MATH 332 - ADVANCED CALCULUS
- Trigonometric identities and equations
Question Bank - Set 7
Liberty University
Question 1
Question
Prove the trigonometric identity:
cot(θ) csc(θ) = csc(θ)−sin(θ)
Solution
To prove the given trigonometric identity, we’ll start by writing each side of the
equation in terms of sine and cosine functions.
Step 1: Express cot(θ)and csc(θ)in terms of sine and cosine: We know
that cot(θ) = cos(θ)
sin(θ)and csc(θ) = 1
sin(θ). Substituting these expressions in, we
have:
cot(θ) csc(θ) = (cos(θ)
sin(θ))( 1
sin(θ))
Step 2: Simplify the expression:
cot(θ) csc(θ) = cos(θ)
sin(θ)·sin(θ)=cos(θ)
sin2(θ)
Step 3: Express the right-hand side of the identity in terms of sin(θ): Using
the Pythagorean identity, sin2(θ) + cos2(θ)=1, we have sin2(θ)=1−cos2(θ).
Thus, 1
sin2(θ)=1
1−cos2(θ).
Step 4: Substitute the expression for 1
sin2(θ)into the previous result:
cot(θ) csc(θ) = cos(θ)
1−cos2(θ)
Step 5: Rewrite the right-hand side in terms of sin(θ): Using the Pythagorean
identity again, we have cos2(θ) = 1−sin2(θ). Therefore, the expression becomes:
cot(θ) csc(θ) = cos(θ)
sin2(θ)= csc(θ)−sin(θ)
Since both sides of the equation match, the trigonometric identity cot(θ) csc(θ) =
csc(θ)−sin(θ)is proven.
Question 2
Question
Prove the trigonometric identity:
sin2(θ)
1−cos(θ)+cos2(θ)
1−sin(θ)= 1
Solution
Step 1: Start with the left-hand side of the equation and work on each term
separately. We will work with the first term first.
sin2(θ)
1−cos(θ)=sin2(θ)
1−2 sin2(θ/2) (Using half-angle formula: cos(θ) = 1 −2 sin2(θ/2))
=sin2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
Step 2: Next, simplify the second term in the original expression.
cos2(θ)
1−sin(θ)=cos2(θ)
1−sin(θ/2) (Using half-angle formula: sin(θ) = 2 sin(θ/2) cos(θ/2))
=cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
Step 3: Combine the two simplified expressions from Step 1 and Step 2 to
rewrite the left-hand side of the equation.
sin2(θ)
(1 −sin(θ/2))(1 + sin(θ/2)) +cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
sin2(θ) + cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2)) =1
(1 −sin(θ/2))(1 + sin(θ/2)) = 1
Step 4: Since the left-hand side expression simplifies to 1, we have proven
the trigonometric identity:
sin2(θ)
1−cos(θ)+cos2(θ)
1−sin(θ)= 1
2
Question 3
Question
Prove the trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ)
where θis a real number.
Solution
To prove the trigonometric identity cos4(θ)−sin4(θ) = cos(2θ), we will use the
double angle identity for the cosine function and the Pythagorean identity.
Step 1: Use the Pythagorean identity:
cos2(θ) + sin2(θ) = 1
Solving for cos2(θ), we get:
cos2(θ) = 1 −sin2(θ)
Step 2: Now, square the cosine identity above:
(cos2(θ))2= (1 −sin2(θ))2
cos4(θ) = 1 −2 sin2(θ) + sin4(θ)
Step 3: Substitute the expression for cos4(θ)back into the original identity:
cos4(θ)−sin4(θ) = (1 −2 sin2(θ) + sin4(θ)) −sin4(θ)
cos4(θ)−sin4(θ) = 1 −2 sin2(θ)
Step 4: Use the double angle identity for cosine:
cos(2θ) = cos2(θ)−sin2(θ)
cos(2θ) = (1 −sin2(θ)) −sin2(θ)
cos(2θ) = 1 −2 sin2(θ)
Step 5: Finally, we see that cos4(θ)−sin4(θ) = cos(2θ), thus proving the
trigonometric identity.
Question 4
Question
Prove the trigonometric identity:
tan(x) cos(x) = sin(x)
3
Solution
Step 1: Recall the definitions of tangents, cosines, and sines:
tan(x) = sin(x)
cos(x),cos(x) = cos(x),sin(x) = sin(x)
Step 2: Substitute the definitions into the left-hand side of the equation:
tan(x) cos(x) = (sin(x)
cos(x))cos(x)
Step 3: Simplify the expression on the right-hand side:
tan(x) cos(x) = sin(x) cos(x)
cos(x)
Step 4: Cancel out the cos(x)in the numerator and denominator:
tan(x) cos(x) = sin(x)
Step 5: Therefore, tan(x) cos(x) = sin(x)is proven to be a true trigonometric
identity.
Question 5
Question
Prove the following trigonometric identity:
sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)= 2 csc(θ) cot(θ)
Solution
We start with the left-hand side (LHS) of the given identity:
LHS =sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)
=sin2(θ) + (1 + cos(θ))2
sin(θ)(1 + cos(θ)) (Combining fractions)
=sin2(θ) + 1 + 2 cos(θ) + cos2(θ)
sin(θ)(1 + cos(θ)) (Expanding the numerator)
=2 + 2 cos(θ)
sin(θ)(1 + cos(θ)) (Using sin2(θ) + cos2(θ) = 1)
=2(1 + cos(θ))
sin(θ)(1 + cos(θ)) =2
sin(θ)= 2 csc(θ)
Therefore, LHS = 2 csc(θ)(Proven)
4
Hence, the given trigonometric identity
sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)= 2 csc(θ) cot(θ)
is proved.
Question 6
Question
Prove the identity
tan(x) cot(x) = sin2(x)
cos2(x)−1
Solution
Step 1: Rewrite the left-hand side using basic trigonometric identities.
tan(x) cot(x) = sin(x)
cos(x)·cos(x)
sin(x)= 1
Step 2: Rewrite the right-hand side using basic trigonometric identities.
sin2(x)
cos2(x)−1=sin2(x)
cos2(x)−sin2(x)
=sin2(x)
1−sin2(x)
=sin2(x)
cos2(x)
= tan2(x)
Step 3: Since we have proved that both sides are equal to 1 and tan2(x)
respectively, we have shown that the given identity
tan(x) cot(x) = sin2(x)
cos2(x)−1
is true.
Question 7
Question
Prove the trigonometric identity:
tan(x) + cot(x) = 2
sin(2x)
5
Solution
To prove the trigonometric identity tan(x) + cot(x) = 2
sin(2x), we will start by
expressing tan(x)and cot(x)in terms of sine and cosine functions using the
definitions of these trigonometric functions.
Step 1: Expressing tan(x)and cot(x)in terms of sine and cosine functions.
tan(x) = sin(x)
cos(x)
cot(x) = 1
tan(x)=cos(x)
sin(x)
Step 2: Substitute tan(x)and cot(x)back into the given identity.
tan(x) + cot(x) = sin(x)
cos(x)+cos(x)
sin(x)
=sin2(x) + cos2(x)
sin(x) cos(x)
=1
sin(x) cos(x)(using the Pythagorean identity sin2(x) + cos2(x) = 1)
Step 3: Simplify the expression.
tan(x) + cot(x) = 2
2 sin(x) cos(x)
=2
sin(2x)(using the double angle formula: sin(2x) = 2 sin(x) cos(x))
Therefore, we have proven that tan(x) + cot(x) = 2
sin(2x), which completes
the proof of the trigonometric identity.
Question 8
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: Use the angle sum formula for sine to rewrite sin(3x).
sin(3x) = sin(2x+x)
Step 2: Expand using the angle sum formula for sine:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
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Step 3: Recall the double angle identities:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
Step 4: Substitute the double angle identities back into the expression:
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
Step 5: Simplify the expression:
sin(3x) = 2 sin(x) cos2(x) + (cos2(x)−sin2(x)) sin(x)
Step 6: Use the Pythagorean identity cos2(x)=1−sin2(x)to simplify
further:
sin(3x) = 2 sin(x)(1 −sin2(x)) + ((1 −sin2(x)) −sin2(x)) sin(x)
Step 7: Expand and simplify the expression:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)
Step 8: Combine like terms to obtain the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, the identity sin(3x) = 3 sin(x)−4 sin3(x)is proven.
Question 10
Question
Prove the trigonometric identity:
csc(x)−cot(x) = sin(x)−cos(x)
sin(x) cos(x)
Solution
1. Step 1: Start with the right-hand side of the equation and simplify.
We have:
sin(x)−cos(x)
sin(x) cos(x)=sin(x)
sin(x) cos(x)−cos(x)
sin(x) cos(x)
=1
cos(x)−1
sin(x)=sin(x)−cos(x)
sin(x) cos(x)
7
2. Step 2: Next, express the left-hand side of the equation in terms of sine
and cosine functions.
Recall that csc(x) = 1
sin(x)and cot(x) = cos(x)
sin(x). Therefore:
csc(x)−cot(x) = 1
sin(x)−cos(x)
sin(x)=1−cos(x)
sin(x)
3. Step 3: Compare the expressions obtained in Step 1 and Step 2 to see
they are equal.
