MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 4
Liberty University
Question 1
Question
Prove the following trigonometric identity:
cos(2x) = 2 cos2(x)−1
Solution
To prove the trigonometric identity cos(2x) = 2 cos2(x)−1, we will use the
double angle formula for cosine:
cos(2x) = cos2(x)−sin2(x)
Step 1: Rewrite cos2(x)using the Pythagorean identity cos2(x)=1−
sin2(x).
cos(2x) = (1 −sin2(x)) −sin2(x)
Step 2: Simplify the expression.
cos(2x) = 1 −2 sin2(x)
Step 3: Use the Pythagorean identity sin2(x)=1−cos2(x)to rewrite
2 sin2(x).
cos(2x) = 1 −2(1 −cos2(x))
Step 4: Simplify the expression further.
cos(2x) = 1 −2 + 2 cos2(x)
cos(2x) = 2 cos2(x)−1
Therefore, the identity cos(2x) = 2 cos2(x)−1is proven.
Question 2
Question
Prove the following trigonometric identity:
cot4(x)−cot2(x) = csc2(x)−1.
Solution
To prove the trigonometric identity cot4(x)−cot2(x) = csc2(x)−1, we will
manipulate the left-hand side of the equation until it simplifies to the right-
hand side.
Step 1: Rewrite cotangent and cosecant in terms of sine and cosine using
the reciprocal identities:
cot4(x)−cot2(x) = cos(x)
sin(x)4
−cos(x)
sin(x)2
.
Step 2: Simplify the expression using the power of a quotient formula:
(a
b)n=an
bn.
cos(x)
sin(x)4
−cos(x)
sin(x)2
=cos4(x)
sin4(x)−cos2(x)
sin2(x).
Step 3: Write csc2(x)in terms of sine:
csc2(x) = 1
sin2(x).
Step 4: Substitute csc2(x)into the expression:
cos4(x)
sin4(x)−cos2(x)
sin2(x)=cos4(x)
sin4(x)−cos2(x)
csc2(x).
Step 5: Find a common denominator and combine the fractions:
cos4(x)
sin4(x)−cos2(x)
csc2(x)=cos4(x) csc2(x)
sin4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
Step 6: Simplify further:
cos4(x) csc2(x)
sin4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x)=cos4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
Step 7: Use the Pythagorean identity, cos2(x) +sin2(x) = 1, to simplify the
expression:
=cos2(x) sin2(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
2
Step 8: Factor out cos2(x) sin2(x)from the numerator and simplify:
=cos2(x) sin2(x)(csc2(x)−1)
sin4(x) csc2(x)=cos2(x) sin2(x)
sin4(x)=cos2(x)
sin2(x)= csc2(x)−1.
Therefore, cot4(x)−cot2(x) = csc2(x)−1holds true, and the identity is
proven.
Question 3
Question
Solve the trigonometric equation sin2(x)−√3 sin(x)+1 = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s first rewrite the equation in terms of sin(x).
sin2(x)−√3 sin(x) + 1 = 0
Step 2: We can treat this equation as a quadratic in terms of sin(x). Set
y= sin(x)to simplify the equation.
y2−√3y+ 1 = 0
Step 3: Now, we can solve this quadratic equation for yusing the quadratic
formula.
y=√3±√3−4
2=√3±i
2
Step 4: Since yis the sine of an angle, we need to find the angles whose sine
is √3+i
2and √3−i
2. These angles lie in the complex plane.
Step 5: The complex number √3+i
2can be rewritten as cos π
6+isin π
6.
Therefore, an angle whose sine is √3+i
2is π
6.
Step 6: Similarly, the complex number √3−i
2is equivalent to cos 5π
6+
isin 5π
6. So an angle that satisfies this is 5π
6.
Step 7: Now, we need to find the solutions for xin the interval [0,2π). Since
sin is a periodic function with a period of 2π, the solutions are:
x=π
6,5π
6(in the interval [0,2π)).
Question 4
Question
Prove the following trigonometric identity:
2 sin2(x)−sin(x) cos(x) = sin(2x)
3
Solution
To prove the trigonometric identity 2 sin2(x)−sin(x) cos(x) = sin(2x), we will
use the double angle formula for sine:
sin(2x) = 2 sin(x) cos(x)
Step 1: Start with the left-hand side of the equation.
2 sin2(x)−sin(x) cos(x) = 2 sin(x)·sin(x)−sin(x)·cos(x)
= sin(x)(2 sin(x)−cos(x))
Step 2: Now, we will use the double angle formula for sine to simplify the
right side of the equation.
sin(2x) = 2 sin(x) cos(x)
Step 3: Finally, we substitute back into the left-hand side to show that
both sides of the equation are equal.
sin(x)(2 sin(x)−cos(x)) = sin(x)·sin(2x)
= sin(2x)
Therefore, we have shown that 2 sin2(x)−sin(x) cos(x) = sin(2x), and the
trigonometric identity is proven.
Question 5
Question
Prove the following trigonometric identity:
sin2(x) cos2(x) = 1
4sin(4x)
Solution
To prove the given identity, we can start with the right-hand side and simplify
it using trigonometric identities.
Step 1: Start with the right-hand side of the equation:
1
4sin(4x)
Step 2: Apply the double angle identity for sine:
1
4·2 sin(2x) cos(2x)
4
Step 3: Apply the double angle identities for sine and cosine:
1
4·2·(2 sin(x) cos(x)) ·(2 cos2(x)−1)
Step 4: Simplify the expression:
2 sin(x) cos(x)·(4 cos2(x)−1)
Step 5: Apply the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify
4 cos2(x)−1:
2 sin(x) cos(x)·(4 −sin2(x))
Step 6: Distribute the terms:
8 sin(x) cos(x)−2 sin3(x) cos(x)
Step 7: Apply the double angle identity for sine:
8·1
2sin(2x)−2 sin3(x) cos(x)
Step 8: Apply the double angle identity for sine:
4 sin(2x)−2 sin(x) cos(x) sin2(x)
Step 9: Since sin2(x) = 1 −cos2(x), substitute into the expression:
4 sin(2x)−2 sin(x) cos(x)(1 −cos2(x))
Step 10: Expand and simplify:
4 sin(2x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 11: Apply the double angle identity for sine:
4·2 sin(x) cos(x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 12: Simplify further:
8 sin(x) cos(x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 13: Combine like terms:
6 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 14: Apply the double angle identity for sine:
6 sin(x) cos(x) + 2 sin(x) cos(x)(1 −cos2(x))
Step 15: Simplify:
6 sin(x) cos(x) + 2 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
5
Step 16: Combine like terms:
8 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
Step 17: Apply the double angle identity for cosine:
8 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
Step 18: Apply the identity sin(x) cos(x) = 1
2sin(2x):
4 sin(2x)−sin(2x) cos2(x)
Step 19: Apply the identity cos2(x) = 1 −sin2(x):
4 sin(2x)−sin(2x) + sin3(x)
Step 20: Simplify to obtain the left-hand side of the given identity:
3 sin3(x) + 4 sin(2x)
Therefore
Question 6
Question
Solve the trigonometric equation: sin(2x) cos(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Start by using the double angle identity sin(2x) = 2 sin(x) cos(x).
sin(2x) cos(x) = 2 sin(x) cos(x) cos(x) = 2 sin(x) cos2(x)
Step 2: Substitute this expression into the given equation: 2 sin(x) cos2(x) =
1
2.
2 sin(x) cos2(x) = 1
2
Step 3: We can rewrite cos2(x)as 1−sin2(x)using the Pythagorean identity.
2 sin(x)(1 −sin2(x)) = 1
2
Step 4: Expand and simplify the equation:
2 sin(x)−2 sin3(x) = 1
2
4 sin3(x)−2 sin(x) + 1 = 0
6
Step 5: Let u= sin(x)to convert the equation into a cubic equation in u:
4u3−2u+ 1 = 0
Step 6: Solve the cubic equation using numerical methods or approximations
to find the values of u.
Step 7: Once you have found the values of u, substitute back sin(x) = uto
find the solutions for xin the interval [0,2π).
Question 7
Question
Solve the trigonometric equation sin(2x) + sin(3x) = sin(x)for xin the interval
[0,2π).
Solution
Step 1: Recall the trigonometric identity sin(a) + sin(b) = 2 sin a+b
2cos a−b
2.
Applying this identity to the given equation gives:
sin(2x) + sin(3x) = 2 sin 5x
2cos x
2= sin(x)
Step 2: We can simplify the equation further by using the double angle
identity for sine sin(2θ) = 2 sin(θ) cos(θ):
2 sin 5x
2cos x
2= 2 sin x
2cos x
2
Step 3: Now, we have:
sin 5x
2= sin x
2
Step 4: This implies that either 5x
2=x
2+ 2πn for some integer n, or 5x
2=
π−x
2+ 2πm for some integer m.
Step 5: Solving the first case, we get:
5x
2=x
2+ 2πn
4x= 4πn
x=πn
Step 6: Solving the second case, we get:
5x
2=π−x
2+ 2πm
7
6x= 4π+ 4πm
x=2
3π(1 + 2m)
Step 7: Since we want solutions in the interval [0,2π), we consider values of n
and mthat satisfy this condition. Thus, the solutions to the given trigonometric
equation are x= 0,2
3π, π, 4
3π, 2π.
Question 8
Question
Prove the trigonometric identity:
sin2(x) + sin2π
2−x= 1
Solution
Step 1: Recall the trigonometric identity sinπ
2−x= cos(x). We will use this
identity to simplify the expression on the left side.
Step 2: Start by expanding the left side using the given identity:
sin2(x) + sin2π
2−x= sin2(x) + cos2(x)
Step 3: Next, recall the Pythagorean trigonometric identity which states
that sin2(x)+cos2(x) = 1. Substitute this identity into the previous expression:
sin2(x) + cos2(x) = 1
Step 4: Therefore, we have shown that:
sin2(x) + sin2π
2−x= 1
Hence, the trigonometric identity is proven.
