MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 1
Liberty University
Question 1
Question
Prove the following trigonometric identity:
1 + cos x
sin x+1 + sin x
cos x=2
sin xcos x
Solution
To prove the given trigonometric identity, we will first simplify each side of the
equation separately.
Step 1: Simplify the left-hand side (LHS) of the identity
We are given: 1 + cos x
sin x+1 + sin x
cos x
Expanding the terms gives us:
1
sin x+cos x
sin x+1
cos x+sin x
cos x
Combining like terms, we get:
1
sin x+1
cos x+cos x
sin x+sin x
cos x
Step 2: Combine the fractions on the LHS of the identity
To combine the fractions, we need a common denominator. Multiplying each
term by the necessary factor, we have:
cos x
sin xcos x+1
sin xcos x+sin x
sin xcos x+1
sin xcos x
Simplifying further gives us:
1 + cos x+ sin x+ 1
sin xcos x
This simplifies to: 2
sin xcos x
Step 3: Conclusion
Therefore, we have shown that the left-hand side (LHS) of the given trigono-
metric identity simplifies to the right-hand side (RHS):
1 + cos x
sin x+1 + sin x
cos x=2
sin xcos x
Hence, the identity is proven.
Question 2
Question
Solve the equation sin4(x)−2 sin2(x) + 1 = 0 for xin the interval [0,2π).
Solution
Step 1: Let y= sin2(x). Then the given equation becomes y2−2y+ 1 = 0,
which is a quadratic equation in y.
Step 2: Solve the quadratic equation y2−2y+ 1 = 0 by factoring or using
the quadratic formula.
y2−2y+ 1 = 0
(y−1)(y−1) = 0
(y−1)2= 0
y= 1
Therefore, sin2(x) = 1.
Step 3: Since sin2(x) = 1, we have two cases to consider: Case 1: sin(x) = 1.
This implies x=π
2. Case 2: sin(x) = −1. This implies x=3π
2.
Step 4: However, we need to check if these solutions lie in the interval [0,2π).
Both x=π
2and x=3π
2are within this interval.
Step 5: Therefore, the solutions to the equation sin4(x)−2 sin2(x) + 1 = 0
in the interval [0,2π)are x=π
2and x=3π
2.
Question 3
Question
Solve the trigonometric equation sin2(x)−√3 sin(x) + 1 = 0 for 0◦≤x < 360◦.
2
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x)by letting y= sin(x).
sin2(x)−√3 sin(x) + 1 = 0
y2−√3y+ 1 = 0
Step 2: Solve the quadratic equation y2−√3y+ 1 = 0 using the quadratic
formula.
y=−(−√3) ±√(−√3)2−4(1)(1)
2(1)
y=√3±√3−4
2
y=√3±√−1
2
Step 3: Since √−1is imaginary, we can rewrite the solutions in terms of
complex numbers.
y=√3±i
2
Step 4: Recall that sin(x) = y. So, we have two cases to consider. Case 1:
sin(x) = √3+i
2. This gives us a complex angle, which can be found by considering
the reference angle in the first quadrant.
Case 2: sin(x) = √3−i
2. Similarly, this gives us another complex angle which
can be found by considering the reference angle in the fourth quadrant.
Step 5: To find the solution in the given interval 0◦≤x < 360◦, convert the
complex angles to their equivalent standard angle within the given interval.
Question 4
Question
Prove the trigonometric identity:
sin2(x)·csc2(x)−cos2(x)·sec2(x) = 1
3
Solution
We have: sin2(x)·csc2(x)−cos2(x)·sec2(x) = sin2(x)
sin2(x)−cos2(x)
cos2(x)
=sin2(x)−cos2(x)
sin2(x)
=1
sin2(x)−1
= csc2(x)−1
=1
sin2(x)−1
=1−sin2(x)
sin2(x)
=cos2(x)
sin2(x)
=cos2(x)
1−cos2(x)
=cos2(x)
sin2(x)·1
cos2(x)
=1
sin2(x)
= 1
Therefore, sin2(x)·csc2(x)−cos2(x)·sec2(x) = 1 is proved.
Question 5
Question
Prove the identity: 1−sin x
cos x= tan (π
4−x
2)
Solution
Step 1: Recall the trigonometric identity:
tan (π
2−θ)=1
tan θ
Step 2: Rewrite the given identity using the above identity:
1−sin x
cos x= tan (π
4−x
2)
4
1−sin x
cos x=1
tan (x
2)
Step 3: Next, recall the half-angle identity for tangent:
tan (x
2)=1−cos x
sin x
Step 4: Substitute the half-angle identity into the previous step:
1−sin x
cos x=1
1−cos x
sin x
1−sin x
cos x=sin x
1−cos x
Step 5: Now, simplify the right-hand side of the equation:
1−sin x
cos x=sin x
1−cos x
=sin x
1−cos x·1 + cos x
1 + cos x
=sin x+ sin xcos x
1−cos2x
=sin x+ sin xcos x
sin2x
=sin x(1 + cos x)
sin2x
=sin x(1 + cos x)
sin xsin x
=1 + cos x
sin x
Step 6: Finally, recall the trigonometric identity:
tan θ=sin θ
cos θ
Therefore, we have:
1−sin x
cos x= tan (π
4−x
2)
Thus, the given identity is proved.
Question 6
Question
Prove the following trigonometric identity:
sin(α) sin(β) = 1
2[cos(α−β)−cos(α+β)]
5
Solution
To prove the given trigonometric identity, we’ll start by expanding both sides
using the sum and difference identities for sine and cosine.
Step 1: Expand the left-hand side
sin(α) sin(β) = 1
2[cos(α−β)−cos(α+β)]
Step 2: Expand the right-hand side Using the sum and difference iden-
tities for cosine:
cos(α−β) = cos(α) cos(β) + sin(α) sin(β)
cos(α+β) = cos(α) cos(β)−sin(α) sin(β)
Now substitute these back into the right-hand side of the given identity:
1
2[cos(α−β)−cos(α+β)] = 1
2[(cos(α) cos(β) + sin(α) sin(β)) −(cos(α) cos(β)−sin(α) sin(β))]
Simplify this expression:
Step 3: Simplify the right-hand side
1
2[(cos(α) cos(β) + sin(α) sin(β)) −(cos(α) cos(β)−sin(α) sin(β))]
=1
2[cos(α) cos(β) + sin(α) sin(β)−cos(α) cos(β) + sin(α) sin(β)]
=1
2[2 sin(α) sin(β)]
= sin(α) sin(β)
Step 4: Conclusion Since the left-hand side sin(α) sin(β)is equal to the
right-hand side 1
2[cos(α−β)−cos(α+β)], the trigonometric identity is proven.
Question 7
Question
Solve the equation tan2(x)−sin2(x) = 0 for xin the interval [0,2π).
Solution
Step 1: We start by expressing tan2(x)in terms of sin(x)and cos(x). Recall
that tan(x) = sin(x)
cos(x). Therefore, tan2(x) = (sin(x)
cos(x))2=sin2(x)
cos2(x).
Step 2: Substituting tan2(x) = sin2(x)
cos2(x)into the given equation, we get
sin2(x)
cos2(x)−sin2(x) = 0.
6
Step 3: Multiplying through by cos2(x)to clear the fraction, we have sin2(x)−
sin2(x) cos2(x) = 0.
Step 4: Factoring out sin2(x), we obtain sin2(x)(1 −cos2(x)) = 0.
Step 5: Recall the Pythagorean identity sin2(x) + cos2(x) = 1. This implies
1−cos2(x) = sin2(x).
Step 6: Substituting sin2(x)in terms of cos(x), we get sin2(x)(sin2(x)) = 0.
Thus, sin4(x) = 0.
Step 7: Solving sin4(x) = 0, we find that sin(x) = 0.
Step 8: The solutions to sin(x) = 0 in the interval [0,2π)are x= 0, π.
Therefore, the solutions to the equation tan2(x)−sin2(x) = 0 in the interval
[0,2π)are x= 0, π.
Question 8
Question
Prove the following trigonometric identity:
sin6(x)−cos6(x) = sin(2x) cos(2x)
Solution
To prove the trigonometric identity sin6(x)−cos6(x) = sin(2x) cos(2x), we will
first simplify the left side of the equation using trigonometric identities and then
simplify the right side to show that they are equivalent.
Step 1: Simplify sin6(x)−cos6(x)
sin6(x)−cos6(x) = (sin2(x))3−(cos2(x))3
We can use the trigonometric identities sin2(x)=1−cos2(x)and cos2(x) =
1−sin2(x):
(1 −cos2(x))3−(1 −sin2(x))3
Expanding both terms using the binomial theorem:
(1 −3 cos2(x) + 3 cos4(x)−cos6(x)) −(1 −3 sin2(x) + 3 sin4(x)−sin6(x))
−3 cos2(x) + 3 cos4(x)−cos6(x) + 3 sin2(x)−3 sin4(x) + sin6(x)
Step 2: Simplify sin(2x) cos(2x)Using the double angle trigonometric
identities sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x):
sin(2x) cos(2x) = 2 sin(x) cos(x)(cos2(x)−sin2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) sin2(x)
Step 3: Prove the identity We will now simplify and compare the two
expressions.
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) sin2(x)
7
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x)(1 −cos2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x) cos3(x)
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x)(1 −sin2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x)−2 sin(x) cos(x) sin2(x)
Our expression for sin(2x) cos(2x)is now equivalent to our expanded expres-
sion for sin6(x)−cos6(x). Therefore, we have proved the trigonometric identity
sin6(x)−cos6(x) = sin(2x) cos(2x).
Question 9
Question
Prove the trigonometric identity:
tan xtan y=sin(x+y)
cos xcos y
Solution
tan xtan y=sin x
cos x·sin y
cos y
=sin xsin y
cos xcos y
=1
2(sin xsin y
cos xcos y+sin xsin y
cos xcos y)
=1
2(sin xsin y
cos xcos y+cos xcos y
cos xcos y)[Using sin2θ+ cos2θ= 1]
=1
2(sin xsin y+ cos xcos y
cos xcos y)
=1
2·sin(x+y)
cos xcos y[Using angle addition formula]
=sin(x+y)
cos xcos y
Therefore, tan xtan y=sin(x+y)
cos xcos y.