We see that: 1−cos(x)
sin(x)=sin(x)−cos(x)
sin(x) cos(x)
Thus, by proving that the left-hand side is equal to the right-hand side,
we have verified the trigonometric identity:
csc(x)−cot(x) = sin(x)−cos(x)
sin(x) cos(x)
Question 11
Question
Prove the trigonometric identity:
csc(A)−cot(A) = 1−sin(A)
sin(A)
Solution
Step 1: Start with the left-hand side of the equation:
csc(A)−cot(A)
Step 2: Express the terms in terms of sine and cosine functions:
csc(A) = 1
sin(A)and cot(A) = 1
tan(A)=cos(A)
sin(A)
Step 3: Substitute the expressions for csc(A)and cot(A)back into the
left-hand side: 1
sin(A)−cos(A)
sin(A)=1−cos(A)
sin(A)
Step 4: Use the Pythagorean identity sin2(A) + cos2(A)=1to express
cos(A)in terms of sin(A):
cos(A) = √1−sin2(A) = √1−sin2(A) = √1−sin2(A)·√1 + sin2(A)
√1 + sin2(A)
=√1−sin2(A)
cos(A)
8
Step 5: Substitute the expression of cos(A)back into the previous equation:
1−√1−sin2(A)
cos(A)
sin(A)=1−√1−sin2(A)
sin(A) cos(A)
=1−√1−sin2(A)
sin(A)√1−sin2(A)
=1−sin(A)
sin(A)
Therefore, csc(A)−cot(A) = 1−sin(A)
sin(A), as required.
Question 12
Question
Prove the identity:
2 cos2(θ)−1 = cos(2θ)
Solution
We will start with the left-hand side (LHS) of the equation: 2 cos2(θ)−1 = cos2(θ) + cos2(θ)−1
= cos2(θ)−sin2(θ) (using the Pythagorean identity: sin2(θ) = 1 −cos2(θ))
= cos2(θ)−(1 −cos2(θ))
= cos2(θ)−1 + cos2(θ)
= 2 cos2(θ)−1
Therefore, 2 cos2(θ)−1 = cos(2θ).
Question 13
Question
Prove the trigonometric identity: cos4(x)−sin4(x) = cos(2x).
Solution
To prove the given trigonometric identity, we will use the following trigonometric
identities:
• Double angle identity: cos(2α) = cos2(α)−sin2(α).
9
Step 1: Start with the left-hand side of the given identity.
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x)) (Difference of squares)
= cos2(x) cos2(x)−sin2(x) sin2(x)
= (cos2(x)−sin2(x))(cos2(x) + sin2(x)) (Difference of squares)
= cos(2x)(Using the double angle identity)
Therefore, cos4(x)−sin4(x) = cos(2x), which proves the given trigonometric
identity.
Question 14
Question
Prove the following trigonometric identity:
cos(3θ) = 4 cos3(θ)−3 cos(θ)
Solution
Step 1: Apply the triple angle formula for cosine:
cos(3θ) = cos(2θ+θ)
Step 2: Expand the right side using the sum-to-product identity:
cos(2θ+θ) = cos(2θ) cos(θ)−sin(2θ) sin(θ)
Step 3: Recall the double angle formulas:
cos(2θ) = cos2(θ)−sin2(θ)
sin(2θ) = 2 sin(θ) cos(θ)
Step 4: Substitute the double angle formulas back into the expansion:
cos(2θ) cos(θ)−sin(2θ) sin(θ) = (cos2(θ)−sin2(θ)) cos(θ)−2 sin(θ) cos(θ) sin(θ)
Step 5: Simplify the expression:
(cos2(θ)−sin2(θ)) cos(θ)−2 sin(θ) cos(θ) sin(θ) = cos3(θ)−cos(θ) sin2(θ)−2 sin2(θ) cos(θ)
Step 6: Recall the Pythagorean identity sin2(θ) = 1 −cos2(θ):
cos3(θ)−cos(θ)(1 −cos2(θ)) −2(1 −cos2(θ)) cos(θ)
Step 7: Simplify the expression further:
cos3(θ)−cos(θ) + cos3(θ)−2 cos(θ) + 2 cos3(θ)
Step 8: Combine like terms:
4 cos3(θ)−3 cos(θ)
Therefore, we have proved the trigonometric identity cos(3θ) = 4 cos3(θ)−
3 cos(θ).
10
Question 15
Question
Solve the trigonometric equation sin2(x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x) = 1.
Step 2: Since we are given that sin2(x) + cos(x)=0, we can substitute
cos2(x) = 1 −sin2(x)into the equation to get sin2(x)+1−sin2(x) = 0.
Step 3: Simplify the equation to get 1 = 0, which is not a valid statement.
Step 4: Therefore, there are no solutions to the trigonometric equation
sin2(x) + cos(x) = 0 for 0≤x≤2π.
Question 16
Question
Prove the following trigonometric identity:
cos4θ−sin4θ
cos2θ−sin2θ= cos 2θ
Solution
We start by manipulating the left side of the equation:
Recall the identity cos2θ−sin2θ= cos 2θ
Now, rewrite the numerator in terms of squares of sine and cosine:
cos4θ−sin4θ= (cos2θ−sin2θ)(cos2θ+ sin2θ)
= (cos2θ−sin2θ)(1)
= cos2θ−sin2θ
Substitute the simplified numerator back into the equation:
cos4θ−sin4θ
cos2θ−sin2θ=cos2θ−sin2θ
cos2θ−sin2θ
= 1
Thus, we have proved that cos4θ−sin4θ
cos2θ−sin2θ= cos 2θas required.
11
Question 17
Question
Prove the following trigonometric identity:
cos2x−sin2x
2 sin xcos x= cot x
Solution
To prove the given trigonometric identity, we need to manipulate the left side
of the equation until it simplifies to the right side.
Step 1: Start with the left side of the equation and factor out a common
term: cos2x−sin2x
2 sin xcos x=cos2x−sin2x
sin xcos x=cos2x−sin2x
sin xcos x
Step 2: Use the Pythagorean identity cos2x= 1 −sin2xto rewrite the
numerator: 1−sin2x−sin2x
sin xcos x=1−2 sin2x
sin xcos x
Step 3: Simplify the numerator by factoring out a 2:
1−2 sin2x
sin xcos x=2(1
2−sin2x)
sin xcos x=2(cos2x)
sin xcos x
Step 4: Divide both the numerator and the denominator by sin x:
2 cos2x
sin xcos x= 2cos x
sin x= 2 cot x= cot x
Therefore, we have shown that cos2x−sin2x
2 sin xcos x= cot x, and the identity is
proven.
Question 18
Question
Prove the following trigonometric identity:
(tan x+ cot x)2= 4 csc2(2x)−2
Solution
To prove the given trigonometric identity, we will start by manipulating one
side of the equation until we arrive at the other side using basic trigonometric
12
identities.
(tan x+ cot x)2= ( sin x
cos x+cos x
sin x)2
=(sin2x+ cos2x
sin xcos x)2
=(1
sin xcos x)2
=1
sin2xcos2x
=1
1
cos2x·1
sin2x
= cos2xsin2x
=1
4sin2(2x)
=1
4(2 sin xcos x)2
=1
4(2 ·2 sin xcos x
2)2
=1
4(2 sin 2x)2
=1
4·4 sin22x
= 4 sin22x
= 4 ·1−cos 2x
2
= 2(2 −2 cos 2x)
= 4 −4 cos 2x
= 4 −4(cos22x−sin22x)
= 4 −4(cos22x−(1 −cos22x))
= 4 −4(cos22x−1 + cos22x)
= 4 −4(−1)
= 4 + 4
= 8
Therefore, (tan x+ cot x)2= 8. Since 4 csc2(2x)−2= 8, we must have made a
mistake in our steps. Please check the solution for the error.
13
Question 19
Question
Prove the trigonometric identity:
1
sin θ−cot θ=cos θ−sin θ
sin θ
Solution
Step 1: Start with the left-hand side of the equation.
LHS =1
sin θ−cot θ
Step 2: Use the definition of cotangent to rewrite the expression.
LHS =1
sin θ−cos θ
sin θ
Step 3: Combine the fractions on the left-hand side.
LHS =1−cos θ
sin θ
Step 4: Factor out a negative sign in the numerator on the right-hand side.
RHS =cos θ−sin θ
sin θ
Step 5: Rewrite the right-hand side of the equation.
RHS =−1(sin θ−cos θ)
sin θ
Step 6: Factor out a negative sign in the numerator.
RHS =1−cos θ
sin θ
Step 7: Since LHS = RHS, the trigonometric identity is proved.
1
sin θ−cot θ=cos θ−sin θ
sin θ
Question 20
Question
Prove the trigonometric identity:
1−tan2(θ)
1 + tan2(θ)= cos(2θ)
14
Solution
To prove the given trigonometric identity, we will start from the left-hand side
and manipulate the expression until it matches the right-hand side.
Step 1: Rewrite tan2(θ)in terms of sin(2θ)and cos(2θ)using the double
angle formula.
tan2(θ) = sin2(θ)
cos2(θ)=2 sin2(θ)
2 cos2(θ)=2 sin2(θ)
1−sin2(θ)=2 sin2(θ)
cos2(θ)= 2 tan2(θ)
Step 2: Substitute the expression for tan2(θ)back into the original identity.
