MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 5
Liberty University
Question 1
Question
Prove the trigonometric identity:
sin3(x)−cos3(x) = sin(x)−cos(x)
Solution
We will start by expanding the left side of the given identity using the trigono-
metric identities sin2(x) = 1 −cos2(x)and cos2(x) = 1 −sin2(x).
Step 1: Expand sin3(x)using the identity sin2(x) = 1 −cos2(x).
sin3(x) = sin(x)·sin2(x)
= sin(x)(1 −cos2(x))
= sin(x)−sin(x) cos2(x)
Step 2: Expand cos3(x)using the identity cos2(x) = 1 −sin2(x).
cos3(x) = cos(x)·cos2(x)
= cos(x)(1 −sin2(x))
= cos(x)−cos(x) sin2(x)
Step 3: Substitute the results from Steps 1 and 2 back into the given identity
to simplify.
sin3(x)−cos3(x)
= (sin(x)−sin(x) cos2(x)) −(cos(x)−cos(x) sin2(x))
= sin(x)−sin(x) cos2(x)−cos(x) + cos(x) sin2(x)
= sin(x)−sin(x)(1 −sin2(x)) −cos(x) + cos(x)(1 −cos2(x))
= sin(x)−sin(x) + sin3(x)−cos(x) + cos(x)−cos3(x)
= sin3(x)−cos3(x)
Therefore,
sin3(x)−cos3(x) = sin(x)−cos(x)
as desired.
Question 2
Question
Solve the trigonometric equation: 2 cos2(x)−√3 cos(x) + 1 = 0 for xin the
interval [0,2π).
Solution
Step 1: Let u= cos(x). The equation becomes 2u2−√3u+ 1 = 0.
Step 2: Solve for uusing the quadratic formula:
u=√3±√3−4(2)(1)
4=√3±√−5
4
Step 3: Since the discriminant is negative, we have complex roots:
u=√3±i√5
4
Step 4: Recall that cos(x) = u. To write the solutions in terms of x, we use
Euler’s formula:
cos(x) = eix +e−ix
2
Step 5: Substitute u=√3+i√5
4into cos(x) = eix +e−ix
2to obtain eix.
Step 6: Similarly, substitute u=√3−i√5
4into cos(x) = eix +e−ix
2to obtain
eix.
Therefore, the solutions to the equation 2 cos2(x)−√3 cos(x) + 1 = 0 in the
interval [0,2π)involve complex numbers and are not expressible with elementary
functions.
2
Question 3
Question
Prove the trigonometric identity: sin(2θ) cos(2θ) = 1
2sin(4θ).
Solution
To prove the given trigonometric identity, we can use the double angle formulas
for sine and cosine.
Step 1: Write down the double angle formulas for sine and cosine. The
double angle formulas are:
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Calculate sin(2θ) cos(2θ)using the double angle formulas.
sin(2θ) cos(2θ) = (2 sin(θ) cos(θ))(cos2(θ)−sin2(θ))
= 2 sin(θ) cos(θ) cos2(θ)−2 sin(θ) cos(θ) sin2(θ)
Step 3: Simplify the expression using trigonometric identities.
2 sin(θ) cos(θ) cos2(θ)−2 sin(θ) cos(θ) sin2(θ) = 2 sin(θ) cos(θ)(cos2(θ)−sin2(θ))
= 2 sin(θ) cos(θ) cos(2θ)
= sin(2θ) cos(2θ)
Step 4: Use the double angle formula for sine to rewrite the expression.
sin(2θ) cos(2θ) = 1
2(2 sin(2θ) cos(2θ))
=1
2sin(4θ)
Therefore, we have proved the trigonometric identity: sin(2θ) cos(2θ) =
1
2sin(4θ).
Question 4
Question
Prove the following trigonometric identity:
sin(x)
1−cos(x)= csc(x) + cot(x)
3
Solution
To prove the given trigonometric identity, we will start with the left-hand side
(LHS) and manipulate it to get the right-hand side (RHS).
Step 1: Start with LHS: sin(x)
1−cos(x).
Step 2: Use the reciprocal identity csc(x) = 1
sin(x)to rewrite the LHS:
sin(x)
1−cos(x)=sin(x)
1−cos(x)·1 + cos(x)
1 + cos(x)=sin(x)(1 + cos(x))
1−cos2(x)
Step 3: Apply the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify
the expression:
sin(x)(1 + cos(x))
1−cos2(x)=sin(x)(1 + cos(x))
sin2(x)= csc(x) + cot(x)
Step 4: Therefore, we have shown that sin(x)
1−cos(x)= csc(x) + cot(x), which
completes the proof of the identity.
Question 5
Question
Prove the following trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
Solution
To prove the trigonometric identity cos4(x)−sin4(x) = cos(2x), we will start
by expanding the left-hand side using trigonometric identities.
Step 1: Expand cos4(x)−sin4(x).
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x))
= (cos2(x) + sin2(x))(cos(x) + sin(x))(cos(x)−sin(x))
= cos(x) cos(2x) cos(−x)−sin(x) sin(2x) sin(−x)
= cos(x) cos(2x) cos(x)−sin(x) sin(2x) sin(x)
= cos2(x) cos(2x)−sin2(x) sin(2x)
Step 2: Simplify the expression using double angle trigonometric identities.
Using the double angle identities:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
4
we can rewrite the expression as:
cos2(x) cos(2x)−sin2(x) sin(2x) = cos2(x)(cos2(x)−sin2(x)) −sin2(x)(2 sin(x) cos(x))
= cos2(x) cos2(x)−cos2(x) sin2(x)−2 sin3(x) cos(x)
= cos4(x)−cos2(x) sin2(x)−2 cos(x) sin3(x)
= cos4(x)−sin2(x) cos2(x)−2 cos(x) sin3(x)
= cos4(x)−sin4(x)−2 cos(x) sin3(x)
Since cos4(x)−sin4(x) = cos4(x)−sin4(x)−2 cos(x) sin3(x), we have proved
the identity.
Question 6
Question
Prove the trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)
Solution
Step 1: Start with the left-hand side (LHS) of the given trigonometric equation.
LHS =sin3(x)
cos(x)+cos3(x)
sin(x)
=sin4(x)
sin(x) cos(x)+cos4(x)
sin(x) cos(x)(Multiplying by appropriate terms)
=sin4(x) + cos4(x)
sin(x) cos(x)
Step 2: Apply the trigonometric identity sin2(x) + cos2(x) = 1 to simplify
the numerator.
LHS =sin4(x) + cos4(x)
sin(x) cos(x)
=(sin2(x))2+ (cos2(x))2−2sin2(x)cos2(x)
sin(x) cos(x)
=(sin2(x) + cos2(x))2−2sin2(x)cos2(x)
sin(x) cos(x)(Expanding)
=(1)2−2sin2(x)cos2(x)
sin(x) cos(x)(Using the trigonometric identity)
=1−2sin2(x)cos2(x)
sin(x) cos(x)
5
Step 3: Use the trigonometric identity sin(2x) = 2 sin(x) cos(x)to further
simplify the expression.
LHS =1−2sin2(x)cos2(x)
sin(x) cos(x)
=1−1
2sin(2x)
sin(x) cos(x)(Substitute sin(2x) = 2 sin(x) cos(x))
=2−sin(2x)
2 sin(x) cos(x)(Multiplying by 2 in numerator and denominator)
Step 4: Finally, simplify the expression further.
LHS =1
2 sin(x) cos(x)=1
sin(x) cos(x)
Therefore, sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)is true, which completes the proof.
Question 7
Question
Solve the equation tan2(x) = 2 tan(x)−1for 0≤x < 2π.
Solution
Step 1: Let’s rewrite the given equation in terms of sine and cosine. Recall the
trigonometric identity tan(x) = sin(x)
cos(x).
tan2(x) = 2 tan(x)−1
(sin(x)
cos(x))2
= 2 (sin(x)
cos(x))−1
sin2(x)
cos2(x)=2 sin(x)
cos(x)−1
sin2(x)
cos2(x)= 2 sin(x)
cos(x)−cos2(x)
cos2(x)
sin2(x)
cos2(x)= 2 sin(x)
cos(x)−1
cos2(x)
Step 2: Now, let’s substitute sin(x)
cos(x)with another variable, say u.
u=sin(x)
cos(x)
⇒u2=sin2(x)
cos2(x)
6
Thus, the equation becomes: u2= 2u−1
cos2(x).
Step 3: Rearranging the equation, we get u2−2u+ 1/cos2(x) = 0. This is
a quadratic equation in u.
Step 4: Now solve the quadratic equation u2−2u+ 1/cos2(x) = 0 using the
quadratic formula:
u=2±√(−2)2−4(1)(1/cos2(x))
2(1) = 1 ±√1−1
cos2(x)
Step 5: Recall that u= tan(x),tan(x)=1±√1−1
cos2(x). Therefore, we
have two cases to consider:
tan(x) = 1 + √1−1
cos2(x)
tan(x) = 1 −√1−1
cos2(x)
Step 6: We know that cos2(x) = 1 −sin2(x)and tan(x) = sin(x)
cos(x). By
substituting cos(x) = √1−sin2(x)into the two equations above, we can solve
for x. Remember to check for extraneous solutions in the process.
Question 8
Question
Solve the equation 2 sin2(x)−3 sin(x) = 2 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
2 sin2(x)−3 sin(x)−2 = 0
Step 2: To solve this quadratic equation, let’s factor it:
(2 sin(x) + 1)(sin(x)−2) = 0
Step 3: Setting each factor to zero gives us:
2 sin(x) + 1 = 0 or sin(x)−2 = 0
Step 4: Solve the first equation 2 sin(x) + 1 = 0:
sin(x) = −1
2
7
Step 5: The solutions to sin(x) = −1
2in the interval [0,2π]are x=7π
6and
x=11π
6.
Step 6: Solve the second equation sin(x)−2 = 0:
sin(x) = 2
This is not possible since the range of sine function is [−1,1].
Step 7: Therefore, the solutions to the equation 2 sin2(x)−3 sin(x)=2in
the interval [0,2π]are x=7π
6and x=11π
6.