Question 9
Question
Prove the following trigonometric identity:
cot(θ) + sin(θ) = csc(θ)−cos(θ)
8
Solution
To prove the trigonometric identity cot(θ) + sin(θ) = csc(θ)−cos(θ), we will
manipulate the expressions using the definitions of trigonometric functions and
basic trigonometric identities.
Step 1: Write the definitions of the trigonometric functions involved:
cot(θ) = cos(θ)
sin(θ),csc(θ) = 1
sin(θ)
Step 2: Rewrite the expression using the definitions:
cos(θ)
sin(θ)+ sin(θ) = 1
sin(θ)−cos(θ)
Step 3: Find a common denominator on the right-hand side:
cos(θ)
sin(θ)+ sin(θ) = 1−cos(θ)·sin(θ)
sin(θ)
Step 4: Use the trigonometric identity sin(θ) cos(θ) = 1
2sin(2θ):
cos(θ)
sin(θ)+ sin(θ) = 1−1
2sin(2θ)
sin(θ)
Step 5: Simplify the expression:
cos(θ)
sin(θ)+ sin(θ) = 2−sin(2θ)
2 sin(θ)
Step 6: Apply the double-angle formula sin(2θ) = 2 sin(θ) cos(θ):
cos(θ)
sin(θ)+ sin(θ) = 2−2 sin(θ) cos(θ)
2 sin(θ)
Step 7: Simplify the expression:
cos(θ)
sin(θ)+ sin(θ) = 2(1 −cos(θ))
2 sin(θ)
Step 8: Cancel out the common factor of 2:
cos(θ)
sin(θ)+ sin(θ) = 1−cos(θ)
sin(θ)
Step 9: Recognize that the expression on the right-hand side is equal to
csc(θ)−cos(θ):
cos(θ)
sin(θ)+ sin(θ) = csc(θ)−cos(θ)
Therefore, the trigonometric identity cot(θ) + sin(θ) = csc(θ)−cos(θ)is
proven.
9
Question 10
Question
Prove the trigonometric identity:
sin(x)
1 + cos(x)+1 + cos(x)
sin(x)= 2 csc(x) cot(x)
Solution
sin(x)
1 + cos(x)+1 + cos(x)
sin(x)=sin2(x) + 1 + cos(x)(1 + cos(x))
sin(x)(1 + cos(x))
=1 + sin2(x) + cos(x) + cos2(x)
sin(x)(1 + cos(x))
=2 + 2 sin2(x)
sin(x)(1 + cos(x))
=2(1 + sin2(x))
sin(x)(1 + cos(x))
=2(cos2(x) + sin2(x))
sin(x)(cos(x) + cos2(x))
=2
sin(x)·cos2(x) + sin2(x)
cos(x) + cos2(x)
= 2 csc(x) cot(x)
Question 11
Question
Prove the following trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: We start with the left-hand side of the given equation. Step 2: Rewrite
sin2(x)as (1 −cos2(x)) using the Pythagorean identity sin2(x) + cos2(x) =
1. Step 3: Rewrite cos2(x)as (1 −sin2(x)) using the Pythagorean identity
sin2(x) + cos2(x) = 1. Step 4: Substitute the rewritten expressions into the
left-hand side of the given equation. Step 5: Simplify the expression by com-
bining like terms in the numerator. Step 6: Combine the two fractions in the
left-hand side as a single fraction. Step 7: Find a common denominator for the
10
combined fraction. Step 8: Combine the terms in the numerator over the com-
mon denominator. Step 9: Simplify the expression to show that the left-hand
side equals 1, completing the proof of the trigonometric identity.
Question 12
Question
Prove the trigonometric identity: sin4(θ) + cos4(θ) = 1 −2 sin2(θ) cos2(θ).
Solution
To prove the given trigonometric identity, we will start with the left side of the
equation and manipulate it until it matches the right side.
Step 1: Start with the left side of the identity.
sin4(θ) + cos4(θ) = (sin2(θ))2+ (cos2(θ))2
= (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
= 1(sin2(θ)−cos2(θ))
= sin2(θ)−cos2(θ)(using sin2(θ) + cos2(θ) = 1)
Step 2: Use the double angle formula sin(2θ) = 2 sin(θ) cos(θ).
sin2(θ)−cos2(θ) = sin2(θ)−(1 −sin2(θ))
= sin2(θ)−1 + sin2(θ)
= 2 sin2(θ)−1
Step 3: Substitute the result back into the expression.
sin4(θ) + cos4(θ) = 2 sin2(θ)−1
= 1 + 2 sin2(θ) cos2(θ)−1
= 2 sin2(θ) cos2(θ)
= 1 −2 sin2(θ) cos2(θ)■
Question 13
Question
Prove the following trigonometric identity:
sin2(x) tan2(x) = tan2(x)−cos2(x)
11
Solution
To prove the trigonometric identity sin2(x) tan2(x) = tan2(x)−cos2(x), we will
start with the left-hand side and simplify it step by step until we obtain the
right-hand side.
Step 1: Start with the left-hand side of the given identity:
sin2(x) tan2(x)
Step 2: Recall that tan(x) = sin(x)
cos(x). Therefore:
sin2(x) tan2(x) = sin2(x)sin(x)
cos(x)2
= sin2(x)·sin2(x)
cos2(x)=sin4(x)
cos2(x)
Step 3: Express sin4(x)in terms of cos using the Pythagorean identity
sin2(x) + cos2(x) = 1:
sin4(x) = (sin2(x))2= (1 −cos2(x))2
Step 4: Now, substitute sin4(x) = (1 −cos2(x))2into the expression:
(1 −cos2(x))2
cos2(x)=1−2 cos2(x) + cos4(x)
cos2(x)=1
cos2(x)−2 + cos2(x)
Step 5: Simplify the expression:
1
cos2(x)−2 + cos2(x) = tan2(x)−2 + cos2(x) = tan2(x)−cos2(x)
Hence, we have shown that sin2(x) tan2(x) = tan2(x)−cos2(x), which com-
pletes the proof of the given trigonometric identity.
Question 14
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.
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
12
Next, we will express sin(2x)and cos(2x)in terms of sin(x)and cos(x)using
double angle identities.
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
= 1 −2 sin2(x)
Substitute these expressions back into the previous equation.
sin(3x) = 2(2 sin(x) cos(x)) cos(x) + (1 −2 sin2(x)) sin(x)
= 4 sin(x) cos2(x) + sin(x)−2 sin3(x)
= 4 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
= 4 sin(x)−4 sin3(x) + sin(x)−2 sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, we have proved the trigonometric identity sin(3x) = 3 sin(x)−
4 sin3(x).
Question 15
Question
Solve the trigonometric equation:
2 sin2x+ 3 cos x−1 = 0
for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2x+cos2x= 1.
2 sin2x+ 3 cos x−1 = 0
2(1 −cos2x) + 3 cos x−1 = 0
2−2 cos2x+ 3 cos x−1 = 0
2−2 cos2x+ 3 cos x−1 = 0
−2 cos2x+ 3 cos x+ 1 = 0
Step 2: Try to factorize the quadratic equation −2 cos2x+ 3 cos x+ 1 = 0.
(−2 cos x+ 1)(cos x+ 1) = 0
13
Step 3: Solve for cos xby setting each factor to zero.
−2 cos x+ 1 = 0 or cos x+ 1 = 0
−2 cos x=−1or cos x=−1/2
cos x= 1/2or cos x=−1
Step 4: Find the solutions for xin the interval [0,2π). Since cos x= 1/2
when x=π
3and 5π
3, and cos x=−1when x=π, the solutions to the given
trigonometric equation are x=π
3, π, 5π
3.
Question 16
Question
Prove the trigonometric identity:
1
sin4(x)= csc2(x) + csc4(x)
Solution
To prove the trigonometric identity, we will start with the right side of the
equation and simplify it to match the left side.
Step 1: Write csc2(x)and csc4(x)in terms of sine and cosine:
csc2(x) = 1
sin2(x)
csc4(x) = 1
sin2(x)2
=1
sin4(x)
Step 2: Add csc2(x)and csc4(x):
csc2(x) + csc4(x) = 1
sin2(x)+1
sin4(x)=sin2(x)+1
sin4(x)
Step 3: Simplify the expression sin2(x)+1
sin4(x):
sin2(x)+1
sin4(x)=1−cos2(x)+1
sin4(x)=2−cos2(x)
sin4(x)
Step 4: Use the Pythagorean identity sin2(x)=1−cos2(x)to simplify
further:
2−cos2(x)
sin4(x)=2−(1 −sin2(x))
sin4(x)=1 + sin2(x)
sin4(x)=1
sin4(x)
14
Step 5: Therefore, we have shown that csc2(x) + csc4(x) = 1
sin4(x). Hence,
the trigonometric identity:
1
sin4(x)= csc2(x) + csc4(x)
has been proven.
Question 17
Question
Prove the trigonometric identity:
sin x−sin y
cos x−cos y= tan x+y
2
Solution
Step 1: Expand the numerator and denominator using the angle difference
identities for sine and cosine:
Since sin(x−y) = sin xcos y−cos xsin yand cos(x−y) = cos xcos y+sin xsin y,
sin x−sin y= 2 sin x−y
2cos x+y
2and cos x−cos y=−2 sin x+y
2sin x−y
2.
Step 2: Substitute the expanded terms back into the original expression:
sin x−sin y
cos x−cos y=2 sin x−y
2cos x+y
2
−2 sin x+y
2sin x−y
2.
Step 3: Simplify the expression by canceling out common terms:
2 sin x−y
2cos x+y
2
−2 sin x+y
2sin x−y
2=−tan x+y
2.
Hence, we have proved the desired trigonometric identity:
sin x−sin y
cos x−cos y= tan x+y
2.
Question 18
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 1 −2 cos2(x)
15
Solution
To prove the given trigonometric identity, we will start with the left hand side
and manipulate it to match the right hand side.
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
= (sin(x)−cos(x))(sin(x) + cos(x))
= sin(x) sin(x) + cos(x) cos(x)
= sin2(x) + cos2(x)
= 1
Therefore, sin4(x)−cos4(x) = 1.
Next, we will simplify the right hand side to match our result.