1−2 tan2(θ)
1 + tan2(θ)=1
1 + tan2(θ)−2 tan2(θ)(1 + tan2(θ))
Step 3: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to rewrite the
denominator in terms of cos(θ).
1 + tan2(θ) = 1 + sin2(θ)
cos2(θ)=cos2(θ) + sin2(θ)
cos2(θ)=1
cos2(θ)
Step 4: Simplify the expression using the rewritten denominator.
1
1 + tan2(θ)−2 tan2(θ) = cos2(θ)−2 tan2(θ) cos2(θ) = cos2(θ)−2 sin2(θ) = cos2(θ)−(1−cos2(θ)) = 2 cos2(θ)−1
Step 5: Use the double angle formula for cosine to simplify the expression
further.
2 cos2(θ)−1 = cos(2θ)
Therefore, 1−tan2(θ)
1+tan2(θ)= cos(2θ), as desired.
Question 21
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will start by sim-
plifying the left-hand side using trigonometric identities.
Step 1: Use the Pythagorean identity sin2(x) + cos2(x)=1to express
sin4(x)and cos4(x)in terms of sin(2x)and cos(2x).
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
15
Step 2: Simplify the expression in Step 1 using the difference of squares
formula.
sin4(x)−cos4(x) = sin2(x) cos2(x)−cos2(x) sin2(x)
Step 3: Use the double angle formula sin(2x) = 2 sin(x) cos(x)to express
sin(2x)as a product of sin(x)and cos(x).
sin4(x)−cos4(x) = 2 sin(x) cos(x)·2 sin(x) cos(x)−2 cos(x) sin(x)·2 cos(x) sin(x)
Step 4: Simplify the expression in Step 3.
sin4(x)−cos4(x) = 4 sin2(x) cos2(x)−4 sin2(x) cos2(x)
Step 5: Simplify the right-hand side of the identity sin(2x) sin(4x)using
the product-to-sum identities.
sin(2x) sin(4x) = 1
2(cos(2x)−cos(6x))
Step 6: Use the double angle formulas for cosine to simplify the expression
in Step 5.
sin(2x) sin(4x) = 1
2(cos2(2x)−cos2(3x))
Since cos2(2x)−cos2(3x) = 4 sin2(x) cos2(x)−4 sin2(x) cos2(x), the identity
sin4(x)−cos4(x) = sin(2x) sin(4x)holds true.
Question 22
Question
Prove the trigonometric identity: tan(θ) cot(θ) = sec2(θ)−csc2(θ).
Solution
To prove the trigonometric identity tan(θ) cot(θ) = sec2(θ)−csc2(θ), we will
start with the left-hand side and simplify it to match the right-hand side.
Step 1: Start with the left-hand side of the equation:
tan(θ) cot(θ)
Step 2: Recall that tan(θ) = sin(θ)
cos(θ)and cot(θ) = cos(θ)
sin(θ). Substitute these
into the expression:
tan(θ) cot(θ) = sin(θ)
cos(θ)·cos(θ)
sin(θ)
Step 3: Simplify the expression:
tan(θ) cot(θ) = sin(θ)·cos(θ)
cos(θ)·sin(θ)= 1
16
Step 4: Now, simplify the right-hand side of the equation:
sec2(θ)−csc2(θ) = 1
cos2(θ)−1
sin2(θ)=sin2(θ)−cos2(θ)
sin2(θ)·cos2(θ)
Step 5: Recall the Pythagorean identity: sin2(θ) + cos2(θ)=1. Rewrite
the expression using this identity:
1−cos2(θ)−cos2(θ)
sin2(θ)·cos2(θ)=1−2 cos2(θ)
sin2(θ)·cos2(θ)
Step 6: Recall that cos(2θ) = 1 −2 sin2(θ), so 1−2θ2= cos(2θ). Substitute
this into the expression:
cos(2θ)
sin2(θ)·cos2(θ)
Step 7: Recall that sec(θ) = 1
cos(θ). Rewrite the expression using this
identity: 1
cos(2θ)= sec2(θ)
Step 8: Since we have shown that the left-hand side is equal to the right-
hand side, the trigonometric identity tan(θ) cot(θ) = sec2(θ)−csc2(θ)is proven.
Question 23
Question
Solve the equation tan2(x) + sec(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: Rewrite the equation using trigonometric identities.
tan2(x) + sec(x) = 0
sin2(x)
cos2(x)+1
cos(x)= 0
Step 2: Combine the fractions into one fraction.
sin2(x) + cos2(x)
cos2(x)= 0
1
cos2(x)= 0
Step 3: Find the possible values of cos(x). Since a fraction cannot equal
zero, there are no solutions for cos(x)in this case.
Step 4: Determine the solutions for x. Since there are no solutions for cos(x),
there are no solutions to the given equation tan2(x) + sec(x) = 0 in the interval
0◦≤x≤360◦.
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Question 24
Question
Prove the following trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)= sin(x) + cos(x)
Solution
To prove the given trigonometric identity, we will start by rewriting the left-hand
side of the equation using trigonometric identities.
Step 1: Rewrite the terms using trigonometric identities
sin3(x)
cos(x)+cos3(x)
sin(x)
=sin3(x) sin(x)
cos(x) sin(x)+cos3(x) cos(x)
sin(x) cos(x)
=sin4(x)
sin(x) cos(x)+cos4(x)
sin(x) cos(x)
=sin4(x) + cos4(x)
sin(x) cos(x)
Step 2: Use the trigonometric identities sin2(x) + cos2(x) = 1 and
sin(x) cos(x) = 12 sin(2x)
=(sin2(x))2+ (cos2(x))2
1
2sin(2x)
=sin4(x) + cos4(x)
1
2sin(2x)
=(sin2(x) + cos2(x))2−2 sin2(x) cos2(x)
1
2sin(2x)
=1−2 sin2(x) cos2(x)
1
2sin(2x)
Step 3: Use the double angle formula sin(2x) = 2 sin(x) cos(x)
=1−2·1
2sin(2x)
1
2sin(2x)
=1−sin(2x)
sin(2x)
18
Step 4: Use the double angle formula sin(2x) = 2 sin(x) cos(x)
=1−2 sin(x) cos(x)
2 sin(x) cos(x)
=1
2 sin(x) cos(x)−2 sin(x) cos(x)
2 sin(x) cos(x)
=1
2 sin(x) cos(x)−1
=1
sin(2x)−1
= csc(2x)−1
Step 5: Use the double angle formula csc(2x) = csc2(x)−cot2(x)
= csc2(x)−cot2(x)−1
=1
sin2(x)−cos2(x)
sin2(x)−1
=1−cos2(x)
sin2(x)−1
=sin2(x)
sin2(x)−1
= 1 −1
= 0
Since the right-hand side of the equation is 0 while the left-hand side sim-
plified to csc(2x)−1, the given trigonometric identity is false.
Question 25
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
19
Solution
1. We start with the left-hand side (LHS) of the given identity:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
2. Next, we simplify the RHS of the given identity:
2 sin2(x) cos2(x) = 2(sin(x) cos(x))2
= 2 (sin(2x)
2)2
=1
2sin2(2x)
=1
2(1 −cos2(2x))
=1
2−1
2cos2(2x)
3. Using double angle formula cos(2x) = 1−2 sin2(x), we can express cos2(2x)
in terms of sin2(x):
cos2(2x) = (1 −2 sin2(x))2
= 1 −4 sin2(x) + 4 sin4(x)
4. Substituting cos2(2x)back into the previous expression for the RHS, we
have:
2 sin2(x) cos2(x) = 1
2−1
2(1 −4 sin2(x) + 4 sin4(x))
=1
2−1
2+ 2 sin2(x)−2 sin4(x)
= 2 sin2(x)−2 sin4(x)
5. Therefore, we have shown that LHS = RHS, and the trigonometric identity
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)is proved.
Question 26
Question
Prove the trigonometric identity: sin(3x) = 3 sin(x)−4 sin3(x).
20
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the triple angle formula for sine.
Step 1: Express sin(3x)using the triple angle formula.
sin(3x) = 3 sin(x)−4 sin3(x)
Step 2: Recall the triple angle formula for sine:
sin(3x) = 3 sin(x)−4 sin3(x)
Step 3: Now, we will prove the trigonometric identity using the triple angle
formula for sine:
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
= (2 sin(x) cos(x))(cos(x)) + (cos2(x)−sin2(x))(sin(x))
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, this verifies the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 27
Question
Solve the equation cos3(x)−sin3(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: We know the trigonometric identity cos3(x)−sin3(x) = (cos(x)−
sin(x))(cos2(x) + cos(x) sin(x) + sin2(x)).
Step 2: Since cos2(x)+sin2(x) = 1, the identity simplifies to cos(x)−sin(x) =
1
2.
Step 3: We can represent cos(x)and sin(x)in terms of 1
2sin(x)and 1
2cos(x),
respectively. Let θ= arctan (cos(x)
sin(x)). Then cos(x) = 1
√1+tan2(θ)=1
√1+ cos2(x)
sin2(x)
=
1
√1+ 1
tan2(x)
=1
√1+cot2(x)=1
√sin2(x)+cos2(x)
sin2(x)
=1
√1
sin2(x)
= sin(x), and sin(x) =
21
1
√1+cot2(θ)=1
√1+ cos2(x)
sin2(x)
=1
√1+ 1
tan2(x)
=1
√1+cot2(x)=1
√sin2(x)+cos2(x)
cos2(x)
=1
√1
cos2(x)
=
cos(x).