Question 9
Question
Prove the following trigonometric identity:
1
1−sin θ=1 + sin θ
cos2θ
Solution
Step 1: To prove the given trigonometric identity, start with the left-hand side
of the equation. 1
1−sin θ
Step 2: Using the compound angle formula for sine, we can rewrite the
denominator as follows:
1−sin θ=2 cos2θ
2
2−2 sin θ
2cos θ
2
2
Step 3: Simplifying the expression, we get:
1−sin θ= 2 cos2θ
2−2 sin θ
2cos θ
2
Step 4: Using the double angle formula for cosine, we can simplify further:
1−sin θ= 2 cos2θ
2−sin θ
Step 5: Dividing both the numerator and denominator by cos θin the original
expression, we get:
1
1−sin θ=
1
cos θ
1−sin θ
cos θ
Step 6: Using trigonometric identities, simplify the expression to get:
1
1−sin θ=sec θ
cos θ−sin θ
8
Step 7: Now, simplify the right-hand side of the given equation:
1 + sin θ
cos2θ=cos θ+ sin θ
cos2θ
Step 8: Rewrite the numerator of the right-hand side in terms of sine and
cosine: 1 + sin θ
cos2θ=cos θ+ sin θ
cos θcos θ
Step 9: Simplify the expression to get:
1 + sin θ
cos2θ= sec θ(cos θ+ sin θ)
Step 10: Finally, simplify the expression to get:
1 + sin θ
cos2θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ
Since both the left-hand side and the right-hand side are equal, the trigono-
metric identity is proved.
Question 10
Question
Prove the trigonometric identity: tan(θ) + sec(θ) = 1+sin(θ)
cos(θ).
Solution
To prove the identity tan(θ)+sec(θ) = 1+sin(θ)
cos(θ), we will manipulate the left-hand
side to obtain the expression on the right-hand side.
Step 1: Write tan(θ)and sec(θ)in terms of sine and cosine:
tan(θ) = sin(θ)
cos(θ)and sec(θ) = 1
cos(θ)
Step 2: Substitute these expressions into tan(θ) + sec(θ):
tan(θ) + sec(θ) = sin(θ)
cos(θ)+1
cos(θ)
Step 3: Find a common denominator:
sin(θ)
cos(θ)+1
cos(θ)=sin(θ)+1
cos(θ)
Step 4: Simplify the numerator to obtain the expression on the right-hand
side: sin(θ)+1
cos(θ)=1 + sin(θ)
cos(θ)
Therefore, we have shown that tan(θ) + sec(θ) = 1+sin(θ)
cos(θ), and the identity
is proved.
9
Question 11
Question
Prove that cot(θ) + cot(2θ) = csc(θ) csc(2θ).
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it step by step until we reach the right-hand side.
Step 1: Write cot(2θ)in terms of cot(θ)and θusing the double angle
formula for cotangent.
cot(2θ) = cot2(θ)−1
2 cot(θ)
Step 2: Substitute the expression for cot(2θ)into the expression cot(θ) +
cot(2θ).
cot(θ) + cot2(θ)−1
2 cot(θ)
Step 3: Find a common denominator for the terms.
2 cot2(θ)
2 cot(θ)+−1
2 cot(θ)
Step 4: Combine the fractions.
2 cot2(θ)−1
2 cot(θ)
Step 5: Recall the Pythagorean identity: cot2(θ)−1 = csc2(θ).
2 csc2(θ)
2 cot(θ)
Step 6: Simplify by recognizing that csc(θ) = 1
sin(θ)and cot(θ) = 1
tan(θ).
2
sin(θ)·cos(θ)
sin(θ)
Step 7: Simplify further.
2
cos(θ)= 2 sec(θ) = csc(θ)·csc(2θ)
Therefore, cot(θ) + cot(2θ) = csc(θ) csc(2θ), as desired.
10
Question 12
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x))
Now, we will simplify the right hand side:
2 sin2(x) cos2(x) = 2(sin(x) cos(x))2
= 2 (1
2sin(2x))2
=1
2sin2(2x)
=1
2(1 −cos2(2x))
=1
2(1 −(1 −2 sin2(x))2)
=1
2(1 −1 + 4 sin4(x)−4 sin2(x))
= 2 sin4(x)−2 sin2(x)
Therefore,
2 sin4(x)−2 sin2(x) = (sin(x) + cos(x))(sin(x)−cos(x))
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Thus, the given identity has been proven.
Question 13
Question
Solve the equation cos(3x) = sin(2x)for xin the interval [0,2π].
11
Solution
Step 1: We can use the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x)
and sin(2x) = 2 sin(x) cos(x). So, the given equation becomes:
4 cos3(x)−3 cos(x) = 2 sin(x) cos(x)
4 cos2(x)−3 = 2 sin(x)
4(1 −sin2(x)) −3 = 2 sin(x)
4−4 sin2(x)−3 = 2 sin(x)
−4 sin2(x) + 2 sin(x) + 1 = 0
−4 sin2(x) + 4 sin(x)−2 sin(x) + 1 = 0
−2 sin(x)(2 sin(x)−2) −(2 sin(x)−1) = 0
(2 sin(x)−1)(1 −2 sin(x)) = 0
Step 2: Now we have two cases to consider: Case 1: 2 sin(x)−1 = 0 =⇒
sin(x) = 1
2This occurs when x=π
6,5π
6.
Case 2: 1−2 sin(x) = 0 =⇒sin(x) = 1
2This does not give us solutions in
the interval [0,2π].
Therefore, the solutions to the equation cos(3x) = sin(2x)in the interval
[0,2π]are x=π
6,5π
6.
Question 14
Question
Prove the following trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x)
Solution
To prove the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x), we will use
the triple angle formula for cosine:
cos(3x) = cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x)
Step 1: Write cos(2x)and sin(2x)in terms of cos(x)and sin(x)using double
angle formulas:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the expressions for cos(2x)and sin(2x)into the triple
angle formula:
cos(3x) = (cos2(x)−sin2(x)) cos(x)−(2 sin(x) cos(x)) sin(x)
12
Step 3: Simplify the expression:
cos(3x) = cos3(x)−sin2(x) cos(x)−2 sin2(x) cos(x)
Step 4: Use the Pythagorean identity sin2(x) = 1 −cos2(x)to replace
sin2(x):
cos(3x) = cos3(x)−(1 −cos2(x)) cos(x)−2(1 −cos2(x)) cos(x)
Step 5: Expand and simplify the expression:
cos(3x) = cos3(x)−cos(x) + cos3(x)−2 cos(x) + 2 cos3(x)
cos(3x) = 4 cos3(x)−3 cos(x)
Therefore, we have proved the trigonometric identity cos(3x) = 4 cos3(x)−
3 cos(x).
Question 15
Question
Prove the identity:
cos3(x)−sin3(x) = cos(3x)
Solution
To prove the identity cos3(x)−sin3(x) = cos(3x), we will use the sum-to-product
identities and double angle formulas for cosine and sine functions.
Step 1: Express cos3(x)−sin3(x)in terms of cos(3x)using the sum-to-
product identity cos(A)−cos(B) = −2 sin (A+B
2)sin (A−B
2).
cos3(x)−sin3(x) = (cos(x)−sin(x))(cos2(x) + cos(x) sin(x) + sin2(x))
= (cos(x)−sin(x))(1 + cos(x) sin(x))
Then, we can use the double-angle formula for cos(2x) = 2 cos2(x)−1to
rewrite cos2(x)in the expression above.
cos3(x)−sin3(x) = (cos(x)−sin(x))(1 + cos(x) sin(x))
= (cos(x)−sin(x))(1 + 1
2sin(2x))
Step 2: Expand the expression on the right-hand side and simplify.
= cos(x)−sin(x) + 1
2(cos(x) sin(2x)−sin(x) sin(2x))
13
= cos(x)−sin(x) + 1
2(2 cos(x) sin(x) cos(x)−sin(x) sin(x)(2 cos(x)))
= cos(x)−sin(x) + cos(x)(cos(2x)) −sin(x)(sin(2x))
= cos(x)−sin(x) + cos(x)(2 cos2(x)−1) −sin(x)(2 sin(x) cos(x))
= cos(x)−sin(x) + 2 cos3(x)−cos(x)−2 sin2(x) cos(x)
Step 3: Simplify the expression further.
= 2 cos3(x)−2 sin2(x) cos(x)−sin(x)
Now, we will rewrite cos(3x)using the triple angle formula for cosine:
cos(3x) = 4 cos3(x)−3 cos(x)
Step 4: Substitute cos3(x)−sin3(x)and cos(3x)back into the equation.
2 cos3(x)−2 sin2(x) cos(x)−sin(x) = 4 cos3(x)−3 cos(x)
Step 5: Simplify the equation.
2 cos3(x)−2 sin2(x) cos(x)−sin(x) = 4 cos3(x)−3 cos(x)
2 cos3(x)−4 cos3(x) + 2 sin2(x) cos(x)−sin(x) + 3 cos(x) = 0
−2 cos3(x) + 2 sin2(x) cos(x) + 3 cos(x)−sin(x) = 0
Therefore, we have shown that cos3(x)−sin3(x) = cos(3x).
Question 16
Question
Prove the trigonometric identity:
cot(x)−tan(x) = 2 cot(2x)
Solution
We will start by expressing the right-hand side of the given identity in terms of
sines and cosines using the double angle formula for cotangent:
2 cot(2x) = 2 (cos(2x)
sin(2x))
Next, we will apply the double angle formulas for cosine and sine to further
simplify the expression:
2 cot(2x) = 2 (cos2(x)−sin2(x)
2 sin(x) cos(x))
14
2 cot(2x) = 2 cos2(x)−2 sin2(x)
2 sin(x) cos(x)
2 cot(2x) = 2 cos2(x)−2 sin2(x)
2 sin(x) cos(x)
2 cot(2x) = 2(cos2(x)−sin2(x))
2 sin(x) cos(x)
2 cot(2x) = 2 cos(2x)
2 sin(2x)
2 cot(2x) = cot(2x)
Thus, we have shown that the right-hand side of the given identity simplifies
to cot(2x). Now, we will show that the left-hand side of the given identity
simplifies to the same expression.
cot(x)−tan(x) = cos(x)
sin(x)−sin(x)
cos(x)
cot(x)−tan(x) = cos2(x)−sin2(x)
sin(x) cos(x)
But using the double angle formula for cosine (cos(2x) = cos2(x)−sin2(x)),
we have:
cot(x)−tan(x) = cos(2x)
sin(2x)= cot(2x)
Therefore, we have successfully proven the trigonometric identity cot(x)−
tan(x) = 2 cot(2x).