1−2 cos2(x) = 1 −2(1 −sin2(x))
= 1 −2 + 2 sin2(x)
= 2 sin2(x)−1
= sin2(x) + cos2(x)−cos2(x)
= sin2(x)−cos2(x)
= 1
Therefore, the identity sin4(x)−cos4(x) = 1 −2 cos2(x)is verified.
Question 19
Question
Prove the trigonometric identity:
sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
Solution
To prove the given trigonometric identity, we will expand sin(5x)using the angle
addition formula for sine repeatedly.
Step 1: Using the double-angle formula sin(2θ) = 2 sin(θ) cos(θ), we have:
sin(5x) = sin(3x+ 2x)
= sin(3x) cos(2x) + cos(3x) sin(2x)
16
Step 2: Expanding cos(2x)and sin(2x)using the double-angle formulas, we
get:
sin(5x) = sin(3x)(2 cos2(x)−1) + cos(3x)(2 sin(x) cos(x))
Step 3: Expanding further and using the angle addition formulas for sine
and cosine, we get:
sin(5x) = 2 sin(3x) cos2(x)−sin(3x) + 2 cos(3x) sin(x) cos(x)
Step 4: Expanding sin(3x)and cos(3x)using the angle addition formulas,
we obtain:
sin(5x) = 2(sin(x) cos(x)(2 cos2(x)−1))−sin(3x)+2(cos3(x)−3 cos(x) sin2(x)) sin(x)
Step 5: Simplifying the above expression gives us:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−sin(3x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 6: Substitute sin(3x) = 3 sin(x)−4 sin3(x)into the equation above:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−(3 sin(x)−4 sin3(x))+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 7: Simplify the expression by expanding and combining like terms:
sin(5x) = (4 cos2(x)−2) sin(x) cos(x)−3 sin(x)+4 sin3(x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 8: Using sin(2x) = 2 sin(x) cos(x), we can further simplify the expres-
sion:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−3 sin(x)+4 sin3(x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 9: After simplification, we arrive at the right-hand side of the given
trigonometric identity:
16 sin5(x)−20 sin3(x) + 5 sin(x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
Therefore, we have proven the trigonometric identity sin(5x) = 16 sin5(x)−
20 sin3(x) + 5 sin(x).
Question 20
Question
Solve the trigonometric equation sin2(x)+cos(x) = 1 for xin the interval [0,2π).
17
Solution
Step 1: Notice that we can rewrite sin2(x)as 1−cos2(x)using the Pythagorean
identity sin2(x) + cos2(x)=1. Step 2: Substitute this new expression into the
original equation: 1−cos2(x) + cos(x) = 1. Step 3: Rearrange the terms to
get −cos2(x) + cos(x)=0. Step 4: Factor out a cos(x)from the equation:
cos(x)(cos(x)−1) = 0. Step 5: Set each factor equal to 0: cos(x) = 0 or
cos(x)−1 = 0. Step 6: Solve cos(x) = 0 for xin the interval [0,2π):x=π
2,3π
2.
Step 7: Solve cos(x)−1 = 0 for xin the interval [0,2π):x= 2π. Step 8:
Therefore, the solutions to the equation sin2(x) + cos(x)=1in the interval
[0,2π)are x=π
2,3π
2,2π.
Question 21
Question
Prove the following trigonometric identity:
sin(θ)
cos(θ)−1=1
sec(θ) + tan(θ)
Solution
Step 1: We will start with the left-hand side of the equation and try to simplify
it using trigonometric identities.
sin(θ)
cos(θ)−1=sin(θ)
cos(θ)−cos(θ)
cos(θ)
[Using the identity sec(θ) = 1
cos(θ)]
=sin(θ)
cos(θ)−cos(θ)
cos(θ)
=sin(θ)
0
=undefined,
since division by zero is undefined. Hence, the left-hand side cannot be simplified
as it is undefined.
Step 2: Next, we will simplify the right-hand side of the equation using
trigonometric identities.
1
sec(θ) + tan(θ)=1
1
cos(θ)+sin(θ)
cos(θ)
=1
1+sin(θ)
cos(θ)
=cos(θ)
1 + sin(θ).
18
Since the right-hand side has been simplified and is defined, but the left-
hand side is undefined, we conclude that the given trigonometric identity is not
valid.
Question 22
Question
Prove the trigonometric identity:
2 sin2(x) + 3 cos2(x) = 5 −cos(2x)
Solution
Step 1: Recall the double angle formula for cosine: cos(2x) = cos2(x)−sin2(x).
Step 2: We will start by expanding the right side of the identity.
Right side = 5 −cos(2x) = 5 −(cos2(x)−sin2(x))
Step 3: Substitute cos2(x) = 1 −sin2(x)into the equation above.
Right side = 5 −(1 −sin2(x)−sin2(x)) = 5 −1 + 2 sin2(x)
Step 4: Simplify the right side and compare it with the left side of the
original identity.
Right side = 4 + 2 sin2(x) = 2(2 + sin2(x))
Step 5: Now, we simplify the left side of the original identity.
2 sin2(x) + 3 cos2(x) = 2 sin2(x) + 3(1 −sin2(x)) = 2 sin2(x)+3−3 sin2(x)
Step 6: Combine like terms in the left side of the original identity.
2 sin2(x) + 3 cos2(x) = −sin2(x) + 3 = −sin2(x)+3
Step 7: Since the right and left sides are not equal, the identity
2 sin2(x) + 3 cos2(x) = 5 −cos(2x)
is not true for all xvalues.
Question 23
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
19
Solution
Step 1: Recall that the double angle identity for sine is sin(2x) = 2 sin(x) cos(x).
Step 2: Substituting the double angle identity into the equation, we have
2 sin(x) cos(x) = cos(x).
Step 3: Dividing both sides by cos(x)(assuming cos(x)= 0), we get 2 sin(x) =
1.
Step 4: Dividing by 2, we find sin(x) = 1
2.
Step 5: The solutions to sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 24
Question
Solve the trigonometric equation cos2(x) = cos(2x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle formula cos(2θ) = 2 cos2(θ)−1.
Step 2: Rewrite the given equation cos2(x) = cos(2x)using the double angle
formula:
cos2(x) = 2 cos2(x)−1.
Step 3: Rearrange the equation by moving all terms to one side:
2 cos2(x)−cos2(x)−1 = 0.
Step 4: Simplify the equation:
cos2(x)−1 = 0.
Step 5: Factor the equation:
(cos(x) + 1)(cos(x)−1) = 0.
Step 6: Set each factor to zero:
cos(x) + 1 = 0
cos(x) = −1
x=π
20
cos(x)−1 = 0
cos(x) = 1
x= 0.
Step 7: Check for extraneous solutions. Since both 0and πare within the
interval [0,2π], both solutions are valid.
Therefore, the solutions to the trigonometric equation cos2(x) = cos(2x)in
the interval [0,2π]are x= 0 and x=π.
Question 25
Question
Prove the following trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
Solution
Let’s start with the left-hand side of the identity and manipulate it step by step
to show that it equals the right-hand side. Step 1: Start with the left-hand side
of the identity.
sin4(θ)−cos4(θ)
Step 2: Rewrite sin4(θ)as (sin2(θ))2and cos4(θ)as (cos2(θ))2to get:
= (sin2(θ))2−(cos2(θ))2
Step 3: Recall the Pythagorean trigonometric identity sin2(θ) + cos2(θ) = 1 and
rewrite the expression using this identity:
= (1 −cos2(θ))2−(cos2(θ))2
Step 4: Expand (1 −cos2(θ))2to get:
= 1 −2 cos2(θ) + (cos2(θ))2−(cos2(θ))2
Step 5: Simplify the expression by canceling out like terms:
= 1 −2 cos2(θ)
Step 6: Now, rewrite sin2(θ) cos2(θ)as 1
4(2 sin θcos θ)2:
= 1 −2(1 −sin2(θ))
Step 7: Rewrite 1−sin2(θ)as cos2(θ):
= 1 −2 + 2 sin2(θ)
21
Step 8: Simplify the expression further to get:
=−1 + 2 sin2(θ)
Step 9: Since −1 = −2·1
2, we can finally rewrite the expression as:
= 2 sin2(θ) cos2(θ)
Step 10: Therefore, we have shown that sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ),
which proves the trigonometric identity.
Question 26
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 formulas for sine and cosine, along with some algebraic ma-
nipulations.
Step 1: Express sin(3x)using the angle addition formula By the
angle addition formula for sine, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x).
Step 2: Express sin(2x)and cos(2x)in terms of sin(x)and cos(x)
Using the double angle formulas for sine and cosine:
sin(2x) = 2 sin(x) cos(x),
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x).
Step 3: Substitute sin(2x)and cos(2x)into the expression for sin(3x)
Substitute the expressions for sin(2x)and cos(2x)into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x).
Step 4: Simplify the expression for sin(3x)
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin3(x).
Step 5: Further simplify the expression for sin(3x)Use the identity
cos2(x) = 1 −sin2(x):
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x).
Step 6: Simplify the expression for sin(3x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x) = 3 sin(x)−4 sin3(x).
Therefore, we have proven the trigonometric identity sin(3x) = 3 sin(x)−
4 sin3(x).
22
Question 27
Question
Solve the trigonometric equation cos2(x) + cos(x)−6 = 0 for 0≤x < 2π.
Solution
Step 1: Let’s denote cos(x)as uto simplify the equation:
u2+u−6 = 0
Step 2: Now we need to solve this quadratic equation for u.
Step 3: Factor the quadratic equation:
(u+ 3)(u−2) = 0
Step 4: Set each factor to zero and solve for u:
u+ 3 = 0 or u−2 = 0
u=−3or u= 2
Step 5: Recall that u= cos(x). Since the cosine function only takes values
between -1 and 1, the solution u=−3is extraneous. Thus, we only consider
u= 2.
Step 6: Solve cos(x) = 2 for 0≤x < 2π: There are no solutions for
cos(x) = 2 since the cosine function only ranges from -1 to 1.
Step 7: Therefore, the trigonometric equation cos2(x) + cos(x)−6=0has
no solutions for 0≤x < 2π.