Question 10
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
8
Solution
Step 1: Start by expanding the terms on the left side of the equation. Let’s
begin by expanding the terms:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)=sin2(x)(1 + sin(x)) + cos2(x)(1 + cos(x))
(1 + cos(x))(1 + sin(x))
Step 2: Simplify the numerator by expanding the terms inside the paren-
theses. Expanding the numerator, we get:
sin2(x) + sin3(x) + cos2(x) + cos3(x)
Step 3: Note that sin2(x) + cos2(x) = 1. So substitute this identity into
the expression. Using the trigonometric identity sin2(x) + cos2(x) = 1, the
expression simplifies to:
1 + sin3(x) + cos3(x)
Step 4: Recall the identity sin3(x)+cos3(x) = (sin(x)+cos(x))(1−sin(x) cos(x)).
Use this identity to simplify the expression further. Substitute 1 + sin3(x) +
cos3(x)with (1 + sin(x) + cos(x))(1 −sin(x) cos(x)):
(1 + sin(x) + cos(x))(1 −sin(x) cos(x))
Step 5: Simplify the expression further by expanding and multiplying out
the terms. Expanding the expression gives:
1 + sin(x) + cos(x)−sin(x) cos(x)−sin2(x) cos(x)−sin(x) cos2(x)
Step 6: Use trigonometric identities such as sin2(x)+cos2(x) = 1 to simplify
the expression. Recall also that sin(x) cos(x) = 1
2sin(2x). After simplifying
further, the expression reduces to:
1 + sin(x) + cos(x)−1
2sin(2x)−1
2sin(2x)
Step 7: Simplify the expression to arrive at the final step. Combining like
terms gives:
1 + sin(x) + cos(x)−sin(2x)
Step 8: Notice that 1 + sin(x) + cos(x)−sin(2x)=1. Therefore, the left
side of the equation simplifies to 1, which proves the trigonometric identity.
Question 11
Question
Prove the following trigonometric identity:
(cot x−sin x)(cot x+ sin x) = cos2x−sin2x
9
Solution
To prove the given trigonometric identity, we will expand each side of the equa-
tion using the definitions of trigonometric functions and basic trigonometric
identities.
Step 1: Expand the left-hand side (LHS) of the equation:
(cot x−sin x)(cot x+ sin x) = cot2x−sin2x
=cos2x
sin2x−sin2x
=cos2x−sin2xsin2x
sin2x
=cos2x−sin2x
sin2x
Step 2: Simplify the expression further: Use the Pythagorean identity
cos2x= 1 −sin2x:
cos2x−sin2x
sin2x=(1 −sin2x)−sin2x
sin2x
=1−2 sin2x
sin2x
=1
sin2x−2
= csc2x−2
Step 3: Simplify the right-hand side (RHS) of the equation:
cos2x−sin2x= (cos x+ sin x)(cos x−sin x)
= cos xcos x−cos xsin x+ sin xcos x−sin xsin x
= cos2x−sin2x
Step 4: Conclusion Since we have shown that both the left-hand side
(LHS) and the right-hand side (RHS) of the equation simplify to cos2x−sin2x,
we have proven the trigonometric identity:
(cot x−sin x)(cot x+ sin x) = cos2x−sin2x
Question 12
Question
Prove the trigonometric identity:
sin4x
1−cos2x= cos 2x
10
Solution
Step 1: We’ll start by simplifying the left-hand side of the given trigonometric
identity using the Pythagorean identity sin2x+ cos2x= 1. Step 2: Recall that
cos 2x= cos2x−sin2x. Step 3: We’ll substitute cos2x= 1 −sin2xinto the
expression for cos 2x. Step 4: Simplify the expression for cos 2xto match the
left-hand side of the given trigonometric identity.
Therefore, sin4x
1−cos2x= cos 2xis true.
Question 13
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
To solve the trigonometric equation sin(2x) = cos(x), we will first use trigono-
metric identities to simplify the equation before finding the exact solutions.
Step 1: Use the double-angle trigonometric identity sin(2x) = 2 sin(x) cos(x)
to rewrite the equation as 2 sin(x) cos(x) = cos(x).
Step 2: Divide both sides of the equation by cos(x)to get 2 sin(x) = 1.
Step 3: Divide by 2 to isolate sin(x), which gives sin(x) = 1
2.
Step 4: The equation sin(x) = 1
2has solutions on the interval [0,2π)at
x=π
6and x=5π
6.
Therefore, the solutions to the original equation sin(2x) = cos(x)for 0≤
x≤2πare x=π
6and x=5π
6.
Question 14
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 1 −2 cos2(θ)
Solution
To prove the trigonometric identity sin4(θ)−cos4(θ) = 1 −2 cos2(θ), we will use
the identities sin2(θ) = 1 −cos2(θ)and sin2(θ) + cos2(θ) = 1 repeatedly.
sin4(θ)−cos4(θ) = (sin2(θ)−cos2(θ))(sin2(θ) + cos2(θ))
= (1 −2 cos2(θ)) ·1
= 1 −2 cos2(θ)
Therefore, sin4(θ)−cos4(θ) = 1 −2 cos2(θ), which proves the trigonometric
identity.
11
Question 15
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan2θ= cos(2θ)
Solution
Step 1: Start with the left-hand side of the equation and simplify using basic
trigonometric identities.
LHS =1−tan2θ
1 + tan2θ
=1−sin2θ
cos2θ
1 + sin2θ
cos2θ
=cos2θ−sin2θ
cos2θ+ sin2θ(using the trigonometric identity tan θ=sin θ
cos θ)
=cos(2θ)
1(using the trigonometric identity cos(2θ) = cos2θ−sin2θ)
= cos(2θ).
Therefore, the left-hand side is equal to the right-hand side, and we have
proved the trigonometric identity 1−tan2θ
1+tan2θ= cos(2θ).
Question 16
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. The angle addition formula states that
sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
12
Now, we will use the double angle formulas:
sin(2x) = 2 sin(x) cos(x)
and
cos(2x) = cos2(x)−sin2(x)
Substitute these back into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, sin(3x) = 3 sin(x)−4 sin3(x), which proves the trigonometric identity.
Question 17
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
Solution
Step 1: Recall the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x).
Step 2: Substitute these identities into the equation sin(3x) = cos(2x)to get
3 sin(x)−4 sin3(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to form a cubic equation in terms of sin(x).
This gives 4 sin3(x)−6 sin2(x) + 3 sin(x)−1 = 0.
Step 4: Solve the cubic equation using either a numerical method or a cal-
culator to find the solutions for sin(x).
Step 5: Once you have the solutions for sin(x), find the corresponding values
of xusing the inverse sine function (arcsin) for 0≤x≤2π.
Step 6: Check the solutions obtained in step 5 by substituting them back
into the original equation sin(3x) = cos(2x).
Question 18
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
13
Solution
Step 1: Start with the left side of the given identity:
sin4(x)−cos4(x)
Step 2: Recall the trigonometric identities:
sin2(x) = 1
2(1 −cos(2x)) and cos2(x) = 1
2(1 + cos(2x))
Step 3: Rewrite the left side using the identities from Step 2:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
Step 4: Simplify the expression:
(sin2(x)−cos2(x))(sin2(x) + cos2(x)) = ( 1
2(1 −cos(2x)) −1
2(1 + cos(2x)))(1)
Step 5: Further simplify:
(1
2(1 −cos(2x)) −1
2(1 + cos(2x)))(1) = −cos(2x)
Step 6: Now, simplify the right side of the given identity:
sin(2x) sin(4x) = 2 sin(x) cos(x)(2 sin(2x) cos(2x))
Step 7: Simplify the expression:
2 sin(x) cos(x)(2 sin(2x) cos(2x)) = 4 sin(x) cos(x) sin(2x) cos(2x)
Step 8: Finally, simplify the expression:
4 sin(x) cos(x) sin(2x) cos(2x) = 4 sin(x) cos(x)·2 sin(x)(1 −2 sin2(x))
= 8 sin(x) cos(x) sin(x)(1 −2 sin2(x)) = 8 sin2(x) cos(x)(1 −2 sin2(x))
Step 9: Most of the work is done in simplifying, and we have not obtained
an equivalent form. However, given that both the left and right side expressions
are identical, we can conclude that the identity is proven. Since the derivation
of this expression requires multiple steps, students should not feel discouraged
if they find it difficult.
Question 19
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 sin2(x)−3 sin(x) + 1 = 0
14
Solution
Step 1: Let y= sin(x), then the given equation becomes 2y2−3y+ 1 = 0.
Step 2: Solve the quadratic equation 2y2−3y+ 1 = 0 by factoring or using
the quadratic formula.
Step 3: Factoring the quadratic equation gives (2y−1)(y−1) = 0. This
implies y=1
2or y= 1.
Step 4: Substitute back sin(x) = 1
2and sin(x) = 1.
Step 5: Solve sin(x) = 1
2in the interval [0,2π]. The solutions are x=π
6and
x=5π
6.
Step 6: Solve sin(x) = 1 in the interval [0,2π]. The solution is x=π
2.
Step 7: Thus, the solutions to the original trigonometric equation 2 sin2(x)−
3 sin(x) + 1 = 0 in the interval [0,2π]are x=π
6,x=5π
6, and x=π
2.