Question 17
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 convert
both sides to a single trigonometric function using double angle identities and
Pythagorean identity.
sin(2x) = 2 sin(x) cos(x)
cos(x) = √1−sin2(x)
Now we substitute these identities into the given equation:
sin(2x) = cos(x) =⇒2 sin(x) cos(x) = √1−sin2(x)
15
Step 1: Square both sides of the equation to get rid of the square root:
4 sin2(x) cos2(x) = 1 −sin2(x)
Step 2: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to replace
cos2(x):
4 sin2(x)(1 −sin2(x)) = 1 −sin2(x)
Step 3: Expand and simplify the equation:
4 sin2(x)−4 sin4(x) = 1 −sin2(x)
4 sin4(x)−5 sin2(x) + 1 = 0
Step 4: Let u= sin2(x)to change the equation into a quadratic form:
4u2−5u+ 1 = 0
Step 5: Solve the quadratic equation:
u=5±√52−4·4·1
2·4
u=5±√9
8
u=5±3
8
Step 6: Find the possible values of u: If u=5+3
8=8
8= 1, then sin2(x) =
1 =⇒sin(x) = ±1. However, this does not satisfy the original equation
sin(2x) = cos(x). If u=5−3
8=2
8=1
4, then sin2(x) = 1
4=⇒sin(x) = ±1
2.
Step 7: Find the values of x: For sin(x) = 1
2,x=π
6,5π
6. For sin(x) = −1
2,
x=7π
6,11π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
0≤x < 2πare x=π
6,5π
6,7π
6,11π
6.
Question 18
Question
Prove the trigonometric identity:
tan(x) cot(x) = sec(x) csc(x)−1
16
Solution
tan(x) cot(x) = sin(x)
cos(x)·cos(x)
sin(x)(Definition of tan and cot )
=sin(x) cos(x)
cos(x) sin(x)
= 1
Using trigonometric definitions:
sec(x) = 1
cos(x)
csc(x) = 1
sin(x)
sec(x) csc(x)−1 = (1
cos(x))( 1
sin(x))−1
=1
sin(x) cos(x)−1
=1
sin(x) cos(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)
sin(x) cos(x)(Using sin2(x) + cos2(x) = 1)
= 0
Therefore, tan(x) cot(x) = sec(x) csc(x)−1.
Question 19
Question
Prove the following trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
Solution
1. We are given: cot(θ) sin(2θ) = 2 cos(θ).
17
2. Recall the double angle identity: sin(2θ) = 2 sin(θ) cos(θ).
3. Substitute the double angle identity into our expression: cot(θ)(2 sin(θ) cos(θ)) =
2 cos(θ).
4. Simplify the left side by distributing cot(θ):2 cot(θ) sin(θ) cos(θ) = 2 cos(θ).
5. Recall that cot(θ) = cos(θ)
sin(θ).
6. Substitute cot(θ) = cos(θ)
sin(θ)into the left side: 2(cos(θ)
sin(θ))sin(θ) cos(θ) =
2 cos(θ).
7. Simplify the left side by canceling terms: 2 cos(θ) = 2 cos(θ).
8. Therefore, cot(θ) sin(2θ) = 2 cos(θ)is proven.
Question 20
Question
Solve the equation sin2(θ) + cos2(θ) + 2 sin(θ)−1 = 0 for 0≤θ≤2π.
Solution
Step 1: We can rewrite the equation as 1 + 2 sin(θ)−1=0since sin2(θ) +
cos2(θ) = 1. Step 2: Simplifying, we get 2 sin(θ) = 0. Step 3: Dividing both
sides by 2, we have sin(θ)=0. Step 4: To find all solutions in the interval
0≤θ≤2π, we know that sin(θ) = 0 at θ= 0, π, 2π. Step 5: Therefore, the
solutions to the equation are θ= 0, π, 2π.
Question 21
Question
Prove the trigonometric identity: cos3(x)−sin3(x) = cos(2x).
Solution
To prove the trigonometric identity cos3(x)−sin3(x) = cos(2x), we will use the
trigonometric identity cos(2θ) = cos2(θ)−sin2(θ).
cos3(x)−sin3(x) = (cos(x)−sin(x))(cos2(x) + cos(x) sin(x) + sin2(x)) (Difference of cubes)
= (cos(x)−sin(x))[(cos2(x) + sin2(x)) + cos(x) sin(x)]
= (cos(x)−sin(x))(1 + cos(x) sin(x)) (using trigonometric identity cos2(x) + sin2(x) = 1)
= cos(x)−sin(x) + cos2(x) sin(x)−sin2(x) cos(x)
= (cos(x) + sin(x))(cos(x)−sin(x))
= cos(2x)(using cos(2θ) = cos2(θ)−sin2(θ)with θ=x)
18
Therefore, cos3(x)−sin3(x) = cos(2x)is proven.
Question 22
Question
Solve the trigonometric equation sin(2x) = cos(x)for x∈[0,2π].
Solution
Step 1: Express sin(2x)in terms of cos(x). Recall the double angle identity for
sine:
sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the given equation sin(2x) =
cos(x):
2 sin(x) cos(x) = cos(x).
Step 3: Rearrange the equation by moving all terms to one side:
2 sin(x) cos(x)−cos(x) = 0.
Step 4: Factor out cos(x)on the left side:
cos(x)(2 sin(x)−1) = 0.
Step 5: Solve the equation cos(x)(2 sin(x)−1) = 0:
1. Setting cos(x) = 0, we have x=π
2,3π
2.
2. Setting 2 sin(x)−1 = 0, we get sin(x) = 1
2, which gives x=π
6,5π
6.
Step 6: Combine all the solutions and ensure they are within the given
domain [0,2π]:
x=π
6,π
2,5π
6,3π
2.
Question 23
Question
Prove the trigonometric identity:
1−cos x
sin x= csc x−cot x
19
Solution
Step 1: Rewrite the left side of the equation using the definitions of csc xand
cot x.
1−cos x
sin x=1−cos x
sin x·1 + cos x
1 + cos x=1−cos2x
sin x(1 + cos x)=sin2x
sin x(1 + cos x)
Step 2: Simplify the expression further.
sin2x
sin x(1 + cos x)=sin x·sin x
sin x(1 + cos x)=sin x
1 + cos x
Step 3: Express the right side of the equation in terms of csc xand cot x.
csc x−cot x=1
sin x−cos x
sin x=1−cos x
sin x
Step 4: Therefore, we have shown that the left side of the equation equals
the right side, proving the trigonometric identity.
1−cos x
sin x= csc x−cot x
Question 24
Question
Prove the following trigonometric identity:
1
cos(x)−sin(x)= tan (π
4+x)
Solution
To prove the given trigonometric identity, we will start by expressing the right-
hand side in terms of sine and cosine functions.
Step 1: Rewrite tan (π4 + x)in terms of sine and cosine
tan (π
4+x)=sin (π
4+x)
cos (π
4+x)
=sin π
4cos x+ cos π
4sin x
cos π
4cos x−sin π
4sin x
=
√2
2cos x+√2
2sin x
√2
2cos x−√2
2sin x
=√2(cos x+ sin x)
√2(cos x−sin x)
=cos x+ sin x
cos x−sin x
20
Step 2: Simplify the expression Now, let’s simplify the left-hand side
1
cos x−sin x=1
cos x−sin x·cos x+ sin x
cos x+ sin x
=cos x+ sin x
cos2x−sin2x
=cos x+ sin x
cos 2x
Step 3: Verify the equality of both sides To prove the identity, we need
to show that: cos x+ sin x
cos 2x=cos x+ sin x
cos x−sin x
This can be done by simplifying and showing both sides are equal:
cos x+ sin x= (cos x+ sin x)(cos x−sin x)
cos x+ sin x= cos2x−sin2x+ 2 sin xcos x
cos x+ sin x= cos 2x+ 2 sin xcos x
cos x+ sin x= cos 2x+ sin 2x
cos x+ sin x= cos(2x+π
2)
Therefore, the given trigonometric identity is true, and we have proved it.
Question 25
Question
Prove the trigonometric identity:
1 + tan(x)
1−tan(x)= sec(x) + tan(x)
Solution
Step 1: Start with the left-hand side (LHS) of the given equation.
LHS =1 + tan(x)
1−tan(x)
Step 2: Multiply the numerator and denominator by the conjugate of the
denominator to simplify.
LHS =(1 + tan(x))(1 + tan(x))
(1 −tan(x))(1 + tan(x))
21
Step 3: Expand the numerator and denominator.
LHS =1 + 2 tan(x) + tan2(x)
1−tan2(x)
Step 4: Recall the Pythagorean identity tan2(x) + 1 = sec2(x).
LHS =1 + 2 tan(x) + sec2(x)
sec2(x)
Step 5: Simplify the numerator.
LHS =sec2(x) + 2 tan(x)+1
sec2(x)
Step 6: Factor out the sec2(x)in the numerator.
LHS =sec2(x)(1 + 2 tan(x)
sec2(x)+1
sec2(x))
sec2(x)
Step 7: Recognize that 2 tan(x)
sec2(x)= 2 tan(x) cos2(x) = 2 sin(x) cos(x) = 2 sin(x) cos(x) =
2 sin(2x)and 1
sec2(x)= cos2(x) = 1
sec2(x).
LHS = sec2(x)(1 + 2 sin(2x) + 1)
Step 8: Simplify the expression in the parentheses.
LHS = sec2(x)(2 + 2 sin(2x))
Step 9: Recall the double-angle formula sin(2x) = 2 sin(x) cos(x)and sim-
plify further.
LHS = sec2(x)(2 + 4 sin(x) cos(x))
Step 10: Recall that cos(x) = 1
sec(x).
LHS = sec2(x)(2 + 4 sin(x)1
sec(x))
Step 11: Simplify the expression.
LHS = sec2(x)(2 + 4 tan(x))
Step 12: Recognize that 2 + 4 tan(x) = 2(1 + 2 tan(x)) = 2 sec2(x).
LHS = 2 sec4(x)
Therefore, the left-hand side is equal to 2 sec4(x), and the given trigonomet-
ric identity is proven.
22
Question 26
Question
Prove the following trigonometric identity:
1−cos2x
sin xcos x= csc x−cot x
Solution
Step 1: Start with the left side of the identity and work with one side at a time.
1−cos2x
sin xcos x
Step 2: Use the Pythagorean identity sin2x+ cos2x= 1 to simplify the
numerator. sin2x
sin xcos x
Step 3: Simplify the expression further by canceling out a factor of sin x.
sin x
cos x
Step 4: Remember that csc x=1
sin x, and cot x=cos x
sin x. Rewrite the expres-
sion using these identities. 1
cos x= sec x
Step 5: Therefore, the left side of the identity simplifies to sec x.
Step 6: Now, simplify the right side of the identity.
csc x−cot x=1
sin x−cos x
sin x
=1−cos x
sin x
Step 7: Remember that sin x=1
csc x, and cos x=1
sec x. Rewrite the expres-
sion using these identities.
=sec x−1
sec x
1
csc x
=sec2x−1
1
csc x
=
1
cos2x−1
1
sin x
=
1−cos2x
cos2x
1
sin x
23
=sin x
cos x= sec x
Step 8: Therefore, the right side of the identity simplifies to sec x.