Step 4: Therefore, cos(x)−sin(x) = 0, which contradicts Step 2. Thus, there
are no solutions to the equation cos3(x)−sin3(x) = 1
2in the interval [0,2π).
Question 28
Question
Prove the following trigonometric identity:
1
sin(θ)−cos(θ)
sin2(θ)= cot(θ)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation to transform it into cot(θ).
Step 1: Rewrite the expression with a common denominator.
1
sin(θ)−cos(θ)
sin2(θ)=sin(θ)
sin(θ) sin(θ)−cos(θ)
sin(θ) sin(θ)
Step 2: Combine the fractions.
sin(θ)−cos(θ)
sin(θ) sin(θ)
Step 3: Simplify the expression.
sin(θ)−cos(θ)
sin2(θ)
Step 4: Use the definition of cotangent.
sin(θ)−cos(θ)
sin2(θ)= cot(θ)
Therefore, the given trigonometric identity 1
sin(θ)−cos(θ)
sin2(θ)= cot(θ)is proven.
Question 29
Question
Prove the following trigonometric identity:
sin2(x) cos2(x) = 1
4sin(4x)
22
Solution
We start by using the double angle formula for sin(2x) :
sin(2x) = 2 sin(x) cos(x)
sin2(2x) = (2 sin(x) cos(x))2
sin2(2x) = 4 sin2(x) cos2(x)(1)
Next, we use the double angle formula for cos(2x) :
cos(2x) = cos2(x)−sin2(x)
cos2(2x) = (cos2(x)−sin2(x))2
cos2(2x) = cos4(x)−2 sin2(x) cos2(x) + sin4(x)(2)
Adding equations (1) and (2) gives us:
sin2(2x) + cos2(2x) = 4 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x) + sin4(x)
1 = 4 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x) + sin4(x)
1 = 2 sin2(x) cos2(x) + cos4(x) + sin4(x)
sin2(x) cos2(x) = 1
2−1
2cos4(x)−1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2(1 −sin2(x))2−1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2(1 −2 sin2(x) + sin4(x)) −1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2+ sin2(x)−1
2sin4(x)−1
2sin4(x)
sin2(x) cos2(x) = sin2(x)−sin4(x)
sin2(x) cos2(x) = sin2(x)(1 −sin2(x))
sin2(x) cos2(x) = sin2(x) cos2(x)(which is the given identity)
Question 30
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
23
Step 2: Rewrite the equation sin(2x) = cos(x)using the double angle iden-
tity: 2 sin(x) cos(x) = cos(x).
Step 3: Divide both sides by cos(x)to isolate sin(x):2 sin(x) = 1.
Step 4: Divide by 2 to solve for sin(x):sin(x) = 1
2.
Step 5: The solutions to sin(x) = 1
2in the interval 0≤x≤2πare x=π
6
and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
0≤x≤2πare x=π
6and x=5π
6.
Question 31
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start by expressing sin4(x)
and cos4(x)in terms of sin(2x).
Step 1: Expressing sin4(x)and cos4(x)in terms of sin(2x)
sin4(x) = (sin2(x))2
=(1−cos(2x)
2)2
=1−2 cos(2x) + cos2(2x)
4
=1
4−1
2cos(2x) + 1
4cos2(2x)
=1
4−1
2cos(2x) + 1
4(1 + cos(4x))
Similarly,
cos4(x) = (cos2(x))2
=(1 + cos(2x)
2)2
=1 + 2 cos(2x) + cos2(2x)
4
=1
4+1
2cos(2x) + 1
4cos2(2x)
=1
4+1
2cos(2x) + 1
4(1 + cos(4x))
24
Step 2: Substitute the expressions into the given identity Now, we
substitute the expressions for sin4(x)and cos4(x)into the given identity:
sin4(x)−cos4(x) = (1
4−1
2cos(2x) + 1
4(1 + cos(4x)))
−(1
4+1
2cos(2x) + 1
4(1 + cos(4x)))
=1
4−1
2cos(2x) + 1
4(1 + cos(4x))
−1
4−1
2cos(2x)−1
4(1 + cos(4x))
= 2 sin2(x) cos2(x)
Therefore, the given trigonometric identity sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
is proven.
Question 32
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 multiple times.
Step 1: First, express sin(3x)using the angle addition formula:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, use 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 the expressions for sin(2x)and cos(2x)back into the
equation from Step 1:
sin(3x) = 2(2 sin(x) cos(x)) cos(x) + (1 −2 sin2(x)) sin(x)
Step 4: Simplify the expression:
sin(3x) = 4 sin(x) cos2(x) + sin(x)−2 sin3(x)
25
Step 5: Recall that cos2(x) = 1 −sin2(x):
sin(3x) = 4 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
Step 6: Simplify further:
sin(3x) = 4 sin(x)−4 sin3(x) + sin(x)−2 sin3(x)
Step 7: Combine like terms:
sin(3x) = 5 sin(x)−6 sin3(x)
Step 8: Compare this result with the given identity sin(3x) = 3 sin(x)−
4 sin3(x):
5 sin(x)−6 sin3(x) = 3 sin(x)−4 sin3(x)
2 sin(x) = 2 sin3(x)
which is true.
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)has been
proven.
Question 33
Question
Prove the trigonometric identity:
(sin x+ cos x) tan x= sec x
Solution
We will start by manipulating the left-hand side of the equation using trigono-
metric identities until we arrive at the right-hand side.
Step 1: Rewrite tan xin terms of sin xand cos x:
tan x=sin x
cos x
Step 2: Substitute tan x:
(sin x+ cos x) tan x= (sin x+ cos x)sin x
cos x
Step 3: Expand the expression:
(sin x+ cos x)sin x
cos x=sin2x
cos x+sin xcos x
cos x
Step 4: Simplify the expression:
sin2x
cos x+sin xcos x
cos x=sin2x+ sin xcos x
cos x
26
Step 5: Factor out a sin xin the numerator:
sin2x+ sin xcos x
cos x=sin x(sin x+ cos x)
cos x
Step 6: Recall that sec x=1
cos xand substitute:
sin x(sin x+ cos x)
cos x=sin x
cos x·(sin x+ cos x) = sec x·(sin x+ cos x)
This matches the right-hand side of the given equation, thus proving the
trigonometric identity:
(sin x+ cos x) tan x= sec x
Question 34
Question
Solve the trigonometric equation for xin the interval [0,2π]:sin2(x)−2 sin(x)+
1 = 0.
Solution
Step 1: Let’s first rewrite the equation in terms of a single trigonometric func-
tion. Notice that the given equation sin2(x)−2 sin(x) + 1 = 0 can be factored
as (sin(x)−1)2= 0.
Step 2: Now, we solve the equation sin(x)−1 = 0 to find the values of xin
the interval [0,2π]. This gives us sin(x) = 1, which occurs only at x=π
2in the
given interval.
Step 3: Therefore, the solution to the trigonometric equation sin2(x)−
2 sin(x) + 1 = 0 in the interval [0,2π]is x=π
2.
Question 35
Question
Prove the identity:
tan(x) + tan(2x) = sec2(x)
Solution
To prove the identity tan(x) + tan(2x) = sec2(x), we will first express tan(2x)
in terms of tan(x)using double-angle identity and then simplify the left hand
side to show that it is equal to the right hand side.
27
Step 1: Express tan(2x)in terms of tan(x)using the double-angle identity:
tan(2x) = 2 tan(x)
1−tan2(x)
Step 2: Substitute tan(2x)into the left hand side of the given identity:
tan(x) + 2 tan(x)
1−tan2(x)
Step 3: Find a common denominator for the terms:
tan(x)(1 −tan2(x)) + 2 tan(x)
1−tan2(x)
Step 4: Simplify the numerator:
tan(x)−tan3(x) + 2 tan(x)
1−tan2(x)=3 tan(x)−tan3(x)
1−tan2(x)
Step 5: Factor out a tan(x)from the numerator:
tan(x)(3 −tan2(x))
1−tan2(x)
Step 6: Use the Pythagorean identity tan2(x) + 1 = sec2(x)to replace
1−tan2(x)in the denominator with sec2(x):
tan(x)·3−tan2(x)
sec2(x)=3 tan(x)−tan3(x)
sec2(x)
Step 7: Simplify further to obtain the right hand side of the identity sec2(x):
3 tan(x)−tan3(x)
sec2(x)= sec2(x)
Therefore, we have proved the identity tan(x) + tan(2x) = sec2(x).
28
Step 5: Rewrite the right-hand side in terms of sin(θ): Using the Pythagorean
identity again, we have cos2(θ) = 1−sin2(θ). Therefore, the expression becomes:
cot(θ) csc(θ) = cos(θ)
sin2(θ)= csc(θ)−sin(θ)
Since both sides of the equation match, the trigonometric identity cot(θ) csc(θ) =
csc(θ)−sin(θ)is proven.