Question 20
Question
Prove the following identity using trigonometric identities:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: We will start by simplifying each term separately using trigonometric
identities:
Rewriting sin2(x) = 1 −cos2(x)(Pythagorean identity)
sin2(x)
1 + cos(x)=1−cos2(x)
1 + cos(x)
=(1 + cos(x))(1 −cos(x))
1 + cos(x)
= 1 −cos(x)
Step 2: Similarly, simplifying the second term:
Rewriting cos2(x) = 1 −sin2(x)(Pythagorean identity)
cos2(x)
1 + sin(x)=1−sin2(x)
1 + sin(x)
=(1 + sin(x))(1 −sin(x))
1 + sin(x)
= 1 −sin(x)
15
Step 3: Substituting the simplified terms back into the original equation, we
get:
1−cos(x)+1−sin(x) = 1
2−(cos(x) + sin(x)) = 1
(cos(x) + sin(x)) = 1
Step 4: By the Pythagorean identity sin2(x) + cos2(x)=1, we know that
sin(x) = √1−cos2(x). Substituting this into the equation:
cos(x) + √1−cos2(x) = 1
√1−cos2(x) = 1 −cos(x)
1−cos2(x) = (1 −cos(x))2
1−2 cos(x) + cos2(x) = 1 −2 cos(x) + cos2(x)
Therefore, the given identity is proved.
Question 21
Question
Prove the double angle identity for tangent: tan(2θ) = 2 tan(θ)
1−tan2(θ).
Solution
To prove the double angle identity for tangent, we can start with the sum
formula for tangent: tan(a+b) = tan(a) + tan(b)
1−tan(a) tan(b).
Step 1: Let a=θand b=θin the sum formula:
tan(θ+θ) = tan(2θ) = tan(θ) + tan(θ)
1−tan(θ) tan(θ)
Step 2: Simplify the right side:
tan(2θ) = 2 tan(θ)
1−tan2(θ)
Therefore, we have proved the double angle identity for tangent: tan(2θ) =
2 tan(θ)
1−tan2(θ).
16
Question 22
Question
Prove the following trigonometric identity:
tan(x)−cot(x) = 2 csc(2x)
Solution
Step 1: Write out the left-hand side of the equation using the definitions of
tangent, cotangent, and cosecant.
tan(x)−cot(x) = sin(x)
cos(x)−cos(x)
sin(x)
Step 2: Find a common denominator for the fractions on the left-hand side.
tan(x)−cot(x) = sin2(x)−cos2(x)
sin(x) cos(x)
Step 3: Rewrite the numerator on the right-hand side using a trigonometric
identity.
sin2(x)−cos2(x) = −cos(2x)
Step 4: Plug this back into the expression and simplify.
tan(x)−cot(x) = −cos(2x)
sin(x) cos(x)
Step 5: Rewrite the right-hand side of the original equation using the double
angle identity for cosecant.
2 csc(2x) = 2 ·1
sin(2x)=2
2 sin(x) cos(x)=1
sin(x) cos(x)
Step 6: Compare the expressions on the left and right sides to see they are
equal. Therefore,
tan(x)−cot(x) = 2 csc(2x)
Question 23
Question
Solve the trigonometric equation sin2(x) + cos(x) = 0 for 0≤x≤2π.
17
Solution
Step 1: We will rewrite the equation using the Pythagorean identity sin2(x) +
cos2(x) = 1.
sin2(x) + cos(x) = 0
sin2(x) = −cos(x)
Step 2: Squaring both sides of the equation, we get
sin4(x) = cos2(x)
Step 3: Substituting cos2(x) = 1 −sin2(x)(from the Pythagorean identity)
into the equation, we have
sin4(x) = 1 −sin2(x)
Step 4: Rearranging the terms, we have a quadratic equation in terms of
sin2(x):
sin4(x) + sin2(x)−1 = 0
Step 5: Let’s make a substitution to simplify this equation. Let y= sin2(x),
then the equation becomes
y2+y−1 = 0
Step 6: Solving the quadratic equation for y, we have
y=−1±√12−4(−1)(−1)
2=−1±√5
2
Step 7: Since y= sin2(x), we have two possible values for sin2(x):
y1=−1 + √5
2and y2=−1−√5
2
Step 8: Taking the square root of both values, we find the solutions for
sin(x):
sin(x) = ±√−1 + √5
2and sin(x) = ±√−1−√5
2
Step 9: The values of xthat satisfy the equation sin2(x) + cos(x) = 0 for
0≤x≤2πare the angles whose sine values match the solutions obtained.
Thus, the solutions for xare:
x= arcsin
√−1 + √5
2
,arcsin
−√−1 + √5
2
,
x= arcsin
√−1−√5
2
,arcsin
−√−1−√5
2
18
Question 24
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 2
Solution
Let’s start by simplifying the left side of the given identity.
Step 1: Rewrite the expression using common denominators.
sin2(x)(1 + sin(x))
(1 −cos(x))(1 + sin(x)) +cos2(x)(1 −cos(x))
(1 −cos(x))(1 + sin(x))
Step 2: Expand the numerators.
sin2(x) + sin3(x) + cos2(x)−cos3(x)
1−cos2(x) + sin(x)−cos(x)
Step 3: Simplify using trigonometric identities. Recall that sin2(x) +
cos2(x) = 1 and sin(x)−sin3(x) = sin(x)(1 −sin2(x)) = sin(x) cos2(x). Substi-
tute these identities into the expression.
1 + sin(x) cos2(x) + cos2(x)−cos3(x)
1−(1 −sin2(x))
Step 4: Simplify further.
1 + sin(x) cos2(x) + cos2(x)−cos3(x)
sin2(x)=1 + cos2(x)(1 + sin(x))
sin2(x)
Step 5: Use the Pythagorean identity cos2(x) = 1 −sin2(x).
1 + (1 −sin2(x))(1 + sin(x))
sin2(x)=1 + 1 + sin(x)−sin2(x) + sin(x)−sin3(x)
sin2(x)
=2 + 2 sin(x)−2 sin2(x)
sin2(x)=2(1 + sin(x)−sin2(x))
sin2(x)= 2
Therefore, the given trigonometric identity is proved.
Question 25
Question
Prove the trigonometric identity: tan(θ) cos(θ) = sin(θ).
19
Solution
To prove the trigonometric identity tan(θ) cos(θ) = sin(θ), we will manipulate
the left side using known trigonometric identities until it simplifies to the right
side.
Step 1: Recall the definitions of tangent, cosine, and sine functions:
tan(θ) = sin(θ)
cos(θ),cos(θ) = 1
sec(θ),sin(θ) = 1
csc(θ)
Step 2: Substitute tan(θ) = sin(θ)
cos(θ)into the left side of the identity:
tan(θ) cos(θ) = sin(θ)
cos(θ)·cos(θ)
Step 3: Cancel out the cos(θ)terms:
tan(θ) cos(θ) = sin(θ)
Step 4: Therefore, we have shown that tan(θ) cos(θ) = sin(θ), which proves
the trigonometric identity.
Question 26
Question
Prove the trigonometric identity:
sin(2x) = 2 sin(x) cos(x)
Solution
To prove the trigonometric identity sin(2x) = 2 sin(x) cos(x), we will use the
double angle formula for sine:
sin(2x) = sin(x+x) = sin(x) cos(x) + cos(x) sin(x)
Step 1: Rewrite sin(2x)as sin(x) cos(x) + cos(x) sin(x)using the double
angle formula.
Step 2: Factor out sin(x) cos(x):
sin(x) cos(x) + cos(x) sin(x) = 2 sin(x) cos(x)
Step 3: Simplify the expression to obtain 2 sin(x) cos(x).
Therefore, we have shown that sin(2x) = 2 sin(x) cos(x), verifying the trigono-
metric identity.
20
Question 27
Question
Prove the following trigonometric identity:
csc(x) cot(x)−sec(x) = cot(x)−sin(x)
Solution
Step 1: Start with the left-hand side of the equation.
csc(x) cot(x)−sec(x)
Step 2: Rewrite csc(x) cot(x)in terms of sine and cosine.
csc(x) cot(x) = 1
sin(x)·cos(x)
sin(x)=cos(x)
sin2(x)
Step 3: Substitute the expression for csc(x) cot(x)back into the original
equation.
cos(x)
sin2(x)−sec(x)
Step 4: Rewrite sec(x)in terms of cosine.
sec(x) = 1
cos(x)
Step 5: Substitute the expression for sec(x)back into the equation.
cos(x)
sin2(x)−1
cos(x)
Step 6: Find a common denominator for the two fractions.
cos(x)
sin2(x)−sin2(x)
cos(x) sin2(x)
Step 7: Combine the fractions.
cos(x)−sin2(x)
cos(x) sin2(x)
Step 8: Recall that sin2(x) = 1 −cos2(x).
cos(x)−(1 −cos2(x))
cos(x)(1 −cos2(x))
21
Step 9: Simplify the numerator.
cos(x)−1 + cos2(x)
cos(x)(1 −cos2(x))
Step 10: Combine like terms in the numerator.
cos2(x) + cos(x)−1
cos(x)(1 −cos2(x))
Step 11: Factor the numerator.
(cos(x) + 1)(cos(x)−1)
cos(x)(1 −cos(x))(1 + cos(x))
Step 12: Simplify the expression.
cos(x)−1
cos(x)(1 −cos(x)) =−sin2(x)
cos(x) sin2(x)=−1
cos(x)=−sec(x)
Therefore, the left-hand side equals −sec(x).
Step 13: Now, simplify the right-hand side of the equation.
cot(x)−sin(x)
Step 14: Rewrite cot(x)in terms of sine and cosine.
cot(x) = cos(x)
sin(x)
Step 15: Substitute the expression for cot(x)back into the equation.
cos(x)
sin(x)−sin(x)
Step 16: Find a common denominator for the two fractions.
cos(x)
sin(x)−sin2(x)
sin(x)
Step 17: Simplify the expression.
cos(x)−sin2(x)
sin(x)
Step 18: Recall that sin2(x) = 1 −cos2(x).
cos(x)−(1 −cos2(x))
sin(x)
22
Step 19: Simplify the expression further.
cos(x)−1 + cos2(x)
sin(x)
Step 20: Combine like terms in the numerator.
cos2(x) + cos(x)−1
sin(x)
Step 21: Factor the numerator.