Step 9: Since the left side simplifies to sec x, and the right side simplifies to
sec x, we have verified that the given trigonometric identity is true.
Question 27
Question
Prove the identity:
2 sin2(x)−tan(x) sin(x) = cos2(x)
Solution
Step 1: Recall the Pythagorean trigonometric identity:
sin2(x) + cos2(x) = 1
Step 2: Start with the left hand side of the given identity and try to simplify
it using the Pythagorean identity:
2 sin2(x)−tan(x) sin(x) = 2 sin2(x)−sin(x)
cos(x)·sin(x)
= 2 sin2(x)−sin2(x)
cos(x)
= sin2(x)(2−1
cos(x))
= sin2(x)·2 cos(x)−1
cos(x)
Step 3: Now, rewrite the right hand side of the identity in terms of sine:
cos2(x) = (1 −sin2(x))
= 1 −sin2(x)
Step 4: Compare the results from Step 2 and Step 3 to check if they are
equal:
sin2(x)·2 cos(x)−1
cos(x)
?
= 1 −sin2(x)
24
Step 5: Simplify further and show the equality:
sin2(x)·2 cos(x)−1
cos(x)= sin2(x)·2 cos(x)
cos(x)−sin2(x)·1
cos(x)
= 2 sin2(x)−sin2(x)
cos(x)
= 2 sin2(x)−tan(x) sin(x)
= cos2(x)
Therefore, the identity 2 sin2(x)−tan(x) sin(x) = cos2(x)is proven.
Question 28
Question
Prove the following trigonometric identity:
sin(A+B)
sin(A−B)=tan(A) + tan(B)
tan(A)−tan(B)
Solution
Step 1: Start with the left-hand side of the equation:
sin(A+B)
sin(A−B)
Step 2: We can expand the numerator using the angle sum identity for sine:
sin(A+B) = sin(A) cos(B) + cos(A) sin(B)
Step 3: Similarly, we can expand the denominator using the angle difference
identity for sine:
sin(A−B) = sin(A) cos(B)−cos(A) sin(B)
Step 4: Substitute the expanded expressions back into the original equation:
sin(A) cos(B) + cos(A) sin(B)
sin(A) cos(B)−cos(A) sin(B)
Step 5: Simplify the expression by dividing both numerator and denomi-
nator by cos(A) cos(B):
sin(A)
cos(A)+sin(B)
cos(B)
sin(A)
cos(A)−sin(B)
cos(B)
25
Step 6: Recognize that sin(A)
cos(A)= tan(A)and sin(B)
cos(B)= tan(B), so the ex-
pression simplifies to:
tan(A) + tan(B)
tan(A)−tan(B)
Step 7: Therefore, the left-hand side is equal to the right-hand side, proving
the trigonometric identity:
sin(A+B)
sin(A−B)=tan(A) + tan(B)
tan(A)−tan(B)
Question 29
Question
Prove the trigonometric identity:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)= csc(θ) + sec(θ)
Solution
Step 1: Start by simplifying the left-hand side of the equation.
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)
Step 2: Find a common denominator for the fractions.
sin2(θ) + (1 −cos(θ))(1 + cos(θ))
sin(θ)(1 −cos(θ))
Step 3: Simplify the numerator.
sin2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 4: Recall the Pythagorean identity sin2(θ) = 1 −cos2(θ).
1−cos2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 5: Simplify the numerator further.
2(1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 6: Use the Pythagorean identity sin2(θ) = 1 −cos2(θ)again.
2(sin2(θ))
sin(θ)(1 −cos(θ))
26
Step 7: Simplify the expression.
2 sin(θ)
1−cos(θ)= 2 csc(θ)
Step 8: Recognize that 2 csc(θ) = csc(θ) + csc(θ).
Step 9: Since 1+cos(θ) = sec(θ), the left-hand side of the equation simplifies
to csc(θ) + sec(θ).
Therefore, the trigonometric identity sin(θ)
1−cos(θ)+1+cos(θ)
sin(θ)= csc(θ) + sec(θ)is
proven.
Question 30
Question
Solve the trigonometric equation tan2(x)−2 tan(x) + 3 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= tan(x). The equation becomes u2−2u+ 3 = 0.
Step 2: Solve the quadratic equation u2−2u+ 3 = 0.
Step 3: To solve the quadratic equation, we can use the quadratic formula:
u=−(−2)±√(−2)2−4(1)(3)
2(1) .
Step 4: Simplifying under the square root, we get u=2±√4−12
2.
Step 5: Further simplifying, we have u=2±√−8
2.
Step 6: Since √−8is not a real number, the equation has no real solutions.
Step 7: Therefore, the trigonometric equation tan2(x)−2 tan(x) + 3 = 0 has
no real solutions in the interval [0,2π].
Question 31
Question
Prove the following trigonometric identity:
sin(θ)
1 + cos(θ)=1−cos(θ)
sin(θ)
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until we reach the right-hand side.
Step 1: Start with the left-hand side of the equation.
sin(θ)
1 + cos(θ)
27
Step 2: Multiply the numerator and denominator by the conjugate of the
denominator.
sin(θ)
1 + cos(θ)·1−cos(θ)
1−cos(θ)=sin(θ)(1 −cos(θ))
1−cos2(θ)
Step 3: Use the Pythagorean identity sin2(θ) = 1 −cos2(θ).
sin(θ)(1 −cos(θ))
sin2(θ)
Step 4: Simplify the expression.
sin(θ)−sin(θ) cos(θ)
sin2(θ)=sin(θ)
sin(θ)−sin(θ) cos(θ)
sin(θ)
Step 5: Simplify further to reach the right-hand side of the equation.
1−cos(θ) = 1−cos(θ)
sin(θ)
Therefore, we have proved the given trigonometric identity:
sin(θ)
1 + cos(θ)=1−cos(θ)
sin(θ)
Question 32
Question
Solve the trigonometric equation sin(x) = cos(2x)for xin the interval [0,2π].
Solution
Let’s solve the trigonometric equation step-by-step.
Step 1: Rewrite the equation using double angle identity.
sin(x) = cos(2x)
sin(x) = cos2(x)−sin2(x)
sin(x) = 1 −2 sin2(x)
Step 2: Rewrite the equation in terms of sin(x)only.
2 sin2(x) + sin(x)−1 = 0
Step 3: Factor the quadratic equation.
(2 sin(x)−1)(sin(x) + 1) = 0
28
Step 4: Solve the equations 2 sin(x)−1=0and sin(x) + 1 = 0. Solving
2 sin(x)−1 = 0 gives sin(x) = 1
2, which implies x=π
6,5π
6.
Solving sin(x) + 1 = 0 gives sin(x) = −1, which implies x=3π
2.
Step 5: Check the solutions in the original equation. - For x=π
6:sin(π
6)=
cos(π
3)=√3
2= cos(π
3), so this solution is extraneous. - For x=5π
6:sin(5π
6)=
cos(5π
3)=−√3
2= cos(5π
3), so this is a valid solution. - For x=3π
2:sin(3π
2)=
−1 = cos(3π) = 1, so this solution is extraneous.
Step 6: Therefore, the solution to the equation sin(x) = cos(2x)in the
interval [0,2π]is x=5π
6.
Question 33
Question
Prove the following trigonometric identity:
sin6(x) + cos6(x) = 1 −3 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to match the right-hand side.
Step 1: Start with the left-hand side sin6(x) + cos6(x).
sin6(x) + cos6(x) = (sin2(x))3+ (cos2(x))3
= (sin2(x) + cos2(x))(sin4(x)−sin2(x) cos2(x) + cos4(x))
Step 2: Recall that sin2(x) + cos2(x)=1. Now substitute this in the
expression.
sin6(x) + cos6(x) = 1(sin4(x)−sin2(x) cos2(x) + cos4(x))
= sin4(x)−sin2(x) cos2(x) + cos4(x)
Step 3: Factor the expression sin4(x)+cos4(x)using the identity sin2(x) cos2(x) =
1
4(sin(2x))2.
sin6(x) + cos6(x) = (sin2(x) + cos2(x))2−3 sin2(x) cos2(x)
= 1 −3 sin2(x) cos2(x)
Therefore, we have shown that sin6(x) + cos6(x) = 1 −3 sin2(x) cos2(x),
which proves the trigonometric identity.
29
Question 34
Question
Prove the following trigonometric identity:
1 + cos(α)
sin(α)+1−cos(α)
cos(α)= 2 csc(α) cot(α)
Solution
Step 1: Begin by simplifying the left-hand side of the equation.
1 + cos(α)
sin(α)+1−cos(α)
cos(α)=1
sin(α)(cos(α)+1)+ 1
cos(α)(1 −cos(α))
=cos(α)+1
sin(α)−cos(α)−1
cos(α)
=cos(α)
sin(α)+1
sin(α)−cos(α)
cos(α)+1
cos(α)
= cot(α) + csc(α)−tan(α) + sec(α)
Step 2: Use trigonometric identities to simplify the expression further.
cot(α) + csc(α)−tan(α) + sec(α) = cos(α)
sin(α)+1
sin(α)−sin(α)
cos(α)+1
cos(α)
=cos2(α)+1−sin2(α)
sin(α) cos(α)
=1−sin2(α)
sin(α) cos(α)
=cos2(α)
sin(α) cos(α)
=cos(α)
sin(α)
= csc(α)
Therefore, 1+cos(α)
sin(α)+1−cos(α)
cos(α)= 2 csc(α) cot(α)is proven to be true.
Question 35
Question
Prove the following trigonometric identity:
sin3(x)−cos3(x) = sin(x)−cos(x)
30
Solution
Step 1: Start with the left-hand side of the given identity:
sin3(x)−cos3(x)
Step 2: Use the formula for the difference of cubes, a3−b3= (a−b)(a2+
ab +b2), with a= sin(x)and b= cos(x):
sin3(x)−cos3(x) = (sin(x)−cos(x))(sin2(x) + sin(x) cos(x) + cos2(x))
Step 3: Recall the trigonometric identity sin2(x) + cos2(x) = 1:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + sin(x) cos(x))
Step 4: Use the trigonometric identity sin(2x) = 2 sin(x) cos(x)to simplify
sin(x) cos(x):
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + 1
2sin(2x))
Step 5: Apply the double angle formula for sine, sin(2x) = 2 sin(x) cos(x),
to get:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + 1
2·2 sin(x) cos(x))
Step 6: Simplify the expression further:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + sin(x))
Step 7: Now, distribute to get:
sin3(x)−cos3(x) = sin(x)−cos(x) + sin2(x)−sin(x) cos(x)
Step 8: Finally, recall the trigonometric identity sin2(x) = 1 −cos2(x):
sin3(x)−cos3(x) = sin(x)−cos(x) + (1 −cos2(x)) −sin(x) cos(x)
Step 9: Simplify the expression further:
sin3(x)−cos3(x) = sin(x)−cos(x)−cos2(x)+1−sin(x) cos(x)
Step 10: Recall the Pythagorean identity sin2(x) + cos2(x) = 1:
sin3(x)−cos3(x) = sin(x)−cos(x)−(1 −sin(x))
Step 11: Simplify the expression to get the right-hand side:
sin3(x)−cos3(x) = sin(x)−cos(x)
Therefore, we have proved the given trigonometric identity sin3(x)−cos3(x) =
sin(x)−cos(x).
