= cos2x−sin2xto
substitute in the expression:
cos2x−sin2x= cos 2x.
Step 4: Therefore, we have shown that cos4x−sin4x= cos 2x, as required.
Question 34
Question
Prove the following trigonometric identity:
cos(2x)·sin(3x)
cos(x)= 2 sin(2x) sin(3x)
29
Solution
We start by simplifying the left side of the equation:
Step 1: cos(2x)·sin(3x)
cos(x)
Step 2: =cos(2x)·(sin(x) cos(2x) + cos(x) sin(2x))
cos(x)(Expand sin(3x)using the angle sum formula)
Step 3: = cos(2x) sin(x) + sin(2x)
Step 4: = 2 cos(x) sin(x)·sin(x) + 2 sin(x) cos(x)(Expand using cos(2x) = 2 cos(x) sin(x))
Step 5: = 2 sin(2x) sin(3x)(Combine terms)
Thus, we have shown that cos(2x)·sin(3x)
cos(x)= 2 sin(2x) sin(3x)
Question 35
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)= cot(θ)−csc(θ)
Solution
To prove the given trigonometric identity, we will first simplify the left-hand side
of the equation using trigonometric identities, and then show that it is equal to
the right-hand side of the equation.
Step 1: Simplify the left-hand side Let’s simplify the left-hand side of
the equation:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)=sin2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
=sin2(θ) + sin2(θ)
sin(θ)(1 −cos(θ))
=2 sin2(θ)
sin(θ)(1 −cos(θ))
=2 sin(θ)
1−cos(θ)
Step 2: Further simplify the left-hand side Using the double angle
identity sin(2θ) = 2 sin(θ) cos(θ), we can express the numerator in terms of
cos(θ):
2 sin(θ)
1−cos(θ)=2 sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)=2 sin(θ)(1 + cos(θ))
1−cos2(θ)=2 sin(θ)(1 + cos(θ))
sin2(θ)
30
Now, the left-hand side of the equation is:
2 sin(θ)(1 + cos(θ))
sin2(θ)= 2 cot(θ) csc(θ) = 2(cot(θ)−csc(θ))
Step 3: Conclude the proof Therefore, we have shown that
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)= 2(cot(θ)−csc(θ)) = cot(θ)−csc(θ)
Hence, the given trigonometric identity has been proved.
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