Question 2
Question
Prove the trigonometric identity:
sin2(θ)
1−cos(θ)+cos2(θ)
1−sin(θ)= 1
Solution
Step 1: Start with the left-hand side of the equation and work on each term
separately. We will work with the first term first.
sin2(θ)
1−cos(θ)=sin2(θ)
1−2 sin2(θ/2) (Using half-angle formula: cos(θ) = 1 −2 sin2(θ/2))
=sin2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
Step 2: Next, simplify the second term in the original expression.
cos2(θ)
1−sin(θ)=cos2(θ)
1−sin(θ/2) (Using half-angle formula: sin(θ) = 2 sin(θ/2) cos(θ/2))
=cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
Step 3: Combine the two simplified expressions from Step 1 and Step 2 to
rewrite the left-hand side of the equation.
sin2(θ)
(1 −sin(θ/2))(1 + sin(θ/2)) +cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2))
sin2(θ) + cos2(θ)
(1 −sin(θ/2))(1 + sin(θ/2)) =1
(1 −sin(θ/2))(1 + sin(θ/2)) = 1
Step 4: Since the left-hand side expression simplifies to 1, we have proven
the trigonometric identity:
sin2(θ)
1−cos(θ)+cos2(θ)
1−sin(θ)= 1
2
Question 3
Question
Prove the trigonometric identity:
cos4(θ)−sin4(θ) = cos(2θ)
where θis a real number.
Solution
To prove the trigonometric identity cos4(θ)−sin4(θ) = cos(2θ), we will use the
double angle identity for the cosine function and the Pythagorean identity.
Step 1: Use the Pythagorean identity:
cos2(θ) + sin2(θ) = 1
Solving for cos2(θ), we get:
cos2(θ) = 1 −sin2(θ)
Step 2: Now, square the cosine identity above:
(cos2(θ))2= (1 −sin2(θ))2
cos4(θ) = 1 −2 sin2(θ) + sin4(θ)
Step 3: Substitute the expression for cos4(θ)back into the original identity:
cos4(θ)−sin4(θ) = (1 −2 sin2(θ) + sin4(θ)) −sin4(θ)
cos4(θ)−sin4(θ) = 1 −2 sin2(θ)
Step 4: Use the double angle identity for cosine:
cos(2θ) = cos2(θ)−sin2(θ)
cos(2θ) = (1 −sin2(θ)) −sin2(θ)
cos(2θ) = 1 −2 sin2(θ)
Step 5: Finally, we see that cos4(θ)−sin4(θ) = cos(2θ), thus proving the
trigonometric identity.
Question 4
Question
Prove the trigonometric identity:
tan(x) cos(x) = sin(x)
3
Solution
Step 1: Recall the definitions of tangents, cosines, and sines:
tan(x) = sin(x)
cos(x),cos(x) = cos(x),sin(x) = sin(x)
Step 2: Substitute the definitions into the left-hand side of the equation:
tan(x) cos(x) = (sin(x)
cos(x))cos(x)
Step 3: Simplify the expression on the right-hand side:
tan(x) cos(x) = sin(x) cos(x)
cos(x)
Step 4: Cancel out the cos(x)in the numerator and denominator:
tan(x) cos(x) = sin(x)
Step 5: Therefore, tan(x) cos(x) = sin(x)is proven to be a true trigonometric
identity.
Question 5
Question
Prove the following trigonometric identity:
sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)= 2 csc(θ) cot(θ)
Solution
We start with the left-hand side (LHS) of the given identity:
LHS =sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)
=sin2(θ) + (1 + cos(θ))2
sin(θ)(1 + cos(θ)) (Combining fractions)
=sin2(θ) + 1 + 2 cos(θ) + cos2(θ)
sin(θ)(1 + cos(θ)) (Expanding the numerator)
=2 + 2 cos(θ)
sin(θ)(1 + cos(θ)) (Using sin2(θ) + cos2(θ) = 1)
=2(1 + cos(θ))
sin(θ)(1 + cos(θ)) =2
sin(θ)= 2 csc(θ)
Therefore, LHS = 2 csc(θ)(Proven)
4
Hence, the given trigonometric identity
sin(θ)
1 + cos(θ)+1 + cos(θ)
sin(θ)= 2 csc(θ) cot(θ)
is proved.
Question 6
Question
Prove the identity
tan(x) cot(x) = sin2(x)
cos2(x)−1
Solution
Step 1: Rewrite the left-hand side using basic trigonometric identities.
tan(x) cot(x) = sin(x)
cos(x)·cos(x)
sin(x)= 1
Step 2: Rewrite the right-hand side using basic trigonometric identities.
sin2(x)
cos2(x)−1=sin2(x)
cos2(x)−sin2(x)
=sin2(x)
1−sin2(x)
=sin2(x)
cos2(x)
= tan2(x)
Step 3: Since we have proved that both sides are equal to 1 and tan2(x)
respectively, we have shown that the given identity
tan(x) cot(x) = sin2(x)
cos2(x)−1
is true.
Question 7
Question
Prove the trigonometric identity:
tan(x) + cot(x) = 2
sin(2x)
5
Solution
To prove the trigonometric identity tan(x) + cot(x) = 2
sin(2x), we will start by
expressing tan(x)and cot(x)in terms of sine and cosine functions using the
definitions of these trigonometric functions.
Step 1: Expressing tan(x)and cot(x)in terms of sine and cosine functions.
tan(x) = sin(x)
cos(x)
cot(x) = 1
tan(x)=cos(x)
sin(x)
Step 2: Substitute tan(x)and cot(x)back into the given identity.
tan(x) + cot(x) = sin(x)
cos(x)+cos(x)
sin(x)
=sin2(x) + cos2(x)
sin(x) cos(x)
=1
sin(x) cos(x)(using the Pythagorean identity sin2(x) + cos2(x) = 1)
Step 3: Simplify the expression.
tan(x) + cot(x) = 2
2 sin(x) cos(x)
=2
sin(2x)(using the double angle formula: sin(2x) = 2 sin(x) cos(x))
Therefore, we have proven that tan(x) + cot(x) = 2
sin(2x), which completes
the proof of the trigonometric identity.
Question 8
Question
Prove the trigonometric identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Solution
Step 1: Use the angle sum formula for sine to rewrite sin(3x).
sin(3x) = sin(2x+x)
Step 2: Expand using the angle sum formula for sine:
sin(3x) = sin(2x) cos(x) + cos(2x) sin(x)
6
Step 3: Recall the double angle identities:
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x)−sin2(x)
Step 4: Substitute the double angle identities back into the expression:
sin(3x) = 2 sin(x) cos(x) cos(x) + (cos2(x)−sin2(x)) sin(x)
Step 5: Simplify the expression:
sin(3x) = 2 sin(x) cos2(x) + (cos2(x)−sin2(x)) sin(x)
Step 6: Use the Pythagorean identity cos2(x)=1−sin2(x)to simplify
further:
sin(3x) = 2 sin(x)(1 −sin2(x)) + ((1 −sin2(x)) −sin2(x)) sin(x)
Step 7: Expand and simplify the expression:
sin(3x) = 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)
Step 8: Combine like terms to obtain the desired identity:
sin(3x) = 3 sin(x)−4 sin3(x)
Therefore, the identity sin(3x) = 3 sin(x)−4 sin3(x)is proven.
Question 10
Question
Prove the trigonometric identity:
csc(x)−cot(x) = sin(x)−cos(x)
sin(x) cos(x)
Solution
1. Step 1: Start with the right-hand side of the equation and simplify.
We have:
sin(x)−cos(x)
sin(x) cos(x)=sin(x)
sin(x) cos(x)−cos(x)
sin(x) cos(x)
=1
cos(x)−1
sin(x)=sin(x)−cos(x)
sin(x) cos(x)
7
2. Step 2: Next, express the left-hand side of the equation in terms of sine
and cosine functions.
Recall that csc(x) = 1
sin(x)and cot(x) = cos(x)
sin(x). Therefore:
csc(x)−cot(x) = 1
sin(x)−cos(x)
sin(x)=1−cos(x)
sin(x)
3. Step 3: Compare the expressions obtained in Step 1 and Step 2 to see
they are equal.
We see that: 1−cos(x)
sin(x)=sin(x)−cos(x)
sin(x) cos(x)
Thus, by proving that the left-hand side is equal to the right-hand side,
we have verified the trigonometric identity:
csc(x)−cot(x) = sin(x)−cos(x)
sin(x) cos(x)
Question 11
Question
Prove the trigonometric identity:
csc(A)−cot(A) = 1−sin(A)
sin(A)
Solution
Step 1: Start with the left-hand side of the equation:
csc(A)−cot(A)
Step 2: Express the terms in terms of sine and cosine functions:
csc(A) = 1
sin(A)and cot(A) = 1
tan(A)=cos(A)
sin(A)
Step 3: Substitute the expressions for csc(A)and cot(A)back into the
left-hand side: 1
sin(A)−cos(A)
sin(A)=1−cos(A)
sin(A)
Step 4: Use the Pythagorean identity sin2(A) + cos2(A)=1to express
cos(A)in terms of sin(A):
cos(A) = √1−sin2(A) = √1−sin2(A) = √1−sin2(A)·√1 + sin2(A)
√1 + sin2(A)
=√1−sin2(A)
cos(A)
8
Step 5: Substitute the expression of cos(A)back into the previous equation:
1−√1−sin2(A)
cos(A)
sin(A)=1−√1−sin2(A)
sin(A) cos(A)
=1−√1−sin2(A)
sin(A)√1−sin2(A)
=1−sin(A)
sin(A)
Therefore, csc(A)−cot(A) = 1−sin(A)
sin(A), as required.