31
Question 3
Question
Prove the trigonometric identity: sin(2θ) cos(2θ) = 1
2sin(4θ).
Solution
To prove the given trigonometric identity, we can use the double angle formulas
for sine and cosine.
Step 1: Write down the double angle formulas for sine and cosine. The
double angle formulas are:
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ)
Step 2: Calculate sin(2θ) cos(2θ)using the double angle formulas.
sin(2θ) cos(2θ) = (2 sin(θ) cos(θ))(cos2(θ)−sin2(θ))
= 2 sin(θ) cos(θ) cos2(θ)−2 sin(θ) cos(θ) sin2(θ)
Step 3: Simplify the expression using trigonometric identities.
2 sin(θ) cos(θ) cos2(θ)−2 sin(θ) cos(θ) sin2(θ) = 2 sin(θ) cos(θ)(cos2(θ)−sin2(θ))
= 2 sin(θ) cos(θ) cos(2θ)
= sin(2θ) cos(2θ)
Step 4: Use the double angle formula for sine to rewrite the expression.
sin(2θ) cos(2θ) = 1
2(2 sin(2θ) cos(2θ))
=1
2sin(4θ)
Therefore, we have proved the trigonometric identity: sin(2θ) cos(2θ) =
1
2sin(4θ).
Question 4
Question
Prove the following trigonometric identity:
sin(x)
1−cos(x)= csc(x) + cot(x)
3
Solution
To prove the given trigonometric identity, we will start with the left-hand side
(LHS) and manipulate it to get the right-hand side (RHS).
Step 1: Start with LHS: sin(x)
1−cos(x).
Step 2: Use the reciprocal identity csc(x) = 1
sin(x)to rewrite the LHS:
sin(x)
1−cos(x)=sin(x)
1−cos(x)·1 + cos(x)
1 + cos(x)=sin(x)(1 + cos(x))
1−cos2(x)
Step 3: Apply the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify
the expression:
sin(x)(1 + cos(x))
1−cos2(x)=sin(x)(1 + cos(x))
sin2(x)= csc(x) + cot(x)
Step 4: Therefore, we have shown that sin(x)
1−cos(x)= csc(x) + cot(x), which
completes the proof of the identity.
Question 5
Question
Prove the following trigonometric identity:
cos4(x)−sin4(x) = cos(2x)
Solution
To prove the trigonometric identity cos4(x)−sin4(x) = cos(2x), we will start
by expanding the left-hand side using trigonometric identities.
Step 1: Expand cos4(x)−sin4(x).
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x))
= (cos2(x) + sin2(x))(cos(x) + sin(x))(cos(x)−sin(x))
= cos(x) cos(2x) cos(−x)−sin(x) sin(2x) sin(−x)
= cos(x) cos(2x) cos(x)−sin(x) sin(2x) sin(x)
= cos2(x) cos(2x)−sin2(x) sin(2x)
Step 2: Simplify the expression using double angle trigonometric identities.
Using the double angle identities:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
4
we can rewrite the expression as:
cos2(x) cos(2x)−sin2(x) sin(2x) = cos2(x)(cos2(x)−sin2(x)) −sin2(x)(2 sin(x) cos(x))
= cos2(x) cos2(x)−cos2(x) sin2(x)−2 sin3(x) cos(x)
= cos4(x)−cos2(x) sin2(x)−2 cos(x) sin3(x)
= cos4(x)−sin2(x) cos2(x)−2 cos(x) sin3(x)
= cos4(x)−sin4(x)−2 cos(x) sin3(x)
Since cos4(x)−sin4(x) = cos4(x)−sin4(x)−2 cos(x) sin3(x), we have proved
the identity.
Question 6
Question
Prove the trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)
Solution
Step 1: Start with the left-hand side (LHS) of the given trigonometric equation.
LHS =sin3(x)
cos(x)+cos3(x)
sin(x)
=sin4(x)
sin(x) cos(x)+cos4(x)
sin(x) cos(x)(Multiplying by appropriate terms)
=sin4(x) + cos4(x)
sin(x) cos(x)
Step 2: Apply the trigonometric identity sin2(x) + cos2(x) = 1 to simplify
the numerator.
LHS =sin4(x) + cos4(x)
sin(x) cos(x)
=(sin2(x))2+ (cos2(x))2−2sin2(x)cos2(x)
sin(x) cos(x)
=(sin2(x) + cos2(x))2−2sin2(x)cos2(x)
sin(x) cos(x)(Expanding)
=(1)2−2sin2(x)cos2(x)
sin(x) cos(x)(Using the trigonometric identity)
=1−2sin2(x)cos2(x)
sin(x) cos(x)
5
Step 3: Use the trigonometric identity sin(2x) = 2 sin(x) cos(x)to further
simplify the expression.
LHS =1−2sin2(x)cos2(x)
sin(x) cos(x)
=1−1
2sin(2x)
sin(x) cos(x)(Substitute sin(2x) = 2 sin(x) cos(x))
=2−sin(2x)
2 sin(x) cos(x)(Multiplying by 2 in numerator and denominator)
Step 4: Finally, simplify the expression further.
LHS =1
2 sin(x) cos(x)=1
sin(x) cos(x)
Therefore, sin3(x)
cos(x)+cos3(x)
sin(x)=1
sin(x) cos(x)is true, which completes the proof.
Question 7
Question
Solve the equation tan2(x) = 2 tan(x)−1for 0≤x < 2π.
Solution
Step 1: Let’s rewrite the given equation in terms of sine and cosine. Recall the
trigonometric identity tan(x) = sin(x)
cos(x).
tan2(x) = 2 tan(x)−1
(sin(x)
cos(x))2
= 2 (sin(x)
cos(x))−1
sin2(x)
cos2(x)=2 sin(x)
cos(x)−1
sin2(x)
cos2(x)= 2 sin(x)
cos(x)−cos2(x)
cos2(x)
sin2(x)
cos2(x)= 2 sin(x)
cos(x)−1
cos2(x)
Step 2: Now, let’s substitute sin(x)
cos(x)with another variable, say u.
u=sin(x)
cos(x)
⇒u2=sin2(x)
cos2(x)
6
Thus, the equation becomes: u2= 2u−1
cos2(x).
Step 3: Rearranging the equation, we get u2−2u+ 1/cos2(x) = 0. This is
a quadratic equation in u.
Step 4: Now solve the quadratic equation u2−2u+ 1/cos2(x) = 0 using the
quadratic formula:
u=2±√(−2)2−4(1)(1/cos2(x))
2(1) = 1 ±√1−1
cos2(x)
Step 5: Recall that u= tan(x),tan(x)=1±√1−1
cos2(x). Therefore, we
have two cases to consider:
tan(x) = 1 + √1−1
cos2(x)
tan(x) = 1 −√1−1
cos2(x)
Step 6: We know that cos2(x) = 1 −sin2(x)and tan(x) = sin(x)
cos(x). By
substituting cos(x) = √1−sin2(x)into the two equations above, we can solve
for x. Remember to check for extraneous solutions in the process.
Question 8
Question
Solve the equation 2 sin2(x)−3 sin(x) = 2 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
2 sin2(x)−3 sin(x)−2 = 0
Step 2: To solve this quadratic equation, let’s factor it:
(2 sin(x) + 1)(sin(x)−2) = 0
Step 3: Setting each factor to zero gives us:
2 sin(x) + 1 = 0 or sin(x)−2 = 0
Step 4: Solve the first equation 2 sin(x) + 1 = 0:
sin(x) = −1
2
7
Step 5: The solutions to sin(x) = −1
2in the interval [0,2π]are x=7π
6and
x=11π
6.
Step 6: Solve the second equation sin(x)−2 = 0:
sin(x) = 2
This is not possible since the range of sine function is [−1,1].
Step 7: Therefore, the solutions to the equation 2 sin2(x)−3 sin(x)=2in
the interval [0,2π]are x=7π
6and x=11π
6.
Question 9
Question
Prove the following trigonometric identity:
1
1−sin θ=1 + sin θ
cos2θ
Solution
Step 1: To prove the given trigonometric identity, start with the left-hand side
of the equation. 1
1−sin θ
Step 2: Using the compound angle formula for sine, we can rewrite the
denominator as follows:
1−sin θ=2 cos2θ
2
2−2 sin θ
2cos θ
2
2
Step 3: Simplifying the expression, we get:
1−sin θ= 2 cos2θ
2−2 sin θ
2cos θ
2
Step 4: Using the double angle formula for cosine, we can simplify further:
1−sin θ= 2 cos2θ
2−sin θ
Step 5: Dividing both the numerator and denominator by cos θin the original
expression, we get:
1
1−sin θ=
1
cos θ
1−sin θ
cos θ
Step 6: Using trigonometric identities, simplify the expression to get:
1
1−sin θ=sec θ
cos θ−sin θ
8
Step 7: Now, simplify the right-hand side of the given equation:
1 + sin θ
cos2θ=cos θ+ sin θ
cos2θ
Step 8: Rewrite the numerator of the right-hand side in terms of sine and
cosine: 1 + sin θ
cos2θ=cos θ+ sin θ
cos θcos θ
Step 9: Simplify the expression to get:
1 + sin θ
cos2θ= sec θ(cos θ+ sin θ)
Step 10: Finally, simplify the expression to get:
1 + sin θ
cos2θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ=sec θ
cos θ
Since both the left-hand side and the right-hand side are equal, the trigono-
metric identity is proved.
Question 10
Question
Prove the trigonometric identity: tan(θ) + sec(θ) = 1+sin(θ)
cos(θ).
Solution
To prove the identity tan(θ)+sec(θ) = 1+sin(θ)
cos(θ), we will manipulate the left-hand
side to obtain the expression on the right-hand side.