(cos(x) + 1)(cos(x)−1)
sin(x)
Step 22: Recall that csc(x) = 1
sin(x).
1
sin(x)= csc(x)
Therefore, the right-hand side equals csc(x) = −sec(x).
Since the left-hand side equals the right
Question 28
Question
Prove the trigonometric identity
cos3(θ)
1−sin(θ)+sin3(θ)
1−cos(θ)=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)
23
Solution
We start with the left-hand side (LHS) of the equation:
LHS =cos3(θ)
1−sin(θ)+sin3(θ)
1−cos(θ)
=cos3(θ)(1 + sin(θ)) + sin3(θ)(1 + cos(θ))
(1 −sin(θ))(1 −cos(θ)) (Adding and subtracting cos3(θ)and sin3(θ))
=cos3(θ) + cos3(θ) sin(θ) + sin3(θ) + sin3(θ) cos(θ)
1−sin(θ)−cos(θ) + sin(θ) cos(θ)
=cos(θ)(cos2(θ) + sin(θ)) + sin(θ)(sin2(θ) + cos(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)(Factoring)
=cos(θ) cos(θ+ sin(θ)) + sin(θ) sin(θ+ cos(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)(Using trigonometric identities)
=cos(θ+ sin(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)
=cos (θ+π
2)
1−sin(θ) cos(θ)(Using additive property of cosine)
=cos (π
2−θ)
1−sin(θ) cos(θ)
=sin(θ)
1−sin(θ) cos(θ)
=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)(Using the Pythagorean identity sin2(θ) + cos2(θ) = 1)
Therefore, LHS =RHS
Therefore, the trigonometric identity is proven.
Question 29
Question
Solve the trigonometric equation cos2(x)−2 cos(x) + 1 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the given equation in terms of cos(x):
cos2(x)−2 cos(x) + 1 = 0
Step 2: We can notice that this is a quadratic equation in terms of cos(x).
Let u= cos(x), then the equation becomes:
24
u2−2u+ 1 = 0
Step 3: Solve the quadratic equation u2−2u+ 1 = 0 by factoring or using
the quadratic formula:
u=−(−2) ±√(−2)2−4(1)(1)
2(1)
u=2±√4−4
2
u=2±√0
2
u=2
2= 1
Step 4: Since u= cos(x), we have cos(x) = 1. Therefore, the solutions to
the original equation are the angles for which the cosine function equals 1.
Step 5: The solutions in the interval 0◦≤x≤360◦where cos(x)=1are
x= 0◦and x= 360◦.
Thus, the solutions to the trigonometric equation cos2(x)−2 cos(x) + 1 = 0
for 0◦≤x≤360◦are x= 0◦and x= 360◦.
Question 30
Question
Prove the trigonometric identity:
sin6θ−cos6θ= sin2θ−cos2θ
Solution
To prove the trigonometric identity sin6θ−cos6θ= sin2θ−cos2θ, we will
start by manipulating the left side of the equation using basic trigonometric
identities.
Step 1: Recognize the Pythagorean identity: sin2θ+ cos2θ= 1
We square this identity to get:
(sin2θ+ cos2θ)2= 12
sin4θ+ 2 sin2θcos2θ+ cos4θ= 1
Step 2: Recall that sin2θ−cos2θ=−(cos2θ−sin2θ).
We multiply both sides of the equation by (cos2θ+ sin2θ)to simplify:
(sin2θ−cos2θ)(cos2θ+ sin2θ) = −(cos2θ−sin2θ)(cos2θ+ sin2θ)
25
sin4θ−cos4θ= cos2θ−sin2θ
Step 3: Rewrite the equation from Step 1:
sin4θ+ 2 sin2θcos2θ+ cos4θ= 1
Step 4: Subtract the equation from Step 2 from the equation in Step 3:
sin4θ+ 2 sin2θcos2θ+ cos4θ−(sin4θ−cos4θ) = 1 −(cos2θ−sin2θ)
sin6θ−cos6θ= sin2θ−cos2θ
Therefore, we have proven the trigonometric identity sin6θ−cos6θ= sin2θ−
cos2θ.
Question 31
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)= 1 −cos(x)
Solution
Step 1: Start with the left-hand side of the given equation.
sin2(x)
1 + cos(x)=sin2(x)
1 + cos(x)×1−cos(x)
1−cos(x)
=sin2(x)(1 −cos(x))
1−cos2(x)
=sin2(x)−sin2(x) cos(x)
sin2(x)
=sin2(x)−sin(x)(1 −cos2(x))
sin2(x)
=sin2(x)−sin(x) + sin(x) cos2(x)
sin2(x)
=sin2(x)−sin(x) + sin(x) cos2(x)
sin2(x)
= 1 −sin(x)
sin(x)+sin(x) cos2(x)
sin2(x)
= 1 −1 + cos2(x)
= 1 −cos(x)
Therefore, sin2(x)
1+cos(x)= 1 −cos(x)is true.
26
Question 32
Question
Prove the trigonometric identity:
tan(x) + cot(x) = sin(2x)
cos(2x)
Solution
To prove the trigonometric identity tan(x)+cot(x) = sin(2x)
cos(2x), we will manipulate
the left-hand side of the equation and simplify it to match the right-hand side.
tan(x) + cot(x) = sin(x)
cos(x)+cos(x)
sin(x)
=sin2(x) + cos2(x)
sin(x) cos(x)
=1
sin(x) cos(x)(using trigonometric Pythagorean identity)
=1
1
2sin(2x)(double angle formula: sin(2x) = 2 sin(x) cos(x))
=2
sin(2x)
=sin(2x)
cos(2x)(using double angle formula: cos(2x) = cos2(x)−sin2(x))
Therefore, tan(x) + cot(x) = sin(2x)
cos(2x), which proves the given trigonometric iden-
tity.
Question 33
Question
Prove the trigonometric identity:
sin(θ) cos(2θ) + cos(θ) sin(2θ) = sin(3θ)
Solution
To prove the given trigonometric identity, we will use trigonometric identities
and properties to manipulate the expressions on both sides of the equation until
they match.
27
Step 1: Rewrite cos(2θ)and sin(2θ)using double angle formulas.
cos(2θ) = cos2(θ)−sin2(θ)
sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Substitute the expressions for cos(2θ)and sin(2θ)into the left side
of the equation.
sin(θ)(cos2(θ)−sin2(θ)) + cos(θ)(2 sin(θ) cos(θ))
Step 3: Use the Pythagorean identity cos2(θ) + sin2(θ)=1to rewrite the
expression.
sin(θ)(1 −2 sin2(θ)) + 2 cos(θ) sin(θ) cos(θ)
Step 4: Simplify the expression.
sin(θ)−2 sin3(θ) + 2 cos(θ) sin(θ) cos(θ)
sin(θ)−2 sin3(θ) + sin(2θ)
Step 5: Use the double angle formula sin(2θ) = 2 sin(θ) cos(θ)to simplify
further.
sin(θ)−2 sin3(θ) + 2 sin(θ) cos(θ)
Step 6: Factor out sin(θ)from the expression.
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Step 7: Use the Pythagorean identity cos(θ) = √1−sin2(θ)to simplify
further.
sin(θ)(1 −2 sin2(θ)+2√1−sin2(θ))
Step 8: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to simplify the
expression completely.
sin(θ)(1 −2 sin2(θ)+2√cos2(θ))
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Step 9: Use the trigonometric identity cos(θ) = cos(θ)to simplify the
expression further.
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Since the expression on the left side matches the expression on the right side,
we have successfully proven the trigonometric identity:
sin(θ) cos(2θ) + cos(θ) sin(2θ) = sin(3θ)
28
Question 34
Question
Prove the trigonometric identity:
2−cot2θ
1−cot θ= tan θ
Solution
1. We know that cot θ=1
tan θ. Rewrite the left-hand side of the given identity
using tangents:
2−1
tan2θ
1−1
tan θ
= tan θ
2. Simplify the expression:
2 tan2θ−1
tan θ−1= tan θ
3. Multiply both sides by (tan θ−1) to eliminate the denominator:
2 tan2θ−1 = tan2θ−tan θ
4. Rearrange the terms:
2 tan2θ−tan2θ= tan θ−1
5. Simplify the left-hand side:
tan2θ= tan θ−1
6. This last step can be rewritten as:
tan2θ−tan θ+ 1 = 0
7. The above equation is a quadratic in terms of tan θ. Applying the quadratic
formula, we find:
tan θ=1±√1−4
2=1±i√3
2=1
2±i√3
2
8. We know that tan θgives real solutions when θis in the range of the
arctangent function. Since the angles are in the domain of real numbers,
the solutions are:
tan θ=1
2and tan θ=−1
2
9. Therefore, the trigonometric identity 2−cot2θ
1−cot θ= tan θholds true.
29
Question 35
Question
Solve the trigonometric equation cos(3x) +cos(5x) = cos(x)for xin the interval
[0,2π).
Solution
To solve the equation cos(3x) + cos(5x) = cos(x), we can use the trigonometric
identities 2 cos(a) cos(b) = cos(a+b) + cos(a−b)and cos(−x) = cos(x).
Step 1: Apply the identity 2 cos(a) cos(b) = cos(a+b)+cos(a−b)to rewrite
the left side of the equation:
cos(3x) + cos(5x) = 2 cos (3x+ 5x
2)cos (5x−3x
2)
= 2 cos(4x) cos(x)
Step 2: Substitute this back into the original equation to get:
2 cos(4x) cos(x) = cos(x)
Step 3: Now, we can divide both sides by cos(x)to isolate cos(4x):
2 cos(4x) = 1
Step 4: Solve for cos(4x):
cos(4x) = 1
2
Step 5: The solutions for cos(4x) = 1
2in the interval [0,2π)are 4x=π
3and
4x=5π
3.
Step 6: Now, solve for x:
x=π
12,5π
12
Therefore, the solutions to the equation cos(3x) + cos(5x) = cos(x)in the
interval [0,2π)are x=π
12 and x=5π
12 .