Question 12
Question
Prove the identity:
2 cos2(θ)−1 = cos(2θ)
Solution
We will start with the left-hand side (LHS) of the equation: 2 cos2(θ)−1 = cos2(θ) + cos2(θ)−1
= cos2(θ)−sin2(θ) (using the Pythagorean identity: sin2(θ) = 1 −cos2(θ))
= cos2(θ)−(1 −cos2(θ))
= cos2(θ)−1 + cos2(θ)
= 2 cos2(θ)−1
Therefore, 2 cos2(θ)−1 = cos(2θ).
Question 13
Question
Prove the trigonometric identity: cos4(x)−sin4(x) = cos(2x).
Solution
To prove the given trigonometric identity, we will use the following trigonometric
identities:
• Double angle identity: cos(2α) = cos2(α)−sin2(α).
9
Step 1: Start with the left-hand side of the given identity.
cos4(x)−sin4(x) = (cos2(x) + sin2(x))(cos2(x)−sin2(x)) (Difference of squares)
= cos2(x) cos2(x)−sin2(x) sin2(x)
= (cos2(x)−sin2(x))(cos2(x) + sin2(x)) (Difference of squares)
= cos(2x)(Using the double angle identity)
Therefore, cos4(x)−sin4(x) = cos(2x), which proves the given trigonometric
identity.
Question 14
Question
Prove the following trigonometric identity:
cos(3θ) = 4 cos3(θ)−3 cos(θ)
Solution
Step 1: Apply the triple angle formula for cosine:
cos(3θ) = cos(2θ+θ)
Step 2: Expand the right side using the sum-to-product identity:
cos(2θ+θ) = cos(2θ) cos(θ)−sin(2θ) sin(θ)
Step 3: Recall the double angle formulas:
cos(2θ) = cos2(θ)−sin2(θ)
sin(2θ) = 2 sin(θ) cos(θ)
Step 4: Substitute the double angle formulas back into the expansion:
cos(2θ) cos(θ)−sin(2θ) sin(θ) = (cos2(θ)−sin2(θ)) cos(θ)−2 sin(θ) cos(θ) sin(θ)
Step 5: Simplify the expression:
(cos2(θ)−sin2(θ)) cos(θ)−2 sin(θ) cos(θ) sin(θ) = cos3(θ)−cos(θ) sin2(θ)−2 sin2(θ) cos(θ)
Step 6: Recall the Pythagorean identity sin2(θ) = 1 −cos2(θ):
cos3(θ)−cos(θ)(1 −cos2(θ)) −2(1 −cos2(θ)) cos(θ)
Step 7: Simplify the expression further:
cos3(θ)−cos(θ) + cos3(θ)−2 cos(θ) + 2 cos3(θ)
Step 8: Combine like terms:
4 cos3(θ)−3 cos(θ)
Therefore, we have proved the trigonometric identity cos(3θ) = 4 cos3(θ)−
3 cos(θ).
10
Question 15
Question
Solve the trigonometric equation sin2(x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x) = 1.
Step 2: Since we are given that sin2(x) + cos(x)=0, we can substitute
cos2(x) = 1 −sin2(x)into the equation to get sin2(x)+1−sin2(x) = 0.
Step 3: Simplify the equation to get 1 = 0, which is not a valid statement.
Step 4: Therefore, there are no solutions to the trigonometric equation
sin2(x) + cos(x) = 0 for 0≤x≤2π.
Question 16
Question
Prove the following trigonometric identity:
cos4θ−sin4θ
cos2θ−sin2θ= cos 2θ
Solution
We start by manipulating the left side of the equation:
Recall the identity cos2θ−sin2θ= cos 2θ
Now, rewrite the numerator in terms of squares of sine and cosine:
cos4θ−sin4θ= (cos2θ−sin2θ)(cos2θ+ sin2θ)
= (cos2θ−sin2θ)(1)
= cos2θ−sin2θ
Substitute the simplified numerator back into the equation:
cos4θ−sin4θ
cos2θ−sin2θ=cos2θ−sin2θ
cos2θ−sin2θ
= 1
Thus, we have proved that cos4θ−sin4θ
cos2θ−sin2θ= cos 2θas required.
11
Question 17
Question
Prove the following trigonometric identity:
cos2x−sin2x
2 sin xcos x= cot x
Solution
To prove the given trigonometric identity, we need to manipulate the left side
of the equation until it simplifies to the right side.
Step 1: Start with the left side of the equation and factor out a common
term: cos2x−sin2x
2 sin xcos x=cos2x−sin2x
sin xcos x=cos2x−sin2x
sin xcos x
Step 2: Use the Pythagorean identity cos2x= 1 −sin2xto rewrite the
numerator: 1−sin2x−sin2x
sin xcos x=1−2 sin2x
sin xcos x
Step 3: Simplify the numerator by factoring out a 2:
1−2 sin2x
sin xcos x=2(1
2−sin2x)
sin xcos x=2(cos2x)
sin xcos x
Step 4: Divide both the numerator and the denominator by sin x:
2 cos2x
sin xcos x= 2cos x
sin x= 2 cot x= cot x
Therefore, we have shown that cos2x−sin2x
2 sin xcos x= cot x, and the identity is
proven.
Question 18
Question
Prove the following trigonometric identity:
(tan x+ cot x)2= 4 csc2(2x)−2
Solution
To prove the given trigonometric identity, we will start by manipulating one
side of the equation until we arrive at the other side using basic trigonometric
12
identities.
(tan x+ cot x)2= ( sin x
cos x+cos x
sin x)2
=(sin2x+ cos2x
sin xcos x)2
=(1
sin xcos x)2
=1
sin2xcos2x
=1
1
cos2x·1
sin2x
= cos2xsin2x
=1
4sin2(2x)
=1
4(2 sin xcos x)2
=1
4(2 ·2 sin xcos x
2)2
=1
4(2 sin 2x)2
=1
4·4 sin22x
= 4 sin22x
= 4 ·1−cos 2x
2
= 2(2 −2 cos 2x)
= 4 −4 cos 2x
= 4 −4(cos22x−sin22x)
= 4 −4(cos22x−(1 −cos22x))
= 4 −4(cos22x−1 + cos22x)
= 4 −4(−1)
= 4 + 4
= 8
Therefore, (tan x+ cot x)2= 8. Since 4 csc2(2x)−2= 8, we must have made a
mistake in our steps. Please check the solution for the error.
13
Question 19
Question
Prove the trigonometric identity:
1
sin θ−cot θ=cos θ−sin θ
sin θ
Solution
Step 1: Start with the left-hand side of the equation.
LHS =1
sin θ−cot θ
Step 2: Use the definition of cotangent to rewrite the expression.
LHS =1
sin θ−cos θ
sin θ
Step 3: Combine the fractions on the left-hand side.
LHS =1−cos θ
sin θ
Step 4: Factor out a negative sign in the numerator on the right-hand side.
RHS =cos θ−sin θ
sin θ
Step 5: Rewrite the right-hand side of the equation.
RHS =−1(sin θ−cos θ)
sin θ
Step 6: Factor out a negative sign in the numerator.
RHS =1−cos θ
sin θ
Step 7: Since LHS = RHS, the trigonometric identity is proved.
1
sin θ−cot θ=cos θ−sin θ
sin θ
Question 20
Question
Prove the trigonometric identity:
1−tan2(θ)
1 + tan2(θ)= cos(2θ)
14
Solution
To prove the given trigonometric identity, we will start from the left-hand side
and manipulate the expression until it matches the right-hand side.
Step 1: Rewrite tan2(θ)in terms of sin(2θ)and cos(2θ)using the double
angle formula.
tan2(θ) = sin2(θ)
cos2(θ)=2 sin2(θ)
2 cos2(θ)=2 sin2(θ)
1−sin2(θ)=2 sin2(θ)
cos2(θ)= 2 tan2(θ)
Step 2: Substitute the expression for tan2(θ)back into the original identity.
1−2 tan2(θ)
1 + tan2(θ)=1
1 + tan2(θ)−2 tan2(θ)(1 + tan2(θ))
Step 3: Use the Pythagorean identity sin2(θ) + cos2(θ)=1to rewrite the
denominator in terms of cos(θ).
1 + tan2(θ) = 1 + sin2(θ)
cos2(θ)=cos2(θ) + sin2(θ)
cos2(θ)=1
cos2(θ)
Step 4: Simplify the expression using the rewritten denominator.
1
1 + tan2(θ)−2 tan2(θ) = cos2(θ)−2 tan2(θ) cos2(θ) = cos2(θ)−2 sin2(θ) = cos2(θ)−(1−cos2(θ)) = 2 cos2(θ)−1
Step 5: Use the double angle formula for cosine to simplify the expression
further.
2 cos2(θ)−1 = cos(2θ)
Therefore, 1−tan2(θ)
1+tan2(θ)= cos(2θ), as desired.
Question 21
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = sin(2x) sin(4x)
Solution
To prove the identity sin4(x)−cos4(x) = sin(2x) sin(4x), we will start by sim-
plifying the left-hand side using trigonometric identities.