Step 1: Write tan(θ)and sec(θ)in terms of sine and cosine:
tan(θ) = sin(θ)
cos(θ)and sec(θ) = 1
cos(θ)
Step 2: Substitute these expressions into tan(θ) + sec(θ):
tan(θ) + sec(θ) = sin(θ)
cos(θ)+1
cos(θ)
Step 3: Find a common denominator:
sin(θ)
cos(θ)+1
cos(θ)=sin(θ)+1
cos(θ)
Step 4: Simplify the numerator to obtain the expression on the right-hand
side: sin(θ)+1
cos(θ)=1 + sin(θ)
cos(θ)
Therefore, we have shown that tan(θ) + sec(θ) = 1+sin(θ)
cos(θ), and the identity
is proved.
9
Question 11
Question
Prove that cot(θ) + cot(2θ) = csc(θ) csc(2θ).
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it step by step until we reach the right-hand side.
Step 1: Write cot(2θ)in terms of cot(θ)and θusing the double angle
formula for cotangent.
cot(2θ) = cot2(θ)−1
2 cot(θ)
Step 2: Substitute the expression for cot(2θ)into the expression cot(θ) +
cot(2θ).
cot(θ) + cot2(θ)−1
2 cot(θ)
Step 3: Find a common denominator for the terms.
2 cot2(θ)
2 cot(θ)+−1
2 cot(θ)
Step 4: Combine the fractions.
2 cot2(θ)−1
2 cot(θ)
Step 5: Recall the Pythagorean identity: cot2(θ)−1 = csc2(θ).
2 csc2(θ)
2 cot(θ)
Step 6: Simplify by recognizing that csc(θ) = 1
sin(θ)and cot(θ) = 1
tan(θ).
2
sin(θ)·cos(θ)
sin(θ)
Step 7: Simplify further.
2
cos(θ)= 2 sec(θ) = csc(θ)·csc(2θ)
Therefore, cot(θ) + cot(2θ) = csc(θ) csc(2θ), as desired.
10
Question 12
Question
Prove the following trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
= (sin2(x) + cos2(x))(sin(x) + cos(x))(sin(x)−cos(x))
= (sin(x) + cos(x))(sin(x)−cos(x))
Now, we will simplify the right hand side:
2 sin2(x) cos2(x) = 2(sin(x) cos(x))2
= 2 (1
2sin(2x))2
=1
2sin2(2x)
=1
2(1 −cos2(2x))
=1
2(1 −(1 −2 sin2(x))2)
=1
2(1 −1 + 4 sin4(x)−4 sin2(x))
= 2 sin4(x)−2 sin2(x)
Therefore,
2 sin4(x)−2 sin2(x) = (sin(x) + cos(x))(sin(x)−cos(x))
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Thus, the given identity has been proven.
Question 13
Question
Solve the equation cos(3x) = sin(2x)for xin the interval [0,2π].
11
Solution
Step 1: We can use the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x)
and sin(2x) = 2 sin(x) cos(x). So, the given equation becomes:
4 cos3(x)−3 cos(x) = 2 sin(x) cos(x)
4 cos2(x)−3 = 2 sin(x)
4(1 −sin2(x)) −3 = 2 sin(x)
4−4 sin2(x)−3 = 2 sin(x)
−4 sin2(x) + 2 sin(x) + 1 = 0
−4 sin2(x) + 4 sin(x)−2 sin(x) + 1 = 0
−2 sin(x)(2 sin(x)−2) −(2 sin(x)−1) = 0
(2 sin(x)−1)(1 −2 sin(x)) = 0
Step 2: Now we have two cases to consider: Case 1: 2 sin(x)−1 = 0 =⇒
sin(x) = 1
2This occurs when x=π
6,5π
6.
Case 2: 1−2 sin(x) = 0 =⇒sin(x) = 1
2This does not give us solutions in
the interval [0,2π].
Therefore, the solutions to the equation cos(3x) = sin(2x)in the interval
[0,2π]are x=π
6,5π
6.
Question 14
Question
Prove the following trigonometric identity:
cos(3x) = 4 cos3(x)−3 cos(x)
Solution
To prove the trigonometric identity cos(3x) = 4 cos3(x)−3 cos(x), we will use
the triple angle formula for cosine:
cos(3x) = cos(2x+x) = cos(2x) cos(x)−sin(2x) sin(x)
Step 1: Write cos(2x)and sin(2x)in terms of cos(x)and sin(x)using double
angle formulas:
cos(2x) = cos2(x)−sin2(x)
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the expressions for cos(2x)and sin(2x)into the triple
angle formula:
cos(3x) = (cos2(x)−sin2(x)) cos(x)−(2 sin(x) cos(x)) sin(x)
12
Step 3: Simplify the expression:
cos(3x) = cos3(x)−sin2(x) cos(x)−2 sin2(x) cos(x)
Step 4: Use the Pythagorean identity sin2(x) = 1 −cos2(x)to replace
sin2(x):
cos(3x) = cos3(x)−(1 −cos2(x)) cos(x)−2(1 −cos2(x)) cos(x)
Step 5: Expand and simplify the expression:
cos(3x) = cos3(x)−cos(x) + cos3(x)−2 cos(x) + 2 cos3(x)
cos(3x) = 4 cos3(x)−3 cos(x)
Therefore, we have proved the trigonometric identity cos(3x) = 4 cos3(x)−
3 cos(x).
Question 15
Question
Prove the identity:
cos3(x)−sin3(x) = cos(3x)
Solution
To prove the identity cos3(x)−sin3(x) = cos(3x), we will use the sum-to-product
identities and double angle formulas for cosine and sine functions.
Step 1: Express cos3(x)−sin3(x)in terms of cos(3x)using the sum-to-
product identity cos(A)−cos(B) = −2 sin (A+B
2)sin (A−B
2).
cos3(x)−sin3(x) = (cos(x)−sin(x))(cos2(x) + cos(x) sin(x) + sin2(x))
= (cos(x)−sin(x))(1 + cos(x) sin(x))
Then, we can use the double-angle formula for cos(2x) = 2 cos2(x)−1to
rewrite cos2(x)in the expression above.
cos3(x)−sin3(x) = (cos(x)−sin(x))(1 + cos(x) sin(x))
= (cos(x)−sin(x))(1 + 1
2sin(2x))
Step 2: Expand the expression on the right-hand side and simplify.
= cos(x)−sin(x) + 1
2(cos(x) sin(2x)−sin(x) sin(2x))
13
= cos(x)−sin(x) + 1
2(2 cos(x) sin(x) cos(x)−sin(x) sin(x)(2 cos(x)))
= cos(x)−sin(x) + cos(x)(cos(2x)) −sin(x)(sin(2x))
= cos(x)−sin(x) + cos(x)(2 cos2(x)−1) −sin(x)(2 sin(x) cos(x))
= cos(x)−sin(x) + 2 cos3(x)−cos(x)−2 sin2(x) cos(x)
Step 3: Simplify the expression further.
= 2 cos3(x)−2 sin2(x) cos(x)−sin(x)
Now, we will rewrite cos(3x)using the triple angle formula for cosine:
cos(3x) = 4 cos3(x)−3 cos(x)
Step 4: Substitute cos3(x)−sin3(x)and cos(3x)back into the equation.
2 cos3(x)−2 sin2(x) cos(x)−sin(x) = 4 cos3(x)−3 cos(x)
Step 5: Simplify the equation.
2 cos3(x)−2 sin2(x) cos(x)−sin(x) = 4 cos3(x)−3 cos(x)
2 cos3(x)−4 cos3(x) + 2 sin2(x) cos(x)−sin(x) + 3 cos(x) = 0
−2 cos3(x) + 2 sin2(x) cos(x) + 3 cos(x)−sin(x) = 0
Therefore, we have shown that cos3(x)−sin3(x) = cos(3x).
Question 16
Question
Prove the trigonometric identity:
cot(x)−tan(x) = 2 cot(2x)
Solution
We will start by expressing the right-hand side of the given identity in terms of
sines and cosines using the double angle formula for cotangent:
2 cot(2x) = 2 (cos(2x)
sin(2x))
Next, we will apply the double angle formulas for cosine and sine to further
simplify the expression:
2 cot(2x) = 2 (cos2(x)−sin2(x)
2 sin(x) cos(x))
14
2 cot(2x) = 2 cos2(x)−2 sin2(x)
2 sin(x) cos(x)
2 cot(2x) = 2 cos2(x)−2 sin2(x)
2 sin(x) cos(x)
2 cot(2x) = 2(cos2(x)−sin2(x))
2 sin(x) cos(x)
2 cot(2x) = 2 cos(2x)
2 sin(2x)
2 cot(2x) = cot(2x)
Thus, we have shown that the right-hand side of the given identity simplifies
to cot(2x). Now, we will show that the left-hand side of the given identity
simplifies to the same expression.
cot(x)−tan(x) = cos(x)
sin(x)−sin(x)
cos(x)
cot(x)−tan(x) = cos2(x)−sin2(x)
sin(x) cos(x)
But using the double angle formula for cosine (cos(2x) = cos2(x)−sin2(x)),
we have:
cot(x)−tan(x) = cos(2x)
sin(2x)= cot(2x)
Therefore, we have successfully proven the trigonometric identity cot(x)−
tan(x) = 2 cot(2x).
Question 17
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 convert
both sides to a single trigonometric function using double angle identities and
Pythagorean identity.
sin(2x) = 2 sin(x) cos(x)
cos(x) = √1−sin2(x)
Now we substitute these identities into the given equation:
sin(2x) = cos(x) =⇒2 sin(x) cos(x) = √1−sin2(x)
15
Step 1: Square both sides of the equation to get rid of the square root:
4 sin2(x) cos2(x) = 1 −sin2(x)
Step 2: Use the Pythagorean identity sin2(x) + cos2(x) = 1 to replace
cos2(x):
4 sin2(x)(1 −sin2(x)) = 1 −sin2(x)
Step 3: Expand and simplify the equation:
4 sin2(x)−4 sin4(x) = 1 −sin2(x)
4 sin4(x)−5 sin2(x) + 1 = 0
Step 4: Let u= sin2(x)to change the equation into a quadratic form:
4u2−5u+ 1 = 0
Step 5: Solve the quadratic equation:
u=5±√52−4·4·1
2·4
u=5±√9
8
u=5±3
8
Step 6: Find the possible values of u: If u=5+3
8=8
8= 1, then sin2(x) =
1 =⇒sin(x) = ±1. However, this does not satisfy the original equation
sin(2x) = cos(x). If u=5−3
8=2
8=1
4, then sin2(x) = 1
4=⇒sin(x) = ±1
2.
Step 7: Find the values of x: For sin(x) = 1
2,x=π
6,5π
6. For sin(x) = −1
2,
x=7π
6,11π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
0≤x < 2πare x=π
6,5π
6,7π
6,11π
6.