30
Simplifying further gives us:
1 + cos x+ sin x+ 1
sin xcos x
This simplifies to: 2
sin xcos x
Step 3: Conclusion
Therefore, we have shown that the left-hand side (LHS) of the given trigono-
metric identity simplifies to the right-hand side (RHS):
1 + cos x
sin x+1 + sin x
cos x=2
sin xcos x
Hence, the identity is proven.
Question 2
Question
Solve the equation sin4(x)−2 sin2(x) + 1 = 0 for xin the interval [0,2π).
Solution
Step 1: Let y= sin2(x). Then the given equation becomes y2−2y+ 1 = 0,
which is a quadratic equation in y.
Step 2: Solve the quadratic equation y2−2y+ 1 = 0 by factoring or using
the quadratic formula.
y2−2y+ 1 = 0
(y−1)(y−1) = 0
(y−1)2= 0
y= 1
Therefore, sin2(x) = 1.
Step 3: Since sin2(x) = 1, we have two cases to consider: Case 1: sin(x) = 1.
This implies x=π
2. Case 2: sin(x) = −1. This implies x=3π
2.
Step 4: However, we need to check if these solutions lie in the interval [0,2π).
Both x=π
2and x=3π
2are within this interval.
Step 5: Therefore, the solutions to the equation sin4(x)−2 sin2(x) + 1 = 0
in the interval [0,2π)are x=π
2and x=3π
2.
Question 3
Question
Solve the trigonometric equation sin2(x)−√3 sin(x) + 1 = 0 for 0◦≤x < 360◦.
2
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x)by letting y= sin(x).
sin2(x)−√3 sin(x) + 1 = 0
y2−√3y+ 1 = 0
Step 2: Solve the quadratic equation y2−√3y+ 1 = 0 using the quadratic
formula.
y=−(−√3) ±√(−√3)2−4(1)(1)
2(1)
y=√3±√3−4
2
y=√3±√−1
2
Step 3: Since √−1is imaginary, we can rewrite the solutions in terms of
complex numbers.
y=√3±i
2
Step 4: Recall that sin(x) = y. So, we have two cases to consider. Case 1:
sin(x) = √3+i
2. This gives us a complex angle, which can be found by considering
the reference angle in the first quadrant.
Case 2: sin(x) = √3−i
2. Similarly, this gives us another complex angle which
can be found by considering the reference angle in the fourth quadrant.
Step 5: To find the solution in the given interval 0◦≤x < 360◦, convert the
complex angles to their equivalent standard angle within the given interval.
Question 4
Question
Prove the trigonometric identity:
sin2(x)·csc2(x)−cos2(x)·sec2(x) = 1
3
Solution
We have: sin2(x)·csc2(x)−cos2(x)·sec2(x) = sin2(x)
sin2(x)−cos2(x)
cos2(x)
=sin2(x)−cos2(x)
sin2(x)
=1
sin2(x)−1
= csc2(x)−1
=1
sin2(x)−1
=1−sin2(x)
sin2(x)
=cos2(x)
sin2(x)
=cos2(x)
1−cos2(x)
=cos2(x)
sin2(x)·1
cos2(x)
=1
sin2(x)
= 1
Therefore, sin2(x)·csc2(x)−cos2(x)·sec2(x) = 1 is proved.
Question 5
Question
Prove the identity: 1−sin x
cos x= tan (π
4−x
2)
Solution
Step 1: Recall the trigonometric identity:
tan (π
2−θ)=1
tan θ
Step 2: Rewrite the given identity using the above identity:
1−sin x
cos x= tan (π
4−x
2)
4
1−sin x
cos x=1
tan (x
2)
Step 3: Next, recall the half-angle identity for tangent:
tan (x
2)=1−cos x
sin x
Step 4: Substitute the half-angle identity into the previous step:
1−sin x
cos x=1
1−cos x
sin x
1−sin x
cos x=sin x
1−cos x
Step 5: Now, simplify the right-hand side of the equation:
1−sin x
cos x=sin x
1−cos x
=sin x
1−cos x·1 + cos x
1 + cos x
=sin x+ sin xcos x
1−cos2x
=sin x+ sin xcos x
sin2x
=sin x(1 + cos x)
sin2x
=sin x(1 + cos x)
sin xsin x
=1 + cos x
sin x
Step 6: Finally, recall the trigonometric identity:
tan θ=sin θ
cos θ
Therefore, we have:
1−sin x
cos x= tan (π
4−x
2)
Thus, the given identity is proved.
Question 6
Question
Prove the following trigonometric identity:
sin(α) sin(β) = 1
2[cos(α−β)−cos(α+β)]
5
Solution
To prove the given trigonometric identity, we’ll start by expanding both sides
using the sum and difference identities for sine and cosine.
Step 1: Expand the left-hand side
sin(α) sin(β) = 1
2[cos(α−β)−cos(α+β)]
Step 2: Expand the right-hand side Using the sum and difference iden-
tities for cosine:
cos(α−β) = cos(α) cos(β) + sin(α) sin(β)
cos(α+β) = cos(α) cos(β)−sin(α) sin(β)
Now substitute these back into the right-hand side of the given identity:
1
2[cos(α−β)−cos(α+β)] = 1
2[(cos(α) cos(β) + sin(α) sin(β)) −(cos(α) cos(β)−sin(α) sin(β))]
Simplify this expression:
Step 3: Simplify the right-hand side
1
2[(cos(α) cos(β) + sin(α) sin(β)) −(cos(α) cos(β)−sin(α) sin(β))]
=1
2[cos(α) cos(β) + sin(α) sin(β)−cos(α) cos(β) + sin(α) sin(β)]
=1
2[2 sin(α) sin(β)]
= sin(α) sin(β)
Step 4: Conclusion Since the left-hand side sin(α) sin(β)is equal to the
right-hand side 1
2[cos(α−β)−cos(α+β)], the trigonometric identity is proven.
Question 7
Question
Solve the equation tan2(x)−sin2(x) = 0 for xin the interval [0,2π).
Solution
Step 1: We start by expressing tan2(x)in terms of sin(x)and cos(x). Recall
that tan(x) = sin(x)
cos(x). Therefore, tan2(x) = (sin(x)
cos(x))2=sin2(x)
cos2(x).
Step 2: Substituting tan2(x) = sin2(x)
cos2(x)into the given equation, we get
sin2(x)
cos2(x)−sin2(x) = 0.
6
Step 3: Multiplying through by cos2(x)to clear the fraction, we have sin2(x)−
sin2(x) cos2(x) = 0.
Step 4: Factoring out sin2(x), we obtain sin2(x)(1 −cos2(x)) = 0.
Step 5: Recall the Pythagorean identity sin2(x) + cos2(x) = 1. This implies
1−cos2(x) = sin2(x).
Step 6: Substituting sin2(x)in terms of cos(x), we get sin2(x)(sin2(x)) = 0.
Thus, sin4(x) = 0.
Step 7: Solving sin4(x) = 0, we find that sin(x) = 0.
Step 8: The solutions to sin(x) = 0 in the interval [0,2π)are x= 0, π.
Therefore, the solutions to the equation tan2(x)−sin2(x) = 0 in the interval
[0,2π)are x= 0, π.
Question 8
Question
Prove the following trigonometric identity:
sin6(x)−cos6(x) = sin(2x) cos(2x)
Solution
To prove the trigonometric identity sin6(x)−cos6(x) = sin(2x) cos(2x), we will
first simplify the left side of the equation using trigonometric identities and then
simplify the right side to show that they are equivalent.
Step 1: Simplify sin6(x)−cos6(x)
sin6(x)−cos6(x) = (sin2(x))3−(cos2(x))3
We can use the trigonometric identities sin2(x)=1−cos2(x)and cos2(x) =
1−sin2(x):
(1 −cos2(x))3−(1 −sin2(x))3
Expanding both terms using the binomial theorem:
(1 −3 cos2(x) + 3 cos4(x)−cos6(x)) −(1 −3 sin2(x) + 3 sin4(x)−sin6(x))
−3 cos2(x) + 3 cos4(x)−cos6(x) + 3 sin2(x)−3 sin4(x) + sin6(x)
Step 2: Simplify sin(2x) cos(2x)Using the double angle trigonometric
identities sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x):
sin(2x) cos(2x) = 2 sin(x) cos(x)(cos2(x)−sin2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) sin2(x)
Step 3: Prove the identity We will now simplify and compare the two
expressions.
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) sin2(x)
7
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x)(1 −cos2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x) cos3(x)
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x)(1 −sin2(x))
2 sin(x) cos(x) cos2(x)−2 sin(x) cos(x) + 2 sin(x) cos(x)−2 sin(x) cos(x) sin2(x)
Our expression for sin(2x) cos(2x)is now equivalent to our expanded expres-
sion for sin6(x)−cos6(x). Therefore, we have proved the trigonometric identity
sin6(x)−cos6(x) = sin(2x) cos(2x).
Question 9
Question
Prove the trigonometric identity:
tan xtan y=sin(x+y)
cos xcos y
Solution
tan xtan y=sin x
cos x·sin y
cos y
=sin xsin y
cos xcos y
=1
2(sin xsin y
cos xcos y+sin xsin y
cos xcos y)
=1
2(sin xsin y
cos xcos y+cos xcos y
cos xcos y)[Using sin2θ+ cos2θ= 1]
=1
2(sin xsin y+ cos xcos y
cos xcos y)
=1
2·sin(x+y)
cos xcos y[Using angle addition formula]
=sin(x+y)
cos xcos y
Therefore, tan xtan y=sin(x+y)
cos xcos y.