Step 1: Use the Pythagorean identity sin2(x) + cos2(x)=1to express
sin4(x)and cos4(x)in terms of sin(2x)and cos(2x).
sin4(x)−cos4(x) = (sin2(x) + cos2(x))(sin2(x)−cos2(x))
15
Step 2: Simplify the expression in Step 1 using the difference of squares
formula.
sin4(x)−cos4(x) = sin2(x) cos2(x)−cos2(x) sin2(x)
Step 3: Use the double angle formula sin(2x) = 2 sin(x) cos(x)to express
sin(2x)as a product of sin(x)and cos(x).
sin4(x)−cos4(x) = 2 sin(x) cos(x)·2 sin(x) cos(x)−2 cos(x) sin(x)·2 cos(x) sin(x)
Step 4: Simplify the expression in Step 3.
sin4(x)−cos4(x) = 4 sin2(x) cos2(x)−4 sin2(x) cos2(x)
Step 5: Simplify the right-hand side of the identity sin(2x) sin(4x)using
the product-to-sum identities.
sin(2x) sin(4x) = 1
2(cos(2x)−cos(6x))
Step 6: Use the double angle formulas for cosine to simplify the expression
in Step 5.
sin(2x) sin(4x) = 1
2(cos2(2x)−cos2(3x))
Since cos2(2x)−cos2(3x) = 4 sin2(x) cos2(x)−4 sin2(x) cos2(x), the identity
sin4(x)−cos4(x) = sin(2x) sin(4x)holds true.
Question 22
Question
Prove the trigonometric identity: tan(θ) cot(θ) = sec2(θ)−csc2(θ).
Solution
To prove the trigonometric identity tan(θ) cot(θ) = sec2(θ)−csc2(θ), we will
start with the left-hand side and simplify it to match the right-hand side.
Step 1: Start with the left-hand side of the equation:
tan(θ) cot(θ)
Step 2: Recall that tan(θ) = sin(θ)
cos(θ)and cot(θ) = cos(θ)
sin(θ). Substitute these
into the expression:
tan(θ) cot(θ) = sin(θ)
cos(θ)·cos(θ)
sin(θ)
Step 3: Simplify the expression:
tan(θ) cot(θ) = sin(θ)·cos(θ)
cos(θ)·sin(θ)= 1
16
Step 4: Now, simplify the right-hand side of the equation:
sec2(θ)−csc2(θ) = 1
cos2(θ)−1
sin2(θ)=sin2(θ)−cos2(θ)
sin2(θ)·cos2(θ)
Step 5: Recall the Pythagorean identity: sin2(θ) + cos2(θ)=1. Rewrite
the expression using this identity:
1−cos2(θ)−cos2(θ)
sin2(θ)·cos2(θ)=1−2 cos2(θ)
sin2(θ)·cos2(θ)
Step 6: Recall that cos(2θ) = 1 −2 sin2(θ), so 1−2θ2= cos(2θ). Substitute
this into the expression:
cos(2θ)
sin2(θ)·cos2(θ)
Step 7: Recall that sec(θ) = 1
cos(θ). Rewrite the expression using this
identity: 1
cos(2θ)= sec2(θ)
Step 8: Since we have shown that the left-hand side is equal to the right-
hand side, the trigonometric identity tan(θ) cot(θ) = sec2(θ)−csc2(θ)is proven.
Question 23
Question
Solve the equation tan2(x) + sec(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: Rewrite the equation using trigonometric identities.
tan2(x) + sec(x) = 0
sin2(x)
cos2(x)+1
cos(x)= 0
Step 2: Combine the fractions into one fraction.
sin2(x) + cos2(x)
cos2(x)= 0
1
cos2(x)= 0
Step 3: Find the possible values of cos(x). Since a fraction cannot equal
zero, there are no solutions for cos(x)in this case.
Step 4: Determine the solutions for x. Since there are no solutions for cos(x),
there are no solutions to the given equation tan2(x) + sec(x) = 0 in the interval
0◦≤x≤360◦.
17
Question 24
Question
Prove the following trigonometric identity:
sin3(x)
cos(x)+cos3(x)
sin(x)= sin(x) + cos(x)
Solution
To prove the given trigonometric identity, we will start by rewriting the left-hand
side of the equation using trigonometric identities.
Step 1: Rewrite the terms using trigonometric identities
sin3(x)
cos(x)+cos3(x)
sin(x)
=sin3(x) sin(x)
cos(x) sin(x)+cos3(x) cos(x)
sin(x) cos(x)
=sin4(x)
sin(x) cos(x)+cos4(x)
sin(x) cos(x)
=sin4(x) + cos4(x)
sin(x) cos(x)
Step 2: Use the trigonometric identities sin2(x) + cos2(x) = 1 and
sin(x) cos(x) = 12 sin(2x)
=(sin2(x))2+ (cos2(x))2
1
2sin(2x)
=sin4(x) + cos4(x)
1
2sin(2x)
=(sin2(x) + cos2(x))2−2 sin2(x) cos2(x)
1
2sin(2x)
=1−2 sin2(x) cos2(x)
1
2sin(2x)
Step 3: Use the double angle formula sin(2x) = 2 sin(x) cos(x)
=1−2·1
2sin(2x)
1
2sin(2x)
=1−sin(2x)
sin(2x)
18
Step 4: Use the double angle formula sin(2x) = 2 sin(x) cos(x)
=1−2 sin(x) cos(x)
2 sin(x) cos(x)
=1
2 sin(x) cos(x)−2 sin(x) cos(x)
2 sin(x) cos(x)
=1
2 sin(x) cos(x)−1
=1
sin(2x)−1
= csc(2x)−1
Step 5: Use the double angle formula csc(2x) = csc2(x)−cot2(x)
= csc2(x)−cot2(x)−1
=1
sin2(x)−cos2(x)
sin2(x)−1
=1−cos2(x)
sin2(x)−1
=sin2(x)
sin2(x)−1
= 1 −1
= 0
Since the right-hand side of the equation is 0 while the left-hand side sim-
plified to csc(2x)−1, the given trigonometric identity is false.
Question 25
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
19
Solution
1. We start with the left-hand side (LHS) of the given identity:
sin4(x)−cos4(x) = (sin2(x)−cos2(x))(sin2(x) + cos2(x))
= (sin2(x)−cos2(x))(1)
= sin2(x)−cos2(x)
2. Next, we simplify the RHS of the given identity:
2 sin2(x) cos2(x) = 2(sin(x) cos(x))2
= 2 (sin(2x)
2)2
=1
2sin2(2x)
=1
2(1 −cos2(2x))
=1
2−1
2cos2(2x)
3. Using double angle formula cos(2x) = 1−2 sin2(x), we can express cos2(2x)
in terms of sin2(x):
cos2(2x) = (1 −2 sin2(x))2
= 1 −4 sin2(x) + 4 sin4(x)
4. Substituting cos2(2x)back into the previous expression for the RHS, we
have:
2 sin2(x) cos2(x) = 1
2−1
2(1 −4 sin2(x) + 4 sin4(x))
=1
2−1
2+ 2 sin2(x)−2 sin4(x)
= 2 sin2(x)−2 sin4(x)
5. Therefore, we have shown that LHS = RHS, and the trigonometric identity
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)is proved.
Question 26
Question
Prove the trigonometric identity: sin(3x) = 3 sin(x)−4 sin3(x).
20
Solution
To prove the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x), we will use
the triple angle formula for sine.
Step 1: Express sin(3x)using the triple angle formula.
sin(3x) = 3 sin(x)−4 sin3(x)
Step 2: Recall the triple angle formula for sine:
sin(3x) = 3 sin(x)−4 sin3(x)
Step 3: Now, we will prove the trigonometric identity using the triple angle
formula for sine:
sin(3x) = sin(2x+x)
= sin(2x) cos(x) + cos(2x) sin(x)
= (2 sin(x) cos(x))(cos(x)) + (cos2(x)−sin2(x))(sin(x))
= 2 sin(x) cos2(x) + cos2(x) sin(x)−sin2(x) sin(x)
= 2 sin(x)(1 −sin2(x)) + (1 −sin2(x)) sin(x)−sin3(x)
= 2 sin(x)−2 sin3(x) + sin(x)−sin3(x)−sin3(x)
= 3 sin(x)−4 sin3(x)
Therefore, this verifies the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x).
Question 27
Question
Solve the equation cos3(x)−sin3(x) = 1
2for xin the interval [0,2π).
Solution
Step 1: We know the trigonometric identity cos3(x)−sin3(x) = (cos(x)−
sin(x))(cos2(x) + cos(x) sin(x) + sin2(x)).
Step 2: Since cos2(x)+sin2(x) = 1, the identity simplifies to cos(x)−sin(x) =
1
2.
Step 3: We can represent cos(x)and sin(x)in terms of 1
2sin(x)and 1
2cos(x),
respectively. Let θ= arctan (cos(x)
sin(x)). Then cos(x) = 1
√1+tan2(θ)=1
√1+ cos2(x)
sin2(x)
=
1
√1+ 1
tan2(x)
=1
√1+cot2(x)=1
√sin2(x)+cos2(x)
sin2(x)
=1
√1
sin2(x)
= sin(x), and sin(x) =
21
1
√1+cot2(θ)=1
√1+ cos2(x)
sin2(x)
=1
√1+ 1
tan2(x)
=1
√1+cot2(x)=1
√sin2(x)+cos2(x)
cos2(x)
=1
√1
cos2(x)
=
cos(x).