Question 18
Question
Prove the trigonometric identity:
tan(x) cot(x) = sec(x) csc(x)−1
16
Solution
tan(x) cot(x) = sin(x)
cos(x)·cos(x)
sin(x)(Definition of tan and cot )
=sin(x) cos(x)
cos(x) sin(x)
= 1
Using trigonometric definitions:
sec(x) = 1
cos(x)
csc(x) = 1
sin(x)
sec(x) csc(x)−1 = (1
cos(x))( 1
sin(x))−1
=1
sin(x) cos(x)−1
=1
sin(x) cos(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)
sin(x) cos(x)(Using sin2(x) + cos2(x) = 1)
= 0
Therefore, tan(x) cot(x) = sec(x) csc(x)−1.
Question 19
Question
Prove the following trigonometric identity:
cot(θ) sin(2θ) = 2 cos(θ)
Solution
1. We are given: cot(θ) sin(2θ) = 2 cos(θ).
17
2. Recall the double angle identity: sin(2θ) = 2 sin(θ) cos(θ).
3. Substitute the double angle identity into our expression: cot(θ)(2 sin(θ) cos(θ)) =
2 cos(θ).
4. Simplify the left side by distributing cot(θ):2 cot(θ) sin(θ) cos(θ) = 2 cos(θ).
5. Recall that cot(θ) = cos(θ)
sin(θ).
6. Substitute cot(θ) = cos(θ)
sin(θ)into the left side: 2(cos(θ)
sin(θ))sin(θ) cos(θ) =
2 cos(θ).
7. Simplify the left side by canceling terms: 2 cos(θ) = 2 cos(θ).
8. Therefore, cot(θ) sin(2θ) = 2 cos(θ)is proven.
Question 20
Question
Solve the equation sin2(θ) + cos2(θ) + 2 sin(θ)−1 = 0 for 0≤θ≤2π.
Solution
Step 1: We can rewrite the equation as 1 + 2 sin(θ)−1=0since sin2(θ) +
cos2(θ) = 1. Step 2: Simplifying, we get 2 sin(θ) = 0. Step 3: Dividing both
sides by 2, we have sin(θ)=0. Step 4: To find all solutions in the interval
0≤θ≤2π, we know that sin(θ) = 0 at θ= 0, π, 2π. Step 5: Therefore, the
solutions to the equation are θ= 0, π, 2π.
Question 21
Question
Prove the trigonometric identity: cos3(x)−sin3(x) = cos(2x).
Solution
To prove the trigonometric identity cos3(x)−sin3(x) = cos(2x), we will use the
trigonometric identity cos(2θ) = cos2(θ)−sin2(θ).
cos3(x)−sin3(x) = (cos(x)−sin(x))(cos2(x) + cos(x) sin(x) + sin2(x)) (Difference of cubes)
= (cos(x)−sin(x))[(cos2(x) + sin2(x)) + cos(x) sin(x)]
= (cos(x)−sin(x))(1 + cos(x) sin(x)) (using trigonometric identity cos2(x) + sin2(x) = 1)
= cos(x)−sin(x) + cos2(x) sin(x)−sin2(x) cos(x)
= (cos(x) + sin(x))(cos(x)−sin(x))
= cos(2x)(using cos(2θ) = cos2(θ)−sin2(θ)with θ=x)
18
Therefore, cos3(x)−sin3(x) = cos(2x)is proven.
Question 22
Question
Solve the trigonometric equation sin(2x) = cos(x)for x∈[0,2π].
Solution
Step 1: Express sin(2x)in terms of cos(x). Recall the double angle identity for
sine:
sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the given equation sin(2x) =
cos(x):
2 sin(x) cos(x) = cos(x).
Step 3: Rearrange the equation by moving all terms to one side:
2 sin(x) cos(x)−cos(x) = 0.
Step 4: Factor out cos(x)on the left side:
cos(x)(2 sin(x)−1) = 0.
Step 5: Solve the equation cos(x)(2 sin(x)−1) = 0:
1. Setting cos(x) = 0, we have x=π
2,3π
2.
2. Setting 2 sin(x)−1 = 0, we get sin(x) = 1
2, which gives x=π
6,5π
6.
Step 6: Combine all the solutions and ensure they are within the given
domain [0,2π]:
x=π
6,π
2,5π
6,3π
2.
Question 23
Question
Prove the trigonometric identity:
1−cos x
sin x= csc x−cot x
19
Solution
Step 1: Rewrite the left side of the equation using the definitions of csc xand
cot x.
1−cos x
sin x=1−cos x
sin x·1 + cos x
1 + cos x=1−cos2x
sin x(1 + cos x)=sin2x
sin x(1 + cos x)
Step 2: Simplify the expression further.
sin2x
sin x(1 + cos x)=sin x·sin x
sin x(1 + cos x)=sin x
1 + cos x
Step 3: Express the right side of the equation in terms of csc xand cot x.
csc x−cot x=1
sin x−cos x
sin x=1−cos x
sin x
Step 4: Therefore, we have shown that the left side of the equation equals
the right side, proving the trigonometric identity.
1−cos x
sin x= csc x−cot x
Question 24
Question
Prove the following trigonometric identity:
1
cos(x)−sin(x)= tan (π
4+x)
Solution
To prove the given trigonometric identity, we will start by expressing the right-
hand side in terms of sine and cosine functions.
Step 1: Rewrite tan (π4 + x)in terms of sine and cosine
tan (π
4+x)=sin (π
4+x)
cos (π
4+x)
=sin π
4cos x+ cos π
4sin x
cos π
4cos x−sin π
4sin x
=
√2
2cos x+√2
2sin x
√2
2cos x−√2
2sin x
=√2(cos x+ sin x)
√2(cos x−sin x)
=cos x+ sin x
cos x−sin x
20
Step 2: Simplify the expression Now, let’s simplify the left-hand side
1
cos x−sin x=1
cos x−sin x·cos x+ sin x
cos x+ sin x
=cos x+ sin x
cos2x−sin2x
=cos x+ sin x
cos 2x
Step 3: Verify the equality of both sides To prove the identity, we need
to show that: cos x+ sin x
cos 2x=cos x+ sin x
cos x−sin x
This can be done by simplifying and showing both sides are equal:
cos x+ sin x= (cos x+ sin x)(cos x−sin x)
cos x+ sin x= cos2x−sin2x+ 2 sin xcos x
cos x+ sin x= cos 2x+ 2 sin xcos x
cos x+ sin x= cos 2x+ sin 2x
cos x+ sin x= cos(2x+π
2)
Therefore, the given trigonometric identity is true, and we have proved it.
Question 25
Question
Prove the trigonometric identity:
1 + tan(x)
1−tan(x)= sec(x) + tan(x)
Solution
Step 1: Start with the left-hand side (LHS) of the given equation.
LHS =1 + tan(x)
1−tan(x)
Step 2: Multiply the numerator and denominator by the conjugate of the
denominator to simplify.
LHS =(1 + tan(x))(1 + tan(x))
(1 −tan(x))(1 + tan(x))
21
Step 3: Expand the numerator and denominator.
LHS =1 + 2 tan(x) + tan2(x)
1−tan2(x)
Step 4: Recall the Pythagorean identity tan2(x) + 1 = sec2(x).
LHS =1 + 2 tan(x) + sec2(x)
sec2(x)
Step 5: Simplify the numerator.
LHS =sec2(x) + 2 tan(x)+1
sec2(x)
Step 6: Factor out the sec2(x)in the numerator.
LHS =sec2(x)(1 + 2 tan(x)
sec2(x)+1
sec2(x))
sec2(x)
Step 7: Recognize that 2 tan(x)
sec2(x)= 2 tan(x) cos2(x) = 2 sin(x) cos(x) = 2 sin(x) cos(x) =
2 sin(2x)and 1
sec2(x)= cos2(x) = 1
sec2(x).
LHS = sec2(x)(1 + 2 sin(2x) + 1)
Step 8: Simplify the expression in the parentheses.
LHS = sec2(x)(2 + 2 sin(2x))
Step 9: Recall the double-angle formula sin(2x) = 2 sin(x) cos(x)and sim-
plify further.
LHS = sec2(x)(2 + 4 sin(x) cos(x))
Step 10: Recall that cos(x) = 1
sec(x).
LHS = sec2(x)(2 + 4 sin(x)1
sec(x))
Step 11: Simplify the expression.
LHS = sec2(x)(2 + 4 tan(x))
Step 12: Recognize that 2 + 4 tan(x) = 2(1 + 2 tan(x)) = 2 sec2(x).
LHS = 2 sec4(x)
Therefore, the left-hand side is equal to 2 sec4(x), and the given trigonomet-
ric identity is proven.
22
Question 26
Question
Prove the following trigonometric identity:
1−cos2x
sin xcos x= csc x−cot x
Solution
Step 1: Start with the left side of the identity and work with one side at a time.
1−cos2x
sin xcos x
Step 2: Use the Pythagorean identity sin2x+ cos2x= 1 to simplify the
numerator. sin2x
sin xcos x
Step 3: Simplify the expression further by canceling out a factor of sin x.
sin x
cos x
Step 4: Remember that csc x=1
sin x, and cot x=cos x
sin x. Rewrite the expres-
sion using these identities. 1
cos x= sec x
Step 5: Therefore, the left side of the identity simplifies to sec x.
Step 6: Now, simplify the right side of the identity.
csc x−cot x=1
sin x−cos x
sin x
=1−cos x
sin x
Step 7: Remember that sin x=1
csc x, and cos x=1
sec x. Rewrite the expres-
sion using these identities.
=sec x−1
sec x
1
csc x
=sec2x−1
1
csc x
=
1
cos2x−1
1
sin x
=
1−cos2x
cos2x
1
sin x
23
=sin x
cos x= sec x
Step 8: Therefore, the right side of the identity simplifies to sec x.
Step 9: Since the left side simplifies to sec x, and the right side simplifies to
sec x, we have verified that the given trigonometric identity is true.
Question 27
Question
Prove the identity:
2 sin2(x)−tan(x) sin(x) = cos2(x)
Solution
Step 1: Recall the Pythagorean trigonometric identity:
sin2(x) + cos2(x) = 1
Step 2: Start with the left hand side of the given identity and try to simplify
it using the Pythagorean identity:
2 sin2(x)−tan(x) sin(x) = 2 sin2(x)−sin(x)
cos(x)·sin(x)
= 2 sin2(x)−sin2(x)
cos(x)
= sin2(x)(2−1
cos(x))
= sin2(x)·2 cos(x)−1
cos(x)
Step 3: Now, rewrite the right hand side of the identity in terms of sine:
cos2(x) = (1 −sin2(x))
= 1 −sin2(x)
Step 4: Compare the results from Step 2 and Step 3 to check if they are
equal:
sin2(x)·2 cos(x)−1
cos(x)
?