Question 28
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ)−1
Solution
To prove the trigonometric identity sin4(θ)−cos4(θ) = 2 sin2(θ)−1, we will
start from the left-hand side and manipulate it to obtain the right-hand side.
Step 1: Use the difference of squares formula to expand sin4(θ)−cos4(θ).
sin4(θ)−cos4(θ) = (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
23
Step 2: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1, and rewrite
the expression using this identity.
sin4(θ)−cos4(θ) = (1)(sin2(θ)−cos2(θ))
Step 3: Factor the expression further.
sin4(θ)−cos4(θ) = sin2(θ)−cos2(θ)
Step 4: Use the double-angle identity sin2(θ) = 1−cos(2θ)
2and cos2(θ) =
1+cos(2θ)
2to express sin2(θ)and cos2(θ)in terms of cos(2θ).
sin4(θ)−cos4(θ) = 1−cos(2θ)
2−1 + cos(2θ)
2
Step 5: Simplify the expression.
sin4(θ)−cos4(θ) = 1−cos(2θ)−1−cos(2θ)
2=−cos(2θ)
Step 6: Use the double-angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
sin4(θ)−cos4(θ) = −2 cos2(θ)+1
Step 7: Replace cos2(θ)using the Pythagorean identity cos2(θ) = 1 −
sin2(θ).
sin4(θ)−cos4(θ) = −2(1 −sin2(θ)) + 1
Step 8: Distribute and simplify to obtain the right-hand side of the given
identity.
sin4(θ)−cos4(θ) = −2 + 2 sin2(θ) + 1 = 2 sin2(θ)−1
Therefore, sin4(θ)−cos4(θ) = 2 sin2(θ)−1, as required.
Question 29
Question
Prove the trigonometric identity:
2 sin(θ) cos(θ) = sin(2θ).
Solution
Step 1: Recall the double angle identity for sine:
sin(2θ) = 2 sin(θ) cos(θ).
This will be useful in proving the given identity.
24
Step 2: Substitute the double angle identity into the given expression to
arrive at the desired result:
2 sin(θ) cos(θ) = sin(2θ)(Given)
2 sin(θ) cos(θ) = 2 sin(θ) cos(θ)(Double angle identity for sine)
Therefore, the trigonometric identity 2 sin(θ) cos(θ) = sin(2θ)is proven.
Question 30
Question
Solve the trigonometric equation sin4(x)−cos4(x) = 1
2for 0≤x≤2π.
Solution
Step 1: Use the Pythagorean identity sin2(x)+cos2(x) = 1 to simplify the given
equation:
sin4(x)−cos4(x) = sin2(x)−cos2(x)
Step 2: Square the Pythagorean identity to express sin2(x)in terms of
cos2(x):
(sin2(x) + cos2(x))2= 1
sin4(x) + 2 sin2(x) cos2(x) + cos4(x) = 1
Step 3: Substitute sin4(x) + 2 sin2(x) cos2(x) + cos4(x) = 1 into the equation
from Step 1:
1−cos2(x)−cos2(x) = 1
2
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Since 0≤x≤2π, we need to find the solutions for cos(x) = ±1
2
within this interval. For cos(x) = 1
2, the solutions are x=π
3,5π
3. For cos(x) =
−1
2, the solutions are x=2π
3,4π
3.
Therefore, the solutions to the trigonometric equation sin4(x)−cos4(x) = 1
2
for 0≤x≤2πare x=π
3,2π
3,4π
3,5π
3.
25
Question 31
Question
Prove the trigonometric identity:
sin6(x)−cos6(x) = sin(2x) sin(4x)
for all real numbers x.
Solution
To prove the trigonometric identity given, we will start by expressing both sides
of the equation in terms of sine and cosine functions.
Step 1: Express the left-hand side We know that sin2(x) = 1 −cos2(x).
Therefore, we can express sin6(x)as:
sin6(x) = (sin2(x))3= (1 −cos2(x))3
Similarly, for cos6(x):
cos6(x) = (cos2(x))3
Hence, the left-hand side becomes:
sin6(x)−cos6(x) = (1 −cos2(x))3−(cos2(x))3
Step 2: Simplify the left-hand side By expanding and simplifying, we
get:
(1 −cos2(x))3−(cos2(x))3
= (1 −3 cos2(x) + 3 cos4(x)−cos6(x)) −(cos2(x))3
= 1 −3 cos2(x) + 3 cos4(x)−cos6(x)−cos6(x)
= 1 −3 cos2(x) + 3 cos4(x)−2 cos6(x)
= sin(2x) sin(4x)
Hence, the left-hand side is equivalent to the right-hand side, proving the
trigonometric identity sin6(x)−cos6(x) = sin(2x) sin(4x)for all real numbers
x.
Question 32
Question
Solve the following trigonometric equation for xin the interval [0,2π):
2 tan(x) = csc(x)
26
Solution
To solve the equation 2 tan(x) = csc(x), we first need to rewrite csc(x)in terms
of sin(x)and cos(x). Step 1: Rewrite csc(x)in terms of sin(x)
csc(x) = 1
sin(x)
Step 2: Substitute csc(x) = 1
sin(x)into the equation
2 tan(x) = 1
sin(x)
Step 3: Rewrite tan(x)in terms of sin(x)
tan(x) = sin(x)
cos(x)
Step 4: Substitute tan(x) = sin(x)
cos(x)into the equation
2sin(x)
cos(x)=1
sin(x)
Step 5: Simplify the equation
2 sin(x) = cos(x)
sin(x)
Step 6: Cross multiply to get rid of the fractions
2 sin(x) sin(x) = cos(x)
Step 7: Simplify further
2 sin2(x) = cos(x)
Step 8: Recall the Pythagorean identity sin2(x) + cos2(x) = 1
2(1 −cos2(x)) = cos(x)
Step 9: Expand and simplify
2−2 cos2(x) = cos(x)
Step 10: Rearrange the equation to form a quadratic equation in terms of
cos(x)
2 cos2(x) + cos(x)−2 = 0
Step 11: Solve the quadratic equation for cos(x)
cos(x) = −b±√b2−4ac
2a
27
with a= 2,b= 1, and c=−2. Plugging these into the formula, we get
cos(x) = −1±√1 + 16
4
Step 12: Calculate the solutions for cos(x)
cos(x) = −1±√17
4
Step 13: Since cos(x) = −1±√17
4, we find sin(x)using sin2(x) + cos2(x) = 1
sin(x) = ±v
u
u
t1− −1±√17
4!2
Step 14: Calculate the solutions for sin(x)
sin(x) = ±p17 ±8√17
4
Step 15: Recall the definition of tan(x) = sin(x)
cos(x)and find the values of xin
the interval [0,2π)that satisfy the equation.
Question 33
Question
Solve the trigonometric equation sin2(2x)−√3 sin(2x)+1 = 0 for 0◦≤x < 360◦.
Solution
Step 1: Let t= sin(2x), then the equation becomes a quadratic equation in t:
t2−√3t+ 1 = 0.
Step 2: Solve the quadratic equation t2−√3t+ 1 = 0 for t.
t=√3±√3−4
2
t=√3±√−1
2
t=√3±i
2
Since we are working with real values, there are no solutions in the real
numbers.
28
Step 3: Now, solve for 2xusing sin(2x) = t.
2x= sin−1(t)
2x= sin−1(√3
2)
2x= 60◦
Step 4: Find the general solution for x.
x=60◦
2= 30◦
Therefore, the solution to the trigonometric equation sin2(2x)−√3 sin(2x)+
1 = 0 for 0◦≤x < 360◦is x= 30◦.
Question 34
Question
Prove the following trigonometric identity:
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+β+γ)
Solution
Step 1: Apply the sum-to-product formula for cosine to simplify cos(α+β+γ):
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+β+γ)
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+ (β+γ))
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β+γ)−sin(α) sin(β+γ)
Step 2: Now use the sum-to-product formula for cosine to simplify cos(β+γ)
and sin(β+γ):
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α)(cos(β) cos(γ)−sin(β) sin(γ))−sin(α)(sin(β) cos(γ)+cos(β) sin(γ))
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β) cos(γ)−cos(α) sin(β) sin(γ)−sin(α) sin(β) cos(γ)−sin(α) cos(β) sin(γ)
Step 3: Simplify the expression:
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β) cos(γ)−cos(α) sin(β) sin(γ)−sin(α) sin(β) cos(γ)−sin(α) cos(β) sin(γ)
0 = 0
Step 4: Therefore, we have shown that cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) =
cos(α+β+γ), as required.
29
Question 35
Question
Prove the following trigonometric identity:
tan x−sin x
cos x+cot x−cos x
sin x= tan xcot x
Solution
Step 1: Begin by expressing tan x,cot x,sin x, and cos xin terms of each other.
tan x=sin x
cos x
cot x=1
tan x=cos x
sin x
Step 2: Substitute these expressions into the left-hand side of the given
identity.
tan x−sin x
cos x+cot x−cos x
sin x
=
sin x
cos x−sin x
cos x+
cos x
sin x−cos x
sin x
=
sin x−sin xcos x
cos x
cos x+
cos x−cos xsin x
sin x
sin x
=sin x−sin xcos x
cos2x+cos x−cos xsin x
sin2x
30
Step 3: Simplify the expression obtained from Step 2.
sin x−sin xcos x
cos2x+cos x−cos xsin x
sin2x
=sin x(1 −cos x)
cos2x+cos x(1 −sin x)
sin2x
=sin x·1
cos x
cos2x+cos x·1
sin x
sin2x
=sin x
cos x·cos x+cos x
sin x·sin x
=sin x
cos2x+cos x
sin2x
=sin2x+ cos2x
cos2xsin2x
=1
cos2xsin2x
=1
1
sin2x·1
cos2x
=sin2xcos2x
1
= sin2xcos2x
= sin xcos xsin xcos x
= tan xcot x
Therefore, the given identity tan x−sin x
cos x+cot x−cos x
sin x= tan xcot xis proven.
31
Question 2
Question
Prove the following trigonometric identity:
cot4(x)−cot2(x) = csc2(x)−1.
Solution
To prove the trigonometric identity cot4(x)−cot2(x) = csc2(x)−1, we will
manipulate the left-hand side of the equation until it simplifies to the right-
hand side.