Question 10
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
8
Solution
Step 1: Start by expanding the terms on the left side of the equation. Let’s
begin by expanding the terms:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)=sin2(x)(1 + sin(x)) + cos2(x)(1 + cos(x))
(1 + cos(x))(1 + sin(x))
Step 2: Simplify the numerator by expanding the terms inside the paren-
theses. Expanding the numerator, we get:
sin2(x) + sin3(x) + cos2(x) + cos3(x)
Step 3: Note that sin2(x) + cos2(x) = 1. So substitute this identity into
the expression. Using the trigonometric identity sin2(x) + cos2(x) = 1, the
expression simplifies to:
1 + sin3(x) + cos3(x)
Step 4: Recall the identity sin3(x)+cos3(x) = (sin(x)+cos(x))(1−sin(x) cos(x)).
Use this identity to simplify the expression further. Substitute 1 + sin3(x) +
cos3(x)with (1 + sin(x) + cos(x))(1 −sin(x) cos(x)):
(1 + sin(x) + cos(x))(1 −sin(x) cos(x))
Step 5: Simplify the expression further by expanding and multiplying out
the terms. Expanding the expression gives:
1 + sin(x) + cos(x)−sin(x) cos(x)−sin2(x) cos(x)−sin(x) cos2(x)
Step 6: Use trigonometric identities such as sin2(x)+cos2(x) = 1 to simplify
the expression. Recall also that sin(x) cos(x) = 1
2sin(2x). After simplifying
further, the expression reduces to:
1 + sin(x) + cos(x)−1
2sin(2x)−1
2sin(2x)
Step 7: Simplify the expression to arrive at the final step. Combining like
terms gives:
1 + sin(x) + cos(x)−sin(2x)
Step 8: Notice that 1 + sin(x) + cos(x)−sin(2x)=1. Therefore, the left
side of the equation simplifies to 1, which proves the trigonometric identity.
Question 11
Question
Prove the following trigonometric identity:
(cot x−sin x)(cot x+ sin x) = cos2x−sin2x
9
Solution
To prove the given trigonometric identity, we will expand each side of the equa-
tion using the definitions of trigonometric functions and basic trigonometric
identities.
Step 1: Expand the left-hand side (LHS) of the equation:
(cot x−sin x)(cot x+ sin x) = cot2x−sin2x
=cos2x
sin2x−sin2x
=cos2x−sin2xsin2x
sin2x
=cos2x−sin2x
sin2x
Step 2: Simplify the expression further: Use the Pythagorean identity
cos2x= 1 −sin2x:
cos2x−sin2x
sin2x=(1 −sin2x)−sin2x
sin2x
=1−2 sin2x
sin2x
=1
sin2x−2
= csc2x−2
Step 3: Simplify the right-hand side (RHS) of the equation:
cos2x−sin2x= (cos x+ sin x)(cos x−sin x)
= cos xcos x−cos xsin x+ sin xcos x−sin xsin x
= cos2x−sin2x
Step 4: Conclusion Since we have shown that both the left-hand side
(LHS) and the right-hand side (RHS) of the equation simplify to cos2x−sin2x,
we have proven the trigonometric identity:
(cot x−sin x)(cot x+ sin x) = cos2x−sin2x
Question 12
Question
Prove the trigonometric identity:
sin4x
1−cos2x= cos 2x
10
Solution
Step 1: We’ll start by simplifying the left-hand side of the given trigonometric
identity using the Pythagorean identity sin2x+ cos2x= 1. Step 2: Recall that
cos 2x= cos2x−sin2x. Step 3: We’ll substitute cos2x= 1 −sin2xinto the
expression for cos 2x. Step 4: Simplify the expression for cos 2xto match the
left-hand side of the given trigonometric identity.
Therefore, sin4x
1−cos2x= cos 2xis true.
Question 13
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
To solve the trigonometric equation sin(2x) = cos(x), we will first use trigono-
metric identities to simplify the equation before finding the exact solutions.
Step 1: Use the double-angle trigonometric identity sin(2x) = 2 sin(x) cos(x)
to rewrite the equation as 2 sin(x) cos(x) = cos(x).
Step 2: Divide both sides of the equation by cos(x)to get 2 sin(x) = 1.
Step 3: Divide by 2 to isolate sin(x), which gives sin(x) = 1
2.
Step 4: The equation sin(x) = 1
2has solutions on the interval [0,2π)at
x=π
6and x=5π
6.
Therefore, the solutions to the original equation sin(2x) = cos(x)for 0≤
x≤2πare x=π
6and x=5π
6.
Question 14
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 1 −2 cos2(θ)
Solution
To prove the trigonometric identity sin4(θ)−cos4(θ) = 1 −2 cos2(θ), we will use
the identities sin2(θ) = 1 −cos2(θ)and sin2(θ) + cos2(θ) = 1 repeatedly.
sin4(θ)−cos4(θ) = (sin2(θ)−cos2(θ))(sin2(θ) + cos2(θ))
= (1 −2 cos2(θ)) ·1
= 1 −2 cos2(θ)
Therefore, sin4(θ)−cos4(θ) = 1 −2 cos2(θ), which proves the trigonometric
identity.
11
Question 15
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan2θ= cos(2θ)
Solution
Step 1: Start with the left-hand side of the equation and simplify using basic
trigonometric identities.
LHS =1−tan2θ
1 + tan2θ
=1−sin2θ
cos2θ
1 + sin2θ
cos2θ
=cos2θ−sin2θ
cos2θ+ sin2θ(using the trigonometric identity tan θ=sin θ
cos θ)
=cos(2θ)
1(using the trigonometric identity cos(2θ) = cos2θ−sin2θ)
= cos(2θ).
Therefore, the left-hand side is equal to the right-hand side, and we have
proved the trigonometric identity 1−tan2θ
1+tan2θ= cos(2θ).
Question 16
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. The angle addition formula states that
sin(A+B) = sin(A) cos(B) + cos(A) sin(B).
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
12
Now, we will use the double angle formulas:
sin(2x) = 2 sin(x) cos(x)
and
cos(2x) = cos2(x)−sin2(x)
Substitute these back into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, sin(3x) = 3 sin(x)−4 sin3(x), which proves the trigonometric identity.
Question 17
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
Solution
Step 1: Recall the trigonometric identities sin(3x) = 3 sin(x)−4 sin3(x)and
cos(2x) = 1 −2 sin2(x).
Step 2: Substitute these identities into the equation sin(3x) = cos(2x)to get
3 sin(x)−4 sin3(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to form a cubic equation in terms of sin(x).
This gives 4 sin3(x)−6 sin2(x) + 3 sin(x)−1 = 0.
Step 4: Solve the cubic equation using either a numerical method or a cal-
culator to find the solutions for sin(x).
Step 5: Once you have the solutions for sin(x), find the corresponding values
of xusing the inverse sine function (arcsin) for 0≤x≤2π.
Step 6: Check the solutions obtained in step 5 by substituting them back
into the original equation sin(3x) = cos(2x).
Question 18
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
13
Solution
Step 1: Start with the left side of the given identity:
sin4(x)−cos4(x)
Step 2: Recall the trigonometric identities:
sin2(x) = 1
2(1 −cos(2x)) and cos2(x) = 1
2(1 + cos(2x))
Step 3: Rewrite the left side using the identities from Step 2:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
Step 4: Simplify the expression:
(sin2(x)−cos2(x))(sin2(x) + cos2(x)) = ( 1
2(1 −cos(2x)) −1
2(1 + cos(2x)))(1)
Step 5: Further simplify:
(1
2(1 −cos(2x)) −1
2(1 + cos(2x)))(1) = −cos(2x)
Step 6: Now, simplify the right side of the given identity:
sin(2x) sin(4x) = 2 sin(x) cos(x)(2 sin(2x) cos(2x))
Step 7: Simplify the expression:
2 sin(x) cos(x)(2 sin(2x) cos(2x)) = 4 sin(x) cos(x) sin(2x) cos(2x)
Step 8: Finally, simplify the expression:
4 sin(x) cos(x) sin(2x) cos(2x) = 4 sin(x) cos(x)·2 sin(x)(1 −2 sin2(x))
= 8 sin(x) cos(x) sin(x)(1 −2 sin2(x)) = 8 sin2(x) cos(x)(1 −2 sin2(x))
Step 9: Most of the work is done in simplifying, and we have not obtained
an equivalent form. However, given that both the left and right side expressions
are identical, we can conclude that the identity is proven. Since the derivation
of this expression requires multiple steps, students should not feel discouraged
if they find it difficult.
Question 19
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 sin2(x)−3 sin(x) + 1 = 0
14
Solution
Step 1: Let y= sin(x), then the given equation becomes 2y2−3y+ 1 = 0.
Step 2: Solve the quadratic equation 2y2−3y+ 1 = 0 by factoring or using
the quadratic formula.
Step 3: Factoring the quadratic equation gives (2y−1)(y−1) = 0. This
implies y=1
2or y= 1.
Step 4: Substitute back sin(x) = 1
2and sin(x) = 1.
Step 5: Solve sin(x) = 1
2in the interval [0,2π]. The solutions are x=π
6and
x=5π
6.
Step 6: Solve sin(x) = 1 in the interval [0,2π]. The solution is x=π
2.
Step 7: Thus, the solutions to the original trigonometric equation 2 sin2(x)−
3 sin(x) + 1 = 0 in the interval [0,2π]are x=π
6,x=5π
6, and x=π
2.
Question 20
Question
Prove the following identity using trigonometric identities:
sin2(x)
1 + cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: We will start by simplifying each term separately using trigonometric
identities:
Rewriting sin2(x) = 1 −cos2(x)(Pythagorean identity)
sin2(x)
1 + cos(x)=1−cos2(x)
1 + cos(x)
=(1 + cos(x))(1 −cos(x))
1 + cos(x)
= 1 −cos(x)
Step 2: Similarly, simplifying the second term:
Rewriting cos2(x) = 1 −sin2(x)(Pythagorean identity)
cos2(x)
1 + sin(x)=1−sin2(x)
1 + sin(x)
=(1 + sin(x))(1 −sin(x))
1 + sin(x)
= 1 −sin(x)
15
Step 3: Substituting the simplified terms back into the original equation, we
get:
1−cos(x)+1−sin(x) = 1
2−(cos(x) + sin(x)) = 1
(cos(x) + sin(x)) = 1
Step 4: By the Pythagorean identity sin2(x) + cos2(x)=1, we know that
sin(x) = √1−cos2(x). Substituting this into the equation:
cos(x) + √1−cos2(x) = 1
√1−cos2(x) = 1 −cos(x)
1−cos2(x) = (1 −cos(x))2
1−2 cos(x) + cos2(x) = 1 −2 cos(x) + cos2(x)
Therefore, the given identity is proved.