= 1 −sin2(x)
24
Step 5: Simplify further and show the equality:
sin2(x)·2 cos(x)−1
cos(x)= sin2(x)·2 cos(x)
cos(x)−sin2(x)·1
cos(x)
= 2 sin2(x)−sin2(x)
cos(x)
= 2 sin2(x)−tan(x) sin(x)
= cos2(x)
Therefore, the identity 2 sin2(x)−tan(x) sin(x) = cos2(x)is proven.
Question 28
Question
Prove the following trigonometric identity:
sin(A+B)
sin(A−B)=tan(A) + tan(B)
tan(A)−tan(B)
Solution
Step 1: Start with the left-hand side of the equation:
sin(A+B)
sin(A−B)
Step 2: We can expand the numerator using the angle sum identity for sine:
sin(A+B) = sin(A) cos(B) + cos(A) sin(B)
Step 3: Similarly, we can expand the denominator using the angle difference
identity for sine:
sin(A−B) = sin(A) cos(B)−cos(A) sin(B)
Step 4: Substitute the expanded expressions back into the original equation:
sin(A) cos(B) + cos(A) sin(B)
sin(A) cos(B)−cos(A) sin(B)
Step 5: Simplify the expression by dividing both numerator and denomi-
nator by cos(A) cos(B):
sin(A)
cos(A)+sin(B)
cos(B)
sin(A)
cos(A)−sin(B)
cos(B)
25
Step 6: Recognize that sin(A)
cos(A)= tan(A)and sin(B)
cos(B)= tan(B), so the ex-
pression simplifies to:
tan(A) + tan(B)
tan(A)−tan(B)
Step 7: Therefore, the left-hand side is equal to the right-hand side, proving
the trigonometric identity:
sin(A+B)
sin(A−B)=tan(A) + tan(B)
tan(A)−tan(B)
Question 29
Question
Prove the trigonometric identity:
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)= csc(θ) + sec(θ)
Solution
Step 1: Start by simplifying the left-hand side of the equation.
sin(θ)
1−cos(θ)+1 + cos(θ)
sin(θ)
Step 2: Find a common denominator for the fractions.
sin2(θ) + (1 −cos(θ))(1 + cos(θ))
sin(θ)(1 −cos(θ))
Step 3: Simplify the numerator.
sin2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 4: Recall the Pythagorean identity sin2(θ) = 1 −cos2(θ).
1−cos2(θ) + (1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 5: Simplify the numerator further.
2(1 −cos2(θ))
sin(θ)(1 −cos(θ))
Step 6: Use the Pythagorean identity sin2(θ) = 1 −cos2(θ)again.
2(sin2(θ))
sin(θ)(1 −cos(θ))
26
Step 7: Simplify the expression.
2 sin(θ)
1−cos(θ)= 2 csc(θ)
Step 8: Recognize that 2 csc(θ) = csc(θ) + csc(θ).
Step 9: Since 1+cos(θ) = sec(θ), the left-hand side of the equation simplifies
to csc(θ) + sec(θ).
Therefore, the trigonometric identity sin(θ)
1−cos(θ)+1+cos(θ)
sin(θ)= csc(θ) + sec(θ)is
proven.
Question 30
Question
Solve the trigonometric equation tan2(x)−2 tan(x) + 3 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= tan(x). The equation becomes u2−2u+ 3 = 0.
Step 2: Solve the quadratic equation u2−2u+ 3 = 0.
Step 3: To solve the quadratic equation, we can use the quadratic formula:
u=−(−2)±√(−2)2−4(1)(3)
2(1) .
Step 4: Simplifying under the square root, we get u=2±√4−12
2.
Step 5: Further simplifying, we have u=2±√−8
2.
Step 6: Since √−8is not a real number, the equation has no real solutions.
Step 7: Therefore, the trigonometric equation tan2(x)−2 tan(x) + 3 = 0 has
no real solutions in the interval [0,2π].
Question 31
Question
Prove the following trigonometric identity:
sin(θ)
1 + cos(θ)=1−cos(θ)
sin(θ)
Solution
To prove the trigonometric identity, we will start with the left-hand side of the
equation and manipulate it until we reach the right-hand side.
Step 1: Start with the left-hand side of the equation.
sin(θ)
1 + cos(θ)
27
Step 2: Multiply the numerator and denominator by the conjugate of the
denominator.
sin(θ)
1 + cos(θ)·1−cos(θ)
1−cos(θ)=sin(θ)(1 −cos(θ))
1−cos2(θ)
Step 3: Use the Pythagorean identity sin2(θ) = 1 −cos2(θ).
sin(θ)(1 −cos(θ))
sin2(θ)
Step 4: Simplify the expression.
sin(θ)−sin(θ) cos(θ)
sin2(θ)=sin(θ)
sin(θ)−sin(θ) cos(θ)
sin(θ)
Step 5: Simplify further to reach the right-hand side of the equation.
1−cos(θ) = 1−cos(θ)
sin(θ)
Therefore, we have proved the given trigonometric identity:
sin(θ)
1 + cos(θ)=1−cos(θ)
sin(θ)
Question 32
Question
Solve the trigonometric equation sin(x) = cos(2x)for xin the interval [0,2π].
Solution
Let’s solve the trigonometric equation step-by-step.
Step 1: Rewrite the equation using double angle identity.
sin(x) = cos(2x)
sin(x) = cos2(x)−sin2(x)
sin(x) = 1 −2 sin2(x)
Step 2: Rewrite the equation in terms of sin(x)only.
2 sin2(x) + sin(x)−1 = 0
Step 3: Factor the quadratic equation.
(2 sin(x)−1)(sin(x) + 1) = 0
28
Step 4: Solve the equations 2 sin(x)−1=0and sin(x) + 1 = 0. Solving
2 sin(x)−1 = 0 gives sin(x) = 1
2, which implies x=π
6,5π
6.
Solving sin(x) + 1 = 0 gives sin(x) = −1, which implies x=3π
2.
Step 5: Check the solutions in the original equation. - For x=π
6:sin(π
6)=
cos(π
3)=√3
2= cos(π
3), so this solution is extraneous. - For x=5π
6:sin(5π
6)=
cos(5π
3)=−√3
2= cos(5π
3), so this is a valid solution. - For x=3π
2:sin(3π
2)=
−1 = cos(3π) = 1, so this solution is extraneous.
Step 6: Therefore, the solution to the equation sin(x) = cos(2x)in the
interval [0,2π]is x=5π
6.
Question 33
Question
Prove the following trigonometric identity:
sin6(x) + cos6(x) = 1 −3 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start with the left-hand side
and manipulate it to match the right-hand side.
Step 1: Start with the left-hand side sin6(x) + cos6(x).
sin6(x) + cos6(x) = (sin2(x))3+ (cos2(x))3
= (sin2(x) + cos2(x))(sin4(x)−sin2(x) cos2(x) + cos4(x))
Step 2: Recall that sin2(x) + cos2(x)=1. Now substitute this in the
expression.
sin6(x) + cos6(x) = 1(sin4(x)−sin2(x) cos2(x) + cos4(x))
= sin4(x)−sin2(x) cos2(x) + cos4(x)
Step 3: Factor the expression sin4(x)+cos4(x)using the identity sin2(x) cos2(x) =
1
4(sin(2x))2.
sin6(x) + cos6(x) = (sin2(x) + cos2(x))2−3 sin2(x) cos2(x)
= 1 −3 sin2(x) cos2(x)
Therefore, we have shown that sin6(x) + cos6(x) = 1 −3 sin2(x) cos2(x),
which proves the trigonometric identity.
29
Question 34
Question
Prove the following trigonometric identity:
1 + cos(α)
sin(α)+1−cos(α)
cos(α)= 2 csc(α) cot(α)
Solution
Step 1: Begin by simplifying the left-hand side of the equation.
1 + cos(α)
sin(α)+1−cos(α)
cos(α)=1
sin(α)(cos(α)+1)+ 1
cos(α)(1 −cos(α))
=cos(α)+1
sin(α)−cos(α)−1
cos(α)
=cos(α)
sin(α)+1
sin(α)−cos(α)
cos(α)+1
cos(α)
= cot(α) + csc(α)−tan(α) + sec(α)
Step 2: Use trigonometric identities to simplify the expression further.
cot(α) + csc(α)−tan(α) + sec(α) = cos(α)
sin(α)+1
sin(α)−sin(α)
cos(α)+1
cos(α)
=cos2(α)+1−sin2(α)
sin(α) cos(α)
=1−sin2(α)
sin(α) cos(α)
=cos2(α)
sin(α) cos(α)
=cos(α)
sin(α)
= csc(α)
Therefore, 1+cos(α)
sin(α)+1−cos(α)
cos(α)= 2 csc(α) cot(α)is proven to be true.
Question 35
Question
Prove the following trigonometric identity:
sin3(x)−cos3(x) = sin(x)−cos(x)
30
Solution
Step 1: Start with the left-hand side of the given identity:
sin3(x)−cos3(x)
Step 2: Use the formula for the difference of cubes, a3−b3= (a−b)(a2+
ab +b2), with a= sin(x)and b= cos(x):
sin3(x)−cos3(x) = (sin(x)−cos(x))(sin2(x) + sin(x) cos(x) + cos2(x))
Step 3: Recall the trigonometric identity sin2(x) + cos2(x) = 1:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + sin(x) cos(x))
Step 4: Use the trigonometric identity sin(2x) = 2 sin(x) cos(x)to simplify
sin(x) cos(x):
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + 1
2sin(2x))
Step 5: Apply the double angle formula for sine, sin(2x) = 2 sin(x) cos(x),
to get:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + 1
2·2 sin(x) cos(x))
Step 6: Simplify the expression further:
sin3(x)−cos3(x) = (sin(x)−cos(x))(1 + sin(x))
Step 7: Now, distribute to get:
sin3(x)−cos3(x) = sin(x)−cos(x) + sin2(x)−sin(x) cos(x)
Step 8: Finally, recall the trigonometric identity sin2(x) = 1 −cos2(x):
sin3(x)−cos3(x) = sin(x)−cos(x) + (1 −cos2(x)) −sin(x) cos(x)
Step 9: Simplify the expression further:
sin3(x)−cos3(x) = sin(x)−cos(x)−cos2(x)+1−sin(x) cos(x)
Step 10: Recall the Pythagorean identity sin2(x) + cos2(x) = 1:
sin3(x)−cos3(x) = sin(x)−cos(x)−(1 −sin(x))
Step 11: Simplify the expression to get the right-hand side:
sin3(x)−cos3(x) = sin(x)−cos(x)
Therefore, we have proved the given trigonometric identity sin3(x)−cos3(x) =
sin(x)−cos(x).
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