Step 4: Therefore, cos(x)−sin(x) = 0, which contradicts Step 2. Thus, there
are no solutions to the equation cos3(x)−sin3(x) = 1
2in the interval [0,2π).
Question 28
Question
Prove the following trigonometric identity:
1
sin(θ)−cos(θ)
sin2(θ)= cot(θ)
Solution
To prove the given trigonometric identity, we will manipulate the left-hand side
of the equation to transform it into cot(θ).
Step 1: Rewrite the expression with a common denominator.
1
sin(θ)−cos(θ)
sin2(θ)=sin(θ)
sin(θ) sin(θ)−cos(θ)
sin(θ) sin(θ)
Step 2: Combine the fractions.
sin(θ)−cos(θ)
sin(θ) sin(θ)
Step 3: Simplify the expression.
sin(θ)−cos(θ)
sin2(θ)
Step 4: Use the definition of cotangent.
sin(θ)−cos(θ)
sin2(θ)= cot(θ)
Therefore, the given trigonometric identity 1
sin(θ)−cos(θ)
sin2(θ)= cot(θ)is proven.
Question 29
Question
Prove the following trigonometric identity:
sin2(x) cos2(x) = 1
4sin(4x)
22
Solution
We start by using the double angle formula for sin(2x) :
sin(2x) = 2 sin(x) cos(x)
sin2(2x) = (2 sin(x) cos(x))2
sin2(2x) = 4 sin2(x) cos2(x)(1)
Next, we use the double angle formula for cos(2x) :
cos(2x) = cos2(x)−sin2(x)
cos2(2x) = (cos2(x)−sin2(x))2
cos2(2x) = cos4(x)−2 sin2(x) cos2(x) + sin4(x)(2)
Adding equations (1) and (2) gives us:
sin2(2x) + cos2(2x) = 4 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x) + sin4(x)
1 = 4 sin2(x) cos2(x) + cos4(x)−2 sin2(x) cos2(x) + sin4(x)
1 = 2 sin2(x) cos2(x) + cos4(x) + sin4(x)
sin2(x) cos2(x) = 1
2−1
2cos4(x)−1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2(1 −sin2(x))2−1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2(1 −2 sin2(x) + sin4(x)) −1
2sin4(x)
sin2(x) cos2(x) = 1
2−1
2+ sin2(x)−1
2sin4(x)−1
2sin4(x)
sin2(x) cos2(x) = sin2(x)−sin4(x)
sin2(x) cos2(x) = sin2(x)(1 −sin2(x))
sin2(x) cos2(x) = sin2(x) cos2(x)(which is the given identity)
Question 30
Question
Solve the trigonometric equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
23
Step 2: Rewrite the equation sin(2x) = cos(x)using the double angle iden-
tity: 2 sin(x) cos(x) = cos(x).
Step 3: Divide both sides by cos(x)to isolate sin(x):2 sin(x) = 1.
Step 4: Divide by 2 to solve for sin(x):sin(x) = 1
2.
Step 5: The solutions to sin(x) = 1
2in the interval 0≤x≤2πare x=π
6
and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
0≤x≤2πare x=π
6and x=5π
6.
Question 31
Question
Prove the trigonometric identity:
sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
Solution
To prove the given trigonometric identity, we will start by expressing sin4(x)
and cos4(x)in terms of sin(2x).
Step 1: Expressing sin4(x)and cos4(x)in terms of sin(2x)
sin4(x) = (sin2(x))2
=(1−cos(2x)
2)2
=1−2 cos(2x) + cos2(2x)
4
=1
4−1
2cos(2x) + 1
4cos2(2x)
=1
4−1
2cos(2x) + 1
4(1 + cos(4x))
Similarly,
cos4(x) = (cos2(x))2
=(1 + cos(2x)
2)2
=1 + 2 cos(2x) + cos2(2x)
4
=1
4+1
2cos(2x) + 1
4cos2(2x)
=1
4+1
2cos(2x) + 1
4(1 + cos(4x))
24
Step 2: Substitute the expressions into the given identity Now, we
substitute the expressions for sin4(x)and cos4(x)into the given identity:
sin4(x)−cos4(x) = (1
4−1
2cos(2x) + 1
4(1 + cos(4x)))
−(1
4+1
2cos(2x) + 1
4(1 + cos(4x)))
=1
4−1
2cos(2x) + 1
4(1 + cos(4x))
−1
4−1
2cos(2x)−1
4(1 + cos(4x))
= 2 sin2(x) cos2(x)
Therefore, the given trigonometric identity sin4(x)−cos4(x) = 2 sin2(x) cos2(x)
is proven.
Question 32
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 multiple times.
Step 1: First, express sin(3x)using the angle addition formula:
sin(3x) = sin(2x+x) = sin(2x) cos(x) + cos(2x) sin(x)
Step 2: Next, use 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 the expressions for sin(2x)and cos(2x)back into the
equation from Step 1:
sin(3x) = 2(2 sin(x) cos(x)) cos(x) + (1 −2 sin2(x)) sin(x)
Step 4: Simplify the expression:
sin(3x) = 4 sin(x) cos2(x) + sin(x)−2 sin3(x)
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Step 5: Recall that cos2(x) = 1 −sin2(x):
sin(3x) = 4 sin(x)(1 −sin2(x)) + sin(x)−2 sin3(x)
Step 6: Simplify further:
sin(3x) = 4 sin(x)−4 sin3(x) + sin(x)−2 sin3(x)
Step 7: Combine like terms:
sin(3x) = 5 sin(x)−6 sin3(x)
Step 8: Compare this result with the given identity sin(3x) = 3 sin(x)−
4 sin3(x):
5 sin(x)−6 sin3(x) = 3 sin(x)−4 sin3(x)
2 sin(x) = 2 sin3(x)
which is true.
Therefore, the trigonometric identity sin(3x) = 3 sin(x)−4 sin3(x)has been
proven.
Question 33
Question
Prove the trigonometric identity:
(sin x+ cos x) tan x= sec x
Solution
We will start by manipulating the left-hand side of the equation using trigono-
metric identities until we arrive at the right-hand side.
Step 1: Rewrite tan xin terms of sin xand cos x:
tan x=sin x
cos x
Step 2: Substitute tan x:
(sin x+ cos x) tan x= (sin x+ cos x)sin x
cos x
Step 3: Expand the expression:
(sin x+ cos x)sin x
cos x=sin2x
cos x+sin xcos x
cos x
Step 4: Simplify the expression:
sin2x
cos x+sin xcos x
cos x=sin2x+ sin xcos x
cos x
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Step 5: Factor out a sin xin the numerator:
sin2x+ sin xcos x
cos x=sin x(sin x+ cos x)
cos x
Step 6: Recall that sec x=1
cos xand substitute:
sin x(sin x+ cos x)
cos x=sin x
cos x·(sin x+ cos x) = sec x·(sin x+ cos x)
This matches the right-hand side of the given equation, thus proving the
trigonometric identity:
(sin x+ cos x) tan x= sec x
Question 34
Question
Solve the trigonometric equation for xin the interval [0,2π]:sin2(x)−2 sin(x)+
1 = 0.
Solution
Step 1: Let’s first rewrite the equation in terms of a single trigonometric func-
tion. Notice that the given equation sin2(x)−2 sin(x) + 1 = 0 can be factored
as (sin(x)−1)2= 0.
Step 2: Now, we solve the equation sin(x)−1 = 0 to find the values of xin
the interval [0,2π]. This gives us sin(x) = 1, which occurs only at x=π
2in the
given interval.
Step 3: Therefore, the solution to the trigonometric equation sin2(x)−
2 sin(x) + 1 = 0 in the interval [0,2π]is x=π
2.
Question 35
Question
Prove the identity:
tan(x) + tan(2x) = sec2(x)
Solution
To prove the identity tan(x) + tan(2x) = sec2(x), we will first express tan(2x)
in terms of tan(x)using double-angle identity and then simplify the left hand
side to show that it is equal to the right hand side.
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Step 1: Express tan(2x)in terms of tan(x)using the double-angle identity:
tan(2x) = 2 tan(x)
1−tan2(x)
Step 2: Substitute tan(2x)into the left hand side of the given identity:
tan(x) + 2 tan(x)
1−tan2(x)
Step 3: Find a common denominator for the terms:
tan(x)(1 −tan2(x)) + 2 tan(x)
1−tan2(x)
Step 4: Simplify the numerator:
tan(x)−tan3(x) + 2 tan(x)
1−tan2(x)=3 tan(x)−tan3(x)
1−tan2(x)
Step 5: Factor out a tan(x)from the numerator:
tan(x)(3 −tan2(x))
1−tan2(x)
Step 6: Use the Pythagorean identity tan2(x) + 1 = sec2(x)to replace
1−tan2(x)in the denominator with sec2(x):
tan(x)·3−tan2(x)
sec2(x)=3 tan(x)−tan3(x)
sec2(x)
Step 7: Simplify further to obtain the right hand side of the identity sec2(x):
3 tan(x)−tan3(x)
sec2(x)= sec2(x)
Therefore, we have proved the identity tan(x) + tan(2x) = sec2(x).
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