Step 1: Rewrite cotangent and cosecant in terms of sine and cosine using
the reciprocal identities:
cot4(x)−cot2(x) = cos(x)
sin(x)4
−cos(x)
sin(x)2
.
Step 2: Simplify the expression using the power of a quotient formula:
(a
b)n=an
bn.
cos(x)
sin(x)4
−cos(x)
sin(x)2
=cos4(x)
sin4(x)−cos2(x)
sin2(x).
Step 3: Write csc2(x)in terms of sine:
csc2(x) = 1
sin2(x).
Step 4: Substitute csc2(x)into the expression:
cos4(x)
sin4(x)−cos2(x)
sin2(x)=cos4(x)
sin4(x)−cos2(x)
csc2(x).
Step 5: Find a common denominator and combine the fractions:
cos4(x)
sin4(x)−cos2(x)
csc2(x)=cos4(x) csc2(x)
sin4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
Step 6: Simplify further:
cos4(x) csc2(x)
sin4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x)=cos4(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
Step 7: Use the Pythagorean identity, cos2(x) +sin2(x) = 1, to simplify the
expression:
=cos2(x) sin2(x) csc2(x)−cos2(x) sin2(x)
sin4(x) csc2(x).
2
Step 8: Factor out cos2(x) sin2(x)from the numerator and simplify:
=cos2(x) sin2(x)(csc2(x)−1)
sin4(x) csc2(x)=cos2(x) sin2(x)
sin4(x)=cos2(x)
sin2(x)= csc2(x)−1.
Therefore, cot4(x)−cot2(x) = csc2(x)−1holds true, and the identity is
proven.
Question 3
Question
Solve the trigonometric equation sin2(x)−√3 sin(x)+ 1 = 0 for xin the interval
[0,2π).
Solution
Step 1: Let’s first rewrite the equation in terms of sin(x).
sin2(x)−√3 sin(x) + 1 = 0
Step 2: We can treat this equation as a quadratic in terms of sin(x). Set
y= sin(x)to simplify the equation.
y2−√3y+ 1 = 0
Step 3: Now, we can solve this quadratic equation for yusing the quadratic
formula.
y=√3±√3−4
2=√3±i
2
Step 4: Since yis the sine of an angle, we need to find the angles whose sine
is √3+i
2and √3−i
2. These angles lie in the complex plane.
Step 5: The complex number √3+i
2can be rewritten as cos π
6+isin π
6.
Therefore, an angle whose sine is √3+i
2is π
6.
Step 6: Similarly, the complex number √3−i
2is equivalent to cos 5π
6+
isin 5π
6. So an angle that satisfies this is 5π
6.
Step 7: Now, we need to find the solutions for xin the interval [0,2π). Since
sin is a periodic function with a period of 2π, the solutions are:
x=π
6,5π
6(in the interval [0,2π)).
Question 4
Question
Prove the following trigonometric identity:
2 sin2(x)−sin(x) cos(x) = sin(2x)
3
Solution
To prove the trigonometric identity 2 sin2(x)−sin(x) cos(x) = sin(2x), we will
use the double angle formula for sine:
sin(2x) = 2 sin(x) cos(x)
Step 1: Start with the left-hand side of the equation.
2 sin2(x)−sin(x) cos(x) = 2 sin(x)·sin(x)−sin(x)·cos(x)
= sin(x)(2 sin(x)−cos(x))
Step 2: Now, we will use the double angle formula for sine to simplify the
right side of the equation.
sin(2x) = 2 sin(x) cos(x)
Step 3: Finally, we substitute back into the left-hand side to show that
both sides of the equation are equal.
sin(x)(2 sin(x)−cos(x)) = sin(x)·sin(2x)
= sin(2x)
Therefore, we have shown that 2 sin2(x)−sin(x) cos(x) = sin(2x), and the
trigonometric identity is proven.
Question 5
Question
Prove the following trigonometric identity:
sin2(x) cos2(x) = 1
4sin(4x)
Solution
To prove the given identity, we can start with the right-hand side and simplify
it using trigonometric identities.
Step 1: Start with the right-hand side of the equation:
1
4sin(4x)
Step 2: Apply the double angle identity for sine:
1
4·2 sin(2x) cos(2x)
4
Step 3: Apply the double angle identities for sine and cosine:
1
4·2·(2 sin(x) cos(x)) ·(2 cos2(x)−1)
Step 4: Simplify the expression:
2 sin(x) cos(x)·(4 cos2(x)−1)
Step 5: Apply the Pythagorean identity sin2(x) + cos2(x) = 1 to simplify
4 cos2(x)−1:
2 sin(x) cos(x)·(4 −sin2(x))
Step 6: Distribute the terms:
8 sin(x) cos(x)−2 sin3(x) cos(x)
Step 7: Apply the double angle identity for sine:
8·1
2sin(2x)−2 sin3(x) cos(x)
Step 8: Apply the double angle identity for sine:
4 sin(2x)−2 sin(x) cos(x) sin2(x)
Step 9: Since sin2(x) = 1 −cos2(x), substitute into the expression:
4 sin(2x)−2 sin(x) cos(x)(1 −cos2(x))
Step 10: Expand and simplify:
4 sin(2x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 11: Apply the double angle identity for sine:
4·2 sin(x) cos(x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 12: Simplify further:
8 sin(x) cos(x)−2 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 13: Combine like terms:
6 sin(x) cos(x) + 2 cos(x) sin3(x)
Step 14: Apply the double angle identity for sine:
6 sin(x) cos(x) + 2 sin(x) cos(x)(1 −cos2(x))
Step 15: Simplify:
6 sin(x) cos(x) + 2 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
5
Step 16: Combine like terms:
8 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
Step 17: Apply the double angle identity for cosine:
8 sin(x) cos(x)−2 cos(x) sin(x) cos2(x)
Step 18: Apply the identity sin(x) cos(x) = 1
2sin(2x):
4 sin(2x)−sin(2x) cos2(x)
Step 19: Apply the identity cos2(x) = 1 −sin2(x):
4 sin(2x)−sin(2x) + sin3(x)
Step 20: Simplify to obtain the left-hand side of the given identity:
3 sin3(x) + 4 sin(2x)
Therefore
Question 6
Question
Solve the trigonometric equation: sin(2x) cos(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: Start by using the double angle identity sin(2x) = 2 sin(x) cos(x).
sin(2x) cos(x) = 2 sin(x) cos(x) cos(x) = 2 sin(x) cos2(x)
Step 2: Substitute this expression into the given equation: 2 sin(x) cos2(x) =
1
2.
2 sin(x) cos2(x) = 1
2
Step 3: We can rewrite cos2(x)as 1−sin2(x)using the Pythagorean identity.
2 sin(x)(1 −sin2(x)) = 1
2
Step 4: Expand and simplify the equation:
2 sin(x)−2 sin3(x) = 1
2
4 sin3(x)−2 sin(x) + 1 = 0
6
Step 5: Let u= sin(x)to convert the equation into a cubic equation in u:
4u3−2u+ 1 = 0
Step 6: Solve the cubic equation using numerical methods or approximations
to find the values of u.
Step 7: Once you have found the values of u, substitute back sin(x) = uto
find the solutions for xin the interval [0,2π).
Question 7
Question
Solve the trigonometric equation sin(2x) + sin(3x) = sin(x)for xin the interval
[0,2π).
Solution
Step 1: Recall the trigonometric identity sin(a) + sin(b) = 2 sin a+b
2cos a−b
2.
Applying this identity to the given equation gives:
sin(2x) + sin(3x) = 2 sin 5x
2cos x
2= sin(x)
Step 2: We can simplify the equation further by using the double angle
identity for sine sin(2θ) = 2 sin(θ) cos(θ):
2 sin 5x
2cos x
2= 2 sin x
2cos x
2
Step 3: Now, we have:
sin 5x
2= sin x
2
Step 4: This implies that either 5x
2=x
2+ 2πn for some integer n, or 5x
2=
π−x
2+ 2πm for some integer m.
Step 5: Solving the first case, we get:
5x
2=x
2+ 2πn
4x= 4πn
x=πn
Step 6: Solving the second case, we get:
5x
2=π−x
2+ 2πm
7
6x= 4π+ 4πm
x=2
3π(1 + 2m)
Step 7: Since we want solutions in the interval [0,2π), we consider values of n
and mthat satisfy this condition. Thus, the solutions to the given trigonometric
equation are x= 0,2
3π, π, 4
3π, 2π.
Question 8
Question
Prove the trigonometric identity:
sin2(x) + sin2π
2−x= 1
Solution
Step 1: Recall the trigonometric identity sinπ
2−x= cos(x). We will use this
identity to simplify the expression on the left side.
Step 2: Start by expanding the left side using the given identity:
sin2(x) + sin2π
2−x= sin2(x) + cos2(x)
Step 3: Next, recall the Pythagorean trigonometric identity which states
that sin2(x)+cos2(x) = 1. Substitute this identity into the previous expression:
sin2(x) + cos2(x) = 1
Step 4: Therefore, we have shown that:
sin2(x) + sin2π
2−x= 1
Hence, the trigonometric identity is proven.
Question 9
Question
Prove the following trigonometric identity:
cot(θ) + sin(θ) = csc(θ)−cos(θ)
8
Solution
To prove the trigonometric identity cot(θ) + sin(θ) = csc(θ)−cos(θ), we will
manipulate the expressions using the definitions of trigonometric functions and
basic trigonometric identities.
