Question 1

Simplify: cos²(π/8) − sin²(π/8)

Accepted Answer

We recognize this as the double angle formula for cosine.

Step 1: Identify the formula.
The expression cos²α − sin²α is the double angle formula: cos 2α = cos²α − sin²α

Step 2: Apply with α = π/8.
cos²(π/8) − sin²(π/8) = cos(2 · π/8) = cos(π/4)

Step 3: Evaluate cos(π/4).
cos(π/4) = 1/√2 = √2/2

Therefore, cos²(π/8) − sin²(π/8) = √2/2.

Question 2

If sin θ = 3/5 and θ lies in the second quadrant, find the value of tan θ.

Accepted Answer

Given: sin θ = 3/5 and θ is in the second quadrant.

Step 1: Use the Pythagorean identity sin²θ + cos²θ = 1.
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 16/25
cos θ = ±4/5

Step 2: Determine the sign of cos θ. In the second quadrant, sine is positive and cosine is negative.
Therefore, cos θ = -4/5

Step 3: Calculate tan θ = sin θ / cos θ
tan θ = (3/5) / (-4/5) = (3/5) × (-5/4) = -3/4

The answer is -3/4.

Question 3

If cos A = 4/5 and sin B = 5/13, where A and B are acute angles, find sin(A + B).

Accepted Answer

Given: cos A = 4/5 and sin B = 5/13, both A and B are acute angles.

Step 1: Find sin A using sin²A + cos²A = 1.
sin²A = 1 - cos²A = 1 - (4/5)² = 1 - 16/25 = 9/25
sin A = 3/5 (positive since A is acute)

Step 2: Find cos B using sin²B + cos²B = 1.
cos²B = 1 - sin²B = 1 - (5/13)² = 1 - 25/169 = 144/169