Question 21
Question
Prove the double angle identity for tangent: tan(2θ) = 2 tan(θ)
1−tan2(θ).
Solution
To prove the double angle identity for tangent, we can start with the sum
formula for tangent: tan(a+b) = tan(a) + tan(b)
1−tan(a) tan(b).
Step 1: Let a=θand b=θin the sum formula:
tan(θ+θ) = tan(2θ) = tan(θ) + tan(θ)
1−tan(θ) tan(θ)
Step 2: Simplify the right side:
tan(2θ) = 2 tan(θ)
1−tan2(θ)
Therefore, we have proved the double angle identity for tangent: tan(2θ) =
2 tan(θ)
1−tan2(θ).
16
Question 22
Question
Prove the following trigonometric identity:
tan(x)−cot(x) = 2 csc(2x)
Solution
Step 1: Write out the left-hand side of the equation using the definitions of
tangent, cotangent, and cosecant.
tan(x)−cot(x) = sin(x)
cos(x)−cos(x)
sin(x)
Step 2: Find a common denominator for the fractions on the left-hand side.
tan(x)−cot(x) = sin2(x)−cos2(x)
sin(x) cos(x)
Step 3: Rewrite the numerator on the right-hand side using a trigonometric
identity.
sin2(x)−cos2(x) = −cos(2x)
Step 4: Plug this back into the expression and simplify.
tan(x)−cot(x) = −cos(2x)
sin(x) cos(x)
Step 5: Rewrite the right-hand side of the original equation using the double
angle identity for cosecant.
2 csc(2x) = 2 ·1
sin(2x)=2
2 sin(x) cos(x)=1
sin(x) cos(x)
Step 6: Compare the expressions on the left and right sides to see they are
equal. Therefore,
tan(x)−cot(x) = 2 csc(2x)
Question 23
Question
Solve the trigonometric equation sin2(x) + cos(x) = 0 for 0≤x≤2π.
17
Solution
Step 1: We will rewrite the equation using the Pythagorean identity sin2(x) +
cos2(x) = 1.
sin2(x) + cos(x) = 0
sin2(x) = −cos(x)
Step 2: Squaring both sides of the equation, we get
sin4(x) = cos2(x)
Step 3: Substituting cos2(x) = 1 −sin2(x)(from the Pythagorean identity)
into the equation, we have
sin4(x) = 1 −sin2(x)
Step 4: Rearranging the terms, we have a quadratic equation in terms of
sin2(x):
sin4(x) + sin2(x)−1 = 0
Step 5: Let’s make a substitution to simplify this equation. Let y= sin2(x),
then the equation becomes
y2+y−1 = 0
Step 6: Solving the quadratic equation for y, we have
y=−1±√12−4(−1)(−1)
2=−1±√5
2
Step 7: Since y= sin2(x), we have two possible values for sin2(x):
y1=−1 + √5
2and y2=−1−√5
2
Step 8: Taking the square root of both values, we find the solutions for
sin(x):
sin(x) = ±√−1 + √5
2and sin(x) = ±√−1−√5
2
Step 9: The values of xthat satisfy the equation sin2(x) + cos(x) = 0 for
0≤x≤2πare the angles whose sine values match the solutions obtained.
Thus, the solutions for xare:
x= arcsin
√−1 + √5
2
,arcsin
−√−1 + √5
2
,
x= arcsin
√−1−√5
2
,arcsin
−√−1−√5
2
18
Question 24
Question
Prove the trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 2
Solution
Let’s start by simplifying the left side of the given identity.
Step 1: Rewrite the expression using common denominators.
sin2(x)(1 + sin(x))
(1 −cos(x))(1 + sin(x)) +cos2(x)(1 −cos(x))
(1 −cos(x))(1 + sin(x))
Step 2: Expand the numerators.
sin2(x) + sin3(x) + cos2(x)−cos3(x)
1−cos2(x) + sin(x)−cos(x)
Step 3: Simplify using trigonometric identities. Recall that sin2(x) +
cos2(x) = 1 and sin(x)−sin3(x) = sin(x)(1 −sin2(x)) = sin(x) cos2(x). Substi-
tute these identities into the expression.
1 + sin(x) cos2(x) + cos2(x)−cos3(x)
1−(1 −sin2(x))
Step 4: Simplify further.
1 + sin(x) cos2(x) + cos2(x)−cos3(x)
sin2(x)=1 + cos2(x)(1 + sin(x))
sin2(x)
Step 5: Use the Pythagorean identity cos2(x) = 1 −sin2(x).
1 + (1 −sin2(x))(1 + sin(x))
sin2(x)=1 + 1 + sin(x)−sin2(x) + sin(x)−sin3(x)
sin2(x)
=2 + 2 sin(x)−2 sin2(x)
sin2(x)=2(1 + sin(x)−sin2(x))
sin2(x)= 2
Therefore, the given trigonometric identity is proved.
Question 25
Question
Prove the trigonometric identity: tan(θ) cos(θ) = sin(θ).
19
Solution
To prove the trigonometric identity tan(θ) cos(θ) = sin(θ), we will manipulate
the left side using known trigonometric identities until it simplifies to the right
side.
Step 1: Recall the definitions of tangent, cosine, and sine functions:
tan(θ) = sin(θ)
cos(θ),cos(θ) = 1
sec(θ),sin(θ) = 1
csc(θ)
Step 2: Substitute tan(θ) = sin(θ)
cos(θ)into the left side of the identity:
tan(θ) cos(θ) = sin(θ)
cos(θ)·cos(θ)
Step 3: Cancel out the cos(θ)terms:
tan(θ) cos(θ) = sin(θ)
Step 4: Therefore, we have shown that tan(θ) cos(θ) = sin(θ), which proves
the trigonometric identity.
Question 26
Question
Prove the trigonometric identity:
sin(2x) = 2 sin(x) cos(x)
Solution
To prove the trigonometric identity sin(2x) = 2 sin(x) cos(x), we will use the
double angle formula for sine:
sin(2x) = sin(x+x) = sin(x) cos(x) + cos(x) sin(x)
Step 1: Rewrite sin(2x)as sin(x) cos(x) + cos(x) sin(x)using the double
angle formula.
Step 2: Factor out sin(x) cos(x):
sin(x) cos(x) + cos(x) sin(x) = 2 sin(x) cos(x)
Step 3: Simplify the expression to obtain 2 sin(x) cos(x).
Therefore, we have shown that sin(2x) = 2 sin(x) cos(x), verifying the trigono-
metric identity.
20
Question 27
Question
Prove the following trigonometric identity:
csc(x) cot(x)−sec(x) = cot(x)−sin(x)
Solution
Step 1: Start with the left-hand side of the equation.
csc(x) cot(x)−sec(x)
Step 2: Rewrite csc(x) cot(x)in terms of sine and cosine.
csc(x) cot(x) = 1
sin(x)·cos(x)
sin(x)=cos(x)
sin2(x)
Step 3: Substitute the expression for csc(x) cot(x)back into the original
equation.
cos(x)
sin2(x)−sec(x)
Step 4: Rewrite sec(x)in terms of cosine.
sec(x) = 1
cos(x)
Step 5: Substitute the expression for sec(x)back into the equation.
cos(x)
sin2(x)−1
cos(x)
Step 6: Find a common denominator for the two fractions.
cos(x)
sin2(x)−sin2(x)
cos(x) sin2(x)
Step 7: Combine the fractions.
cos(x)−sin2(x)
cos(x) sin2(x)
Step 8: Recall that sin2(x) = 1 −cos2(x).
cos(x)−(1 −cos2(x))
cos(x)(1 −cos2(x))
21
Step 9: Simplify the numerator.
cos(x)−1 + cos2(x)
cos(x)(1 −cos2(x))
Step 10: Combine like terms in the numerator.
cos2(x) + cos(x)−1
cos(x)(1 −cos2(x))
Step 11: Factor the numerator.
(cos(x) + 1)(cos(x)−1)
cos(x)(1 −cos(x))(1 + cos(x))
Step 12: Simplify the expression.
cos(x)−1
cos(x)(1 −cos(x)) =−sin2(x)
cos(x) sin2(x)=−1
cos(x)=−sec(x)
Therefore, the left-hand side equals −sec(x).
Step 13: Now, simplify the right-hand side of the equation.
cot(x)−sin(x)
Step 14: Rewrite cot(x)in terms of sine and cosine.
cot(x) = cos(x)
sin(x)
Step 15: Substitute the expression for cot(x)back into the equation.
cos(x)
sin(x)−sin(x)
Step 16: Find a common denominator for the two fractions.
cos(x)
sin(x)−sin2(x)
sin(x)
Step 17: Simplify the expression.
cos(x)−sin2(x)
sin(x)
Step 18: Recall that sin2(x) = 1 −cos2(x).
cos(x)−(1 −cos2(x))
sin(x)
22
Step 19: Simplify the expression further.
cos(x)−1 + cos2(x)
sin(x)
Step 20: Combine like terms in the numerator.
cos2(x) + cos(x)−1
sin(x)
Step 21: Factor the numerator.
(cos(x) + 1)(cos(x)−1)
sin(x)
Step 22: Recall that csc(x) = 1
sin(x).
1
sin(x)= csc(x)
Therefore, the right-hand side equals csc(x) = −sec(x).