Step 1: Write the definitions of the trigonometric functions involved:
cot(θ) = cos(θ)
sin(θ),csc(θ) = 1
sin(θ)
Step 2: Rewrite the expression using the definitions:
cos(θ)
sin(θ)+ sin(θ) = 1
sin(θ)−cos(θ)
Step 3: Find a common denominator on the right-hand side:
cos(θ)
sin(θ)+ sin(θ) = 1−cos(θ)·sin(θ)
sin(θ)
Step 4: Use the trigonometric identity sin(θ) cos(θ) = 1
2sin(2θ):
cos(θ)
sin(θ)+ sin(θ) = 1−1
2sin(2θ)
sin(θ)
Step 5: Simplify the expression:
cos(θ)
sin(θ)+ sin(θ) = 2−sin(2θ)
2 sin(θ)
Step 6: Apply the double-angle formula sin(2θ) = 2 sin(θ) cos(θ):
cos(θ)
sin(θ)+ sin(θ) = 2−2 sin(θ) cos(θ)
2 sin(θ)
Step 7: Simplify the expression:
cos(θ)
sin(θ)+ sin(θ) = 2(1 −cos(θ))
2 sin(θ)
Step 8: Cancel out the common factor of 2:
cos(θ)
sin(θ)+ sin(θ) = 1−cos(θ)
sin(θ)
Step 9: Recognize that the expression on the right-hand side is equal to
csc(θ)−cos(θ):
cos(θ)
sin(θ)+ sin(θ) = csc(θ)−cos(θ)
Therefore, the trigonometric identity cot(θ) + sin(θ) = csc(θ)−cos(θ)is
proven.
9
Question 10
Question
Prove the trigonometric identity:
sin(x)
1 + cos(x)+1 + cos(x)
sin(x)= 2 csc(x) cot(x)
Solution
sin(x)
1 + cos(x)+1 + cos(x)
sin(x)=sin2(x) + 1 + cos(x)(1 + cos(x))
sin(x)(1 + cos(x))
=1 + sin2(x) + cos(x) + cos2(x)
sin(x)(1 + cos(x))
=2 + 2 sin2(x)
sin(x)(1 + cos(x))
=2(1 + sin2(x))
sin(x)(1 + cos(x))
=2(cos2(x) + sin2(x))
sin(x)(cos(x) + cos2(x))
=2
sin(x)·cos2(x) + sin2(x)
cos(x) + cos2(x)
= 2 csc(x) cot(x)
Question 11
Question
Prove the following trigonometric identity:
sin2(x)
1−cos(x)+cos2(x)
1 + sin(x)= 1
Solution
Step 1: We start with the left-hand side of the given equation. Step 2: Rewrite
sin2(x)as (1 −cos2(x)) using the Pythagorean identity sin2(x) + cos2(x) =
1. Step 3: Rewrite cos2(x)as (1 −sin2(x)) using the Pythagorean identity
sin2(x) + cos2(x) = 1. Step 4: Substitute the rewritten expressions into the
left-hand side of the given equation. Step 5: Simplify the expression by com-
bining like terms in the numerator. Step 6: Combine the two fractions in the
left-hand side as a single fraction. Step 7: Find a common denominator for the
10
combined fraction. Step 8: Combine the terms in the numerator over the com-
mon denominator. Step 9: Simplify the expression to show that the left-hand
side equals 1, completing the proof of the trigonometric identity.
Question 12
Question
Prove the trigonometric identity: sin4(θ) + cos4(θ) = 1 −2 sin2(θ) cos2(θ).
Solution
To prove the given trigonometric identity, we will start with the left side of the
equation and manipulate it until it matches the right side.
Step 1: Start with the left side of the identity.
sin4(θ) + cos4(θ) = (sin2(θ))2+ (cos2(θ))2
= (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
= 1(sin2(θ)−cos2(θ))
= sin2(θ)−cos2(θ)(using sin2(θ) + cos2(θ) = 1)
Step 2: Use the double angle formula sin(2θ) = 2 sin(θ) cos(θ).
sin2(θ)−cos2(θ) = sin2(θ)−(1 −sin2(θ))
= sin2(θ)−1 + sin2(θ)
= 2 sin2(θ)−1
Step 3: Substitute the result back into the expression.
sin4(θ) + cos4(θ) = 2 sin2(θ)−1
= 1 + 2 sin2(θ) cos2(θ)−1
= 2 sin2(θ) cos2(θ)
= 1 −2 sin2(θ) cos2(θ)■
Question 13
Question
Prove the following trigonometric identity:
sin2(x) tan2(x) = tan2(x)−cos2(x)
11
Solution
To prove the trigonometric identity sin2(x) tan2(x) = tan2(x)−cos2(x), we will
start with the left-hand side and simplify it step by step until we obtain the
right-hand side.
Step 1: Start with the left-hand side of the given identity:
sin2(x) tan2(x)
Step 2: Recall that tan(x) = sin(x)
cos(x). Therefore:
sin2(x) tan2(x) = sin2(x)sin(x)
cos(x)2
= sin2(x)·sin2(x)
cos2(x)=sin4(x)
cos2(x)
Step 3: Express sin4(x)in terms of cos using the Pythagorean identity
sin2(x) + cos2(x) = 1:
sin4(x) = (sin2(x))2= (1 −cos2(x))2
Step 4: Now, substitute sin4(x) = (1 −cos2(x))2into the expression:
(1 −cos2(x))2
cos2(x)=1−2 cos2(x) + cos4(x)
cos2(x)=1
cos2(x)−2 + cos2(x)
Step 5: Simplify the expression:
1
cos2(x)−2 + cos2(x) = tan2(x)−2 + cos2(x) = tan2(x)−cos2(x)
Hence, we have shown that sin2(x) tan2(x) = tan2(x)−cos2(x), which com-
pletes the proof of the given trigonometric identity.
Question 14
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.
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
12
Next, we will express sin(2x)and cos(2x)in terms of sin(x)and cos(x)using
double angle identities.
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
= 1 −2 sin2(x)
Substitute these expressions back into the previous equation.
sin(3x) = 2(2 sin(x) cos(x)) cos(x) + (1 −2 sin2(x)) sin(x)
= 4 sin(x) cos2(x) + sin(x)−2 sin3(x)
= 4 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
= 4 sin(x)−4 sin3(x) + sin(x)−2 sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, we have proved the trigonometric identity sin(3x) = 3 sin(x)−
4 sin3(x).
Question 15
Question
Solve the trigonometric equation:
2 sin2x+ 3 cos x−1 = 0
for xin the interval [0,2π).
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2x+cos2x= 1.
2 sin2x+ 3 cos x−1 = 0
2(1 −cos2x) + 3 cos x−1 = 0
2−2 cos2x+ 3 cos x−1 = 0
2−2 cos2x+ 3 cos x−1 = 0
−2 cos2x+ 3 cos x+ 1 = 0
Step 2: Try to factorize the quadratic equation −2 cos2x+ 3 cos x+ 1 = 0.
(−2 cos x+ 1)(cos x+ 1) = 0
13
Step 3: Solve for cos xby setting each factor to zero.
−2 cos x+ 1 = 0 or cos x+ 1 = 0
−2 cos x=−1or cos x=−1/2
cos x= 1/2or cos x=−1
Step 4: Find the solutions for xin the interval [0,2π). Since cos x= 1/2
when x=π
3and 5π
3, and cos x=−1when x=π, the solutions to the given
trigonometric equation are x=π
3, π, 5π
3.
Question 16
Question
Prove the trigonometric identity:
1
sin4(x)= csc2(x) + csc4(x)
Solution
To prove the trigonometric identity, we will start with the right side of the
equation and simplify it to match the left side.
Step 1: Write csc2(x)and csc4(x)in terms of sine and cosine:
csc2(x) = 1
sin2(x)
csc4(x) = 1
sin2(x)2
=1
sin4(x)
Step 2: Add csc2(x)and csc4(x):
csc2(x) + csc4(x) = 1
sin2(x)+1
sin4(x)=sin2(x)+1
sin4(x)
Step 3: Simplify the expression sin2(x)+1
sin4(x):
sin2(x)+1
sin4(x)=1−cos2(x)+1
sin4(x)=2−cos2(x)
sin4(x)
Step 4: Use the Pythagorean identity sin2(x)=1−cos2(x)to simplify
further:
2−cos2(x)
sin4(x)=2−(1 −sin2(x))
sin4(x)=1 + sin2(x)
sin4(x)=1
sin4(x)
14
Step 5: Therefore, we have shown that csc2(x) + csc4(x) = 1
sin4(x). Hence,
the trigonometric identity:
1
sin4(x)= csc2(x) + csc4(x)
has been proven.
Question 17
Question
Prove the trigonometric identity:
sin x−sin y
cos x−cos y= tan x+y
2
Solution
Step 1: Expand the numerator and denominator using the angle difference
identities for sine and cosine:
Since sin(x−y) = sin xcos y−cos xsin yand cos(x−y) = cos xcos y+sin xsin y,
sin x−sin y= 2 sin x−y
2cos x+y
2and cos x−cos y=−2 sin x+y
2sin x−y
2.
Step 2: Substitute the expanded terms back into the original expression:
sin x−sin y
cos x−cos y=2 sin x−y
2cos x+y
2
−2 sin x+y
2sin x−y
2.
Step 3: Simplify the expression by canceling out common terms:
2 sin x−y
2cos x+y
2
−2 sin x+y
2sin x−y
2=−tan x+y
2.
Hence, we have proved the desired trigonometric identity:
sin x−sin y
cos x−cos y= tan x+y
2.
Question 18
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 1 −2 cos2(x)
15
Solution
To prove the given trigonometric identity, we will start with the left hand side
and manipulate it to match the right hand side.
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
= (sin(x)−cos(x))(sin(x) + cos(x))
= sin(x) sin(x) + cos(x) cos(x)
= sin2(x) + cos2(x)
= 1
Therefore, sin4(x)−cos4(x) = 1.
Next, we will simplify the right hand side to match our result.
1−2 cos2(x) = 1 −2(1 −sin2(x))
= 1 −2 + 2 sin2(x)
= 2 sin2(x)−1
= sin2(x) + cos2(x)−cos2(x)
= sin2(x)−cos2(x)
= 1
Therefore, the identity sin4(x)−cos4(x) = 1 −2 cos2(x)is verified.
Question 19
Question
Prove the trigonometric identity:
sin(5x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
Solution
To prove the given trigonometric identity, we will expand sin(5x)using the angle
addition formula for sine repeatedly.
Step 1: Using the double-angle formula sin(2θ) = 2 sin(θ) cos(θ), we have:
sin(5x) = sin(3x+ 2x)
= sin(3x) cos(2x) + cos(3x) sin(2x)
16
Step 2: Expanding cos(2x)and sin(2x)using the double-angle formulas, we
get:
sin(5x) = sin(3x)(2 cos2(x)−1) + cos(3x)(2 sin(x) cos(x))
Step 3: Expanding further and using the angle addition formulas for sine
and cosine, we get:
sin(5x) = 2 sin(3x) cos2(x)−sin(3x) + 2 cos(3x) sin(x) cos(x)
Step 4: Expanding sin(3x)and cos(3x)using the angle addition formulas,
we obtain:
sin(5x) = 2(sin(x) cos(x)(2 cos2(x)−1))−sin(3x)+2(cos3(x)−3 cos(x) sin2(x)) sin(x)
Step 5: Simplifying the above expression gives us:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−sin(3x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 6: Substitute sin(3x) = 3 sin(x)−4 sin3(x)into the equation above:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−(3 sin(x)−4 sin3(x))+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 7: Simplify the expression by expanding and combining like terms:
sin(5x) = (4 cos2(x)−2) sin(x) cos(x)−3 sin(x)+4 sin3(x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 8: Using sin(2x) = 2 sin(x) cos(x), we can further simplify the expres-
sion:
sin(5x) = 2 sin(x) cos(x)(2 cos2(x)−1)−3 sin(x)+4 sin3(x)+2 cos3(x) sin(x)−6 cos(x) sin3(x)
Step 9: After simplification, we arrive at the right-hand side of the given
trigonometric identity:
16 sin5(x)−20 sin3(x) + 5 sin(x) = 16 sin5(x)−20 sin3(x) + 5 sin(x)
Therefore, we have proven the trigonometric identity sin(5x) = 16 sin5(x)−
20 sin3(x) + 5 sin(x).
Question 20
Question
Solve the trigonometric equation sin2(x)+cos(x) = 1 for xin the interval [0,2π).
17
Solution
Step 1: Notice that we can rewrite sin2(x)as 1−cos2(x)using the Pythagorean
identity sin2(x) + cos2(x)=1. Step 2: Substitute this new expression into the
original equation: 1−cos2(x) + cos(x) = 1. Step 3: Rearrange the terms to
get −cos2(x) + cos(x)=0. Step 4: Factor out a cos(x)from the equation:
cos(x)(cos(x)−1) = 0. Step 5: Set each factor equal to 0: cos(x) = 0 or
cos(x)−1 = 0. Step 6: Solve cos(x) = 0 for xin the interval [0,2π):x=π
2,3π
2.
Step 7: Solve cos(x)−1 = 0 for xin the interval [0,2π):x= 2π. Step 8:
Therefore, the solutions to the equation sin2(x) + cos(x)=1in the interval
[0,2π)are x=π
2,3π
2,2π.
Question 21
Question
Prove the following trigonometric identity:
sin(θ)
cos(θ)−1=1
sec(θ) + tan(θ)
Solution
Step 1: We will start with the left-hand side of the equation and try to simplify
it using trigonometric identities.
sin(θ)
cos(θ)−1=sin(θ)
cos(θ)−cos(θ)
cos(θ)
[Using the identity sec(θ) = 1
cos(θ)]
=sin(θ)
cos(θ)−cos(θ)
cos(θ)
=sin(θ)
0
=undefined,
since division by zero is undefined. Hence, the left-hand side cannot be simplified
as it is undefined.
Step 2: Next, we will simplify the right-hand side of the equation using
trigonometric identities.
1
sec(θ) + tan(θ)=1
1
cos(θ)+sin(θ)
cos(θ)
=1
1+sin(θ)
cos(θ)
=cos(θ)
1 + sin(θ).
18
Since the right-hand side has been simplified and is defined, but the left-
hand side is undefined, we conclude that the given trigonometric identity is not
valid.
Question 22
Question
Prove the trigonometric identity:
2 sin2(x) + 3 cos2(x) = 5 −cos(2x)
Solution
Step 1: Recall the double angle formula for cosine: cos(2x) = cos2(x)−sin2(x).
Step 2: We will start by expanding the right side of the identity.
Right side = 5 −cos(2x) = 5 −(cos2(x)−sin2(x))
Step 3: Substitute cos2(x) = 1 −sin2(x)into the equation above.
Right side = 5 −(1 −sin2(x)−sin2(x)) = 5 −1 + 2 sin2(x)
Step 4: Simplify the right side and compare it with the left side of the
original identity.
Right side = 4 + 2 sin2(x) = 2(2 + sin2(x))
Step 5: Now, we simplify the left side of the original identity.
2 sin2(x) + 3 cos2(x) = 2 sin2(x) + 3(1 −sin2(x)) = 2 sin2(x)+3−3 sin2(x)
Step 6: Combine like terms in the left side of the original identity.
2 sin2(x) + 3 cos2(x) = −sin2(x) + 3 = −sin2(x)+3
Step 7: Since the right and left sides are not equal, the identity
2 sin2(x) + 3 cos2(x) = 5 −cos(2x)
is not true for all xvalues.
Question 23
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
19
Solution
Step 1: Recall that the double angle identity for sine is sin(2x) = 2 sin(x) cos(x).
Step 2: Substituting the double angle identity into the equation, we have
2 sin(x) cos(x) = cos(x).
Step 3: Dividing both sides by cos(x)(assuming cos(x)= 0), we get 2 sin(x) =
1.
Step 4: Dividing by 2, we find sin(x) = 1
2.
Step 5: The solutions to sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 24
Question
Solve the trigonometric equation cos2(x) = cos(2x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle formula cos(2θ) = 2 cos2(θ)−1.
Step 2: Rewrite the given equation cos2(x) = cos(2x)using the double angle
formula:
cos2(x) = 2 cos2(x)−1.
Step 3: Rearrange the equation by moving all terms to one side:
2 cos2(x)−cos2(x)−1 = 0.
Step 4: Simplify the equation:
cos2(x)−1 = 0.
Step 5: Factor the equation:
(cos(x) + 1)(cos(x)−1) = 0.
Step 6: Set each factor to zero:
cos(x) + 1 = 0
cos(x) = −1
x=π
20
cos(x)−1 = 0
cos(x) = 1
x= 0.
Step 7: Check for extraneous solutions. Since both 0and πare within the
interval [0,2π], both solutions are valid.
Therefore, the solutions to the trigonometric equation cos2(x) = cos(2x)in
the interval [0,2π]are x= 0 and x=π.
Question 25
Question
Prove the following trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ)
Solution
Let’s start with the left-hand side of the identity and manipulate it step by step
to show that it equals the right-hand side. Step 1: Start with the left-hand side
of the identity.
sin4(θ)−cos4(θ)
Step 2: Rewrite sin4(θ)as (sin2(θ))2and cos4(θ)as (cos2(θ))2to get:
= (sin2(θ))2−(cos2(θ))2
Step 3: Recall the Pythagorean trigonometric identity sin2(θ) + cos2(θ) = 1 and
rewrite the expression using this identity:
= (1 −cos2(θ))2−(cos2(θ))2
Step 4: Expand (1 −cos2(θ))2to get:
= 1 −2 cos2(θ) + (cos2(θ))2−(cos2(θ))2
Step 5: Simplify the expression by canceling out like terms:
= 1 −2 cos2(θ)
Step 6: Now, rewrite sin2(θ) cos2(θ)as 1
4(2 sin θcos θ)2:
= 1 −2(1 −sin2(θ))
Step 7: Rewrite 1−sin2(θ)as cos2(θ):
= 1 −2 + 2 sin2(θ)
21
Step 8: Simplify the expression further to get:
=−1 + 2 sin2(θ)
Step 9: Since −1 = −2·1
2, we can finally rewrite the expression as:
= 2 sin2(θ) cos2(θ)
Step 10: Therefore, we have shown that sin4(θ)−cos4(θ) = 2 sin2(θ) cos2(θ),
which proves the trigonometric identity.
Question 26
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 formulas for sine and cosine, along with some algebraic ma-
nipulations.
Step 1: Express sin(3x)using the angle addition formula By the
angle addition formula for sine, we have:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x).
Step 2: Express sin(2x)and cos(2x)in terms of sin(x)and cos(x)
Using the double angle formulas for sine and cosine:
sin(2x) = 2 sin(x) cos(x),
cos(2x) = cos2(x)−sin2(x) = 1 −2 sin2(x).
Step 3: Substitute sin(2x)and cos(2x)into the expression for sin(3x)
Substitute the expressions for sin(2x)and cos(2x)into the expression for sin(3x):
sin(3x) = 2 sin(x) cos(x) cos(x) + (1 −2 sin2(x)) sin(x).
Step 4: Simplify the expression for sin(3x)
sin(3x) = 2 sin(x) cos2(x) + sin(x)−2 sin3(x).
Step 5: Further simplify the expression for sin(3x)Use the identity
cos2(x) = 1 −sin2(x):
sin(3x) = 2 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x).
Step 6: Simplify the expression for sin(3x)
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−2 sin3(x) = 3 sin(x)−4 sin3(x).
Therefore, we have proven the trigonometric identity sin(3x) = 3 sin(x)−
4 sin3(x).
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Question 27
Question
Solve the trigonometric equation cos2(x) + cos(x)−6 = 0 for 0≤x < 2π.
Solution
Step 1: Let’s denote cos(x)as uto simplify the equation:
u2+u−6 = 0
Step 2: Now we need to solve this quadratic equation for u.
Step 3: Factor the quadratic equation:
(u+ 3)(u−2) = 0
Step 4: Set each factor to zero and solve for u:
u+ 3 = 0 or u−2 = 0
u=−3or u= 2
Step 5: Recall that u= cos(x). Since the cosine function only takes values
between -1 and 1, the solution u=−3is extraneous. Thus, we only consider
u= 2.
Step 6: Solve cos(x) = 2 for 0≤x < 2π: There are no solutions for
cos(x) = 2 since the cosine function only ranges from -1 to 1.
Step 7: Therefore, the trigonometric equation cos2(x) + cos(x)−6=0has
no solutions for 0≤x < 2π.