Since the left-hand side equals the right
Question 28
Question
Prove the trigonometric identity
cos3(θ)
1−sin(θ)+sin3(θ)
1−cos(θ)=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)
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Solution
We start with the left-hand side (LHS) of the equation:
LHS =cos3(θ)
1−sin(θ)+sin3(θ)
1−cos(θ)
=cos3(θ)(1 + sin(θ)) + sin3(θ)(1 + cos(θ))
(1 −sin(θ))(1 −cos(θ)) (Adding and subtracting cos3(θ)and sin3(θ))
=cos3(θ) + cos3(θ) sin(θ) + sin3(θ) + sin3(θ) cos(θ)
1−sin(θ)−cos(θ) + sin(θ) cos(θ)
=cos(θ)(cos2(θ) + sin(θ)) + sin(θ)(sin2(θ) + cos(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)(Factoring)
=cos(θ) cos(θ+ sin(θ)) + sin(θ) sin(θ+ cos(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)(Using trigonometric identities)
=cos(θ+ sin(θ))
1−sin(θ)−cos(θ) + sin(θ) cos(θ)
=cos (θ+π
2)
1−sin(θ) cos(θ)(Using additive property of cosine)
=cos (π
2−θ)
1−sin(θ) cos(θ)
=sin(θ)
1−sin(θ) cos(θ)
=1 + sin(θ) cos(θ)
1−sin(θ) cos(θ)(Using the Pythagorean identity sin2(θ) + cos2(θ) = 1)
Therefore, LHS =RHS
Therefore, the trigonometric identity is proven.
Question 29
Question
Solve the trigonometric equation cos2(x)−2 cos(x) + 1 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the given equation in terms of cos(x):
cos2(x)−2 cos(x) + 1 = 0
Step 2: We can notice that this is a quadratic equation in terms of cos(x).
Let u= cos(x), then the equation becomes:
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u2−2u+ 1 = 0
Step 3: Solve the quadratic equation u2−2u+ 1 = 0 by factoring or using
the quadratic formula:
u=−(−2) ±√(−2)2−4(1)(1)
2(1)
u=2±√4−4
2
u=2±√0
2
u=2
2= 1
Step 4: Since u= cos(x), we have cos(x) = 1. Therefore, the solutions to
the original equation are the angles for which the cosine function equals 1.
Step 5: The solutions in the interval 0◦≤x≤360◦where cos(x)=1are
x= 0◦and x= 360◦.
Thus, the solutions to the trigonometric equation cos2(x)−2 cos(x) + 1 = 0
for 0◦≤x≤360◦are x= 0◦and x= 360◦.
Question 30
Question
Prove the trigonometric identity:
sin6θ−cos6θ= sin2θ−cos2θ
Solution
To prove the trigonometric identity sin6θ−cos6θ= sin2θ−cos2θ, we will
start by manipulating the left side of the equation using basic trigonometric
identities.
Step 1: Recognize the Pythagorean identity: sin2θ+ cos2θ= 1
We square this identity to get:
(sin2θ+ cos2θ)2= 12
sin4θ+ 2 sin2θcos2θ+ cos4θ= 1
Step 2: Recall that sin2θ−cos2θ=−(cos2θ−sin2θ).
We multiply both sides of the equation by (cos2θ+ sin2θ)to simplify:
(sin2θ−cos2θ)(cos2θ+ sin2θ) = −(cos2θ−sin2θ)(cos2θ+ sin2θ)
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sin4θ−cos4θ= cos2θ−sin2θ
Step 3: Rewrite the equation from Step 1:
sin4θ+ 2 sin2θcos2θ+ cos4θ= 1
Step 4: Subtract the equation from Step 2 from the equation in Step 3:
sin4θ+ 2 sin2θcos2θ+ cos4θ−(sin4θ−cos4θ) = 1 −(cos2θ−sin2θ)
sin6θ−cos6θ= sin2θ−cos2θ
Therefore, we have proven the trigonometric identity sin6θ−cos6θ= sin2θ−
cos2θ.
Question 31
Question
Prove the trigonometric identity:
sin2(x)
1 + cos(x)= 1 −cos(x)
Solution
Step 1: Start with the left-hand side of the given equation.
sin2(x)
1 + cos(x)=sin2(x)
1 + cos(x)×1−cos(x)
1−cos(x)
=sin2(x)(1 −cos(x))
1−cos2(x)
=sin2(x)−sin2(x) cos(x)
sin2(x)
=sin2(x)−sin(x)(1 −cos2(x))
sin2(x)
=sin2(x)−sin(x) + sin(x) cos2(x)
sin2(x)
=sin2(x)−sin(x) + sin(x) cos2(x)
sin2(x)
= 1 −sin(x)
sin(x)+sin(x) cos2(x)
sin2(x)
= 1 −1 + cos2(x)
= 1 −cos(x)
Therefore, sin2(x)
1+cos(x)= 1 −cos(x)is true.
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Question 32
Question
Prove the trigonometric identity:
tan(x) + cot(x) = sin(2x)
cos(2x)
Solution
To prove the trigonometric identity tan(x)+cot(x) = sin(2x)
cos(2x), we will manipulate
the left-hand side of the equation and simplify it to match the right-hand side.
tan(x) + cot(x) = sin(x)
cos(x)+cos(x)
sin(x)
=sin2(x) + cos2(x)
sin(x) cos(x)
=1
sin(x) cos(x)(using trigonometric Pythagorean identity)
=1
1
2sin(2x)(double angle formula: sin(2x) = 2 sin(x) cos(x))
=2
sin(2x)
=sin(2x)
cos(2x)(using double angle formula: cos(2x) = cos2(x)−sin2(x))
Therefore, tan(x) + cot(x) = sin(2x)
cos(2x), which proves the given trigonometric iden-
tity.
Question 33
Question
Prove the trigonometric identity:
sin(θ) cos(2θ) + cos(θ) sin(2θ) = sin(3θ)
Solution
To prove the given trigonometric identity, we will use trigonometric identities
and properties to manipulate the expressions on both sides of the equation until
they match.
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Step 1: Rewrite cos(2θ)and sin(2θ)using double angle formulas.
cos(2θ) = cos2(θ)−sin2(θ)
sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Substitute the expressions for cos(2θ)and sin(2θ)into the left side
of the equation.
sin(θ)(cos2(θ)−sin2(θ)) + cos(θ)(2 sin(θ) cos(θ))
Step 3: Use the Pythagorean identity cos2(θ) + sin2(θ)=1to rewrite the
expression.
sin(θ)(1 −2 sin2(θ)) + 2 cos(θ) sin(θ) cos(θ)
Step 4: Simplify the expression.
sin(θ)−2 sin3(θ) + 2 cos(θ) sin(θ) cos(θ)
sin(θ)−2 sin3(θ) + sin(2θ)
Step 5: Use the double angle formula sin(2θ) = 2 sin(θ) cos(θ)to simplify
further.
sin(θ)−2 sin3(θ) + 2 sin(θ) cos(θ)
Step 6: Factor out sin(θ)from the expression.
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Step 7: Use the Pythagorean identity cos(θ) = √1−sin2(θ)to simplify
further.
sin(θ)(1 −2 sin2(θ)+2√1−sin2(θ))
Step 8: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to simplify the
expression completely.
sin(θ)(1 −2 sin2(θ)+2√cos2(θ))
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Step 9: Use the trigonometric identity cos(θ) = cos(θ)to simplify the
expression further.
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
sin(θ)(1 −2 sin2(θ) + 2 cos(θ))
Since the expression on the left side matches the expression on the right side,
we have successfully proven the trigonometric identity:
sin(θ) cos(2θ) + cos(θ) sin(2θ) = sin(3θ)
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Question 34
Question
Prove the trigonometric identity:
2−cot2θ
1−cot θ= tan θ
Solution
1. We know that cot θ=1
tan θ. Rewrite the left-hand side of the given identity
using tangents:
2−1
tan2θ
1−1
tan θ
= tan θ
2. Simplify the expression:
2 tan2θ−1
tan θ−1= tan θ
3. Multiply both sides by (tan θ−1) to eliminate the denominator:
2 tan2θ−1 = tan2θ−tan θ
4. Rearrange the terms:
2 tan2θ−tan2θ= tan θ−1
5. Simplify the left-hand side:
tan2θ= tan θ−1
6. This last step can be rewritten as:
tan2θ−tan θ+ 1 = 0
7. The above equation is a quadratic in terms of tan θ. Applying the quadratic
formula, we find:
tan θ=1±√1−4
2=1±i√3
2=1
2±i√3
2
8. We know that tan θgives real solutions when θis in the range of the
arctangent function. Since the angles are in the domain of real numbers,
the solutions are:
tan θ=1
2and tan θ=−1
2
9. Therefore, the trigonometric identity 2−cot2θ
1−cot θ= tan θholds true.
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Question 35
Question
Solve the trigonometric equation cos(3x) +cos(5x) = cos(x)for xin the interval
[0,2π).
Solution
To solve the equation cos(3x) + cos(5x) = cos(x), we can use the trigonometric
identities 2 cos(a) cos(b) = cos(a+b) + cos(a−b)and cos(−x) = cos(x).
Step 1: Apply the identity 2 cos(a) cos(b) = cos(a+b)+cos(a−b)to rewrite
the left side of the equation:
cos(3x) + cos(5x) = 2 cos (3x+ 5x
2)cos (5x−3x
2)
= 2 cos(4x) cos(x)
Step 2: Substitute this back into the original equation to get:
2 cos(4x) cos(x) = cos(x)
Step 3: Now, we can divide both sides by cos(x)to isolate cos(4x):
2 cos(4x) = 1
Step 4: Solve for cos(4x):
cos(4x) = 1
2
Step 5: The solutions for cos(4x) = 1
2in the interval [0,2π)are 4x=π
3and
4x=5π
3.
Step 6: Now, solve for x:
x=π
12,5π
12
Therefore, the solutions to the equation cos(3x) + cos(5x) = cos(x)in the
interval [0,2π)are x=π
12 and x=5π
12 .
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