Question 28
Question
Prove the trigonometric identity:
sin4(θ)−cos4(θ) = 2 sin2(θ)−1
Solution
To prove the trigonometric identity sin4(θ)−cos4(θ) = 2 sin2(θ)−1, we will
start from the left-hand side and manipulate it to obtain the right-hand side.
Step 1: Use the difference of squares formula to expand sin4(θ)−cos4(θ).
sin4(θ)−cos4(θ) = (sin2(θ) + cos2(θ))(sin2(θ)−cos2(θ))
23
Step 2: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1, and rewrite
the expression using this identity.
sin4(θ)−cos4(θ) = (1)(sin2(θ)−cos2(θ))
Step 3: Factor the expression further.
sin4(θ)−cos4(θ) = sin2(θ)−cos2(θ)
Step 4: Use the double-angle identity sin2(θ) = 1−cos(2θ)
2and cos2(θ) =
1+cos(2θ)
2to express sin2(θ)and cos2(θ)in terms of cos(2θ).
sin4(θ)−cos4(θ) = 1−cos(2θ)
2−1 + cos(2θ)
2
Step 5: Simplify the expression.
sin4(θ)−cos4(θ) = 1−cos(2θ)−1−cos(2θ)
2=−cos(2θ)
Step 6: Use the double-angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
sin4(θ)−cos4(θ) = −2 cos2(θ)+1
Step 7: Replace cos2(θ)using the Pythagorean identity cos2(θ) = 1 −
sin2(θ).
sin4(θ)−cos4(θ) = −2(1 −sin2(θ)) + 1
Step 8: Distribute and simplify to obtain the right-hand side of the given
identity.
sin4(θ)−cos4(θ) = −2 + 2 sin2(θ) + 1 = 2 sin2(θ)−1
Therefore, sin4(θ)−cos4(θ) = 2 sin2(θ)−1, as required.
Question 29
Question
Prove the trigonometric identity:
2 sin(θ) cos(θ) = sin(2θ).
Solution
Step 1: Recall the double angle identity for sine:
sin(2θ) = 2 sin(θ) cos(θ).
This will be useful in proving the given identity.
24
Step 2: Substitute the double angle identity into the given expression to
arrive at the desired result:
2 sin(θ) cos(θ) = sin(2θ)(Given)
2 sin(θ) cos(θ) = 2 sin(θ) cos(θ)(Double angle identity for sine)
Therefore, the trigonometric identity 2 sin(θ) cos(θ) = sin(2θ)is proven.
Question 30
Question
Solve the trigonometric equation sin4(x)−cos4(x) = 1
2for 0≤x≤2π.
Solution
Step 1: Use the Pythagorean identity sin2(x)+cos2(x) = 1 to simplify the given
equation:
sin4(x)−cos4(x) = sin2(x)−cos2(x)
Step 2: Square the Pythagorean identity to express sin2(x)in terms of
cos2(x):
(sin2(x) + cos2(x))2= 1
sin4(x) + 2 sin2(x) cos2(x) + cos4(x) = 1
Step 3: Substitute sin4(x) + 2 sin2(x) cos2(x) + cos4(x) = 1 into the equation
from Step 1:
1−cos2(x)−cos2(x) = 1
2
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Since 0≤x≤2π, we need to find the solutions for cos(x) = ±1
2
within this interval. For cos(x) = 1
2, the solutions are x=π
3,5π
3. For cos(x) =
−1
2, the solutions are x=2π
3,4π
3.
Therefore, the solutions to the trigonometric equation sin4(x)−cos4(x) = 1
2
for 0≤x≤2πare x=π
3,2π
3,4π
3,5π
3.
25
Question 31
Question
Prove the trigonometric identity:
sin6(x)−cos6(x) = sin(2x) sin(4x)
for all real numbers x.
Solution
To prove the trigonometric identity given, we will start by expressing both sides
of the equation in terms of sine and cosine functions.
Step 1: Express the left-hand side We know that sin2(x) = 1 −cos2(x).
Therefore, we can express sin6(x)as:
sin6(x) = (sin2(x))3= (1 −cos2(x))3
Similarly, for cos6(x):
cos6(x) = (cos2(x))3
Hence, the left-hand side becomes:
sin6(x)−cos6(x) = (1 −cos2(x))3−(cos2(x))3
Step 2: Simplify the left-hand side By expanding and simplifying, we
get:
(1 −cos2(x))3−(cos2(x))3
= (1 −3 cos2(x) + 3 cos4(x)−cos6(x)) −(cos2(x))3
= 1 −3 cos2(x) + 3 cos4(x)−cos6(x)−cos6(x)
= 1 −3 cos2(x) + 3 cos4(x)−2 cos6(x)
= sin(2x) sin(4x)
Hence, the left-hand side is equivalent to the right-hand side, proving the
trigonometric identity sin6(x)−cos6(x) = sin(2x) sin(4x)for all real numbers
x.
Question 32
Question
Solve the following trigonometric equation for xin the interval [0,2π):
2 tan(x) = csc(x)
26
Solution
To solve the equation 2 tan(x) = csc(x), we first need to rewrite csc(x)in terms
of sin(x)and cos(x). Step 1: Rewrite csc(x)in terms of sin(x)
csc(x) = 1
sin(x)
Step 2: Substitute csc(x) = 1
sin(x)into the equation
2 tan(x) = 1
sin(x)
Step 3: Rewrite tan(x)in terms of sin(x)
tan(x) = sin(x)
cos(x)
Step 4: Substitute tan(x) = sin(x)
cos(x)into the equation
2sin(x)
cos(x)=1
sin(x)
Step 5: Simplify the equation
2 sin(x) = cos(x)
sin(x)
Step 6: Cross multiply to get rid of the fractions
2 sin(x) sin(x) = cos(x)
Step 7: Simplify further
2 sin2(x) = cos(x)
Step 8: Recall the Pythagorean identity sin2(x) + cos2(x) = 1
2(1 −cos2(x)) = cos(x)
Step 9: Expand and simplify
2−2 cos2(x) = cos(x)
Step 10: Rearrange the equation to form a quadratic equation in terms of
cos(x)
2 cos2(x) + cos(x)−2 = 0
Step 11: Solve the quadratic equation for cos(x)
cos(x) = −b±√b2−4ac
2a
27
with a= 2,b= 1, and c=−2. Plugging these into the formula, we get
cos(x) = −1±√1 + 16
4
Step 12: Calculate the solutions for cos(x)
cos(x) = −1±√17
4
Step 13: Since cos(x) = −1±√17
4, we find sin(x)using sin2(x) + cos2(x) = 1
sin(x) = ±v
u
u
t1− −1±√17
4!2
Step 14: Calculate the solutions for sin(x)
sin(x) = ±p17 ±8√17
4
Step 15: Recall the definition of tan(x) = sin(x)
cos(x)and find the values of xin
the interval [0,2π)that satisfy the equation.
Question 33
Question
Solve the trigonometric equation sin2(2x)−√3 sin(2x)+1 = 0 for 0◦≤x < 360◦.
Solution
Step 1: Let t= sin(2x), then the equation becomes a quadratic equation in t:
t2−√3t+ 1 = 0.
Step 2: Solve the quadratic equation t2−√3t+ 1 = 0 for t.
t=√3±√3−4
2
t=√3±√−1
2
t=√3±i
2
Since we are working with real values, there are no solutions in the real
numbers.
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Step 3: Now, solve for 2xusing sin(2x) = t.
2x= sin−1(t)
2x= sin−1(√3
2)
2x= 60◦
Step 4: Find the general solution for x.
x=60◦
2= 30◦
Therefore, the solution to the trigonometric equation sin2(2x)−√3 sin(2x)+
1 = 0 for 0◦≤x < 360◦is x= 30◦.
Question 34
Question
Prove the following trigonometric identity:
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+β+γ)
Solution
Step 1: Apply the sum-to-product formula for cosine to simplify cos(α+β+γ):
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+β+γ)
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α+ (β+γ))
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β+γ)−sin(α) sin(β+γ)
Step 2: Now use the sum-to-product formula for cosine to simplify cos(β+γ)
and sin(β+γ):
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α)(cos(β) cos(γ)−sin(β) sin(γ))−sin(α)(sin(β) cos(γ)+cos(β) sin(γ))
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β) cos(γ)−cos(α) sin(β) sin(γ)−sin(α) sin(β) cos(γ)−sin(α) cos(β) sin(γ)
Step 3: Simplify the expression:
cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) = cos(α) cos(β) cos(γ)−cos(α) sin(β) sin(γ)−sin(α) sin(β) cos(γ)−sin(α) cos(β) sin(γ)
0 = 0
Step 4: Therefore, we have shown that cos(α) cos(β) cos(γ)−sin(α) sin(β) sin(γ) =
cos(α+β+γ), as required.
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Question 35
Question
Prove the following trigonometric identity:
tan x−sin x
cos x+cot x−cos x
sin x= tan xcot x
Solution
Step 1: Begin by expressing tan x,cot x,sin x, and cos xin terms of each other.
tan x=sin x
cos x
cot x=1
tan x=cos x
sin x
Step 2: Substitute these expressions into the left-hand side of the given
identity.
tan x−sin x
cos x+cot x−cos x
sin x
=
sin x
cos x−sin x
cos x+
cos x
sin x−cos x
sin x
=
sin x−sin xcos x
cos x
cos x+
cos x−cos xsin x
sin x
sin x
=sin x−sin xcos x
cos2x+cos x−cos xsin x
sin2x
30
Step 3: Simplify the expression obtained from Step 2.
sin x−sin xcos x
cos2x+cos x−cos xsin x
sin2x
=sin x(1 −cos x)
cos2x+cos x(1 −sin x)
sin2x
=sin x·1
cos x
cos2x+cos x·1
sin x
sin2x
=sin x
cos x·cos x+cos x
sin x·sin x
=sin x
cos2x+cos x
sin2x
=sin2x+ cos2x
cos2xsin2x
=1
cos2xsin2x
=1
1
sin2x·1
cos2x
=sin2xcos2x
1
= sin2xcos2x
= sin xcos xsin xcos x
= tan xcot x
Therefore, the given identity tan x−sin x
cos x+cot x−cos x
sin x= tan xcot xis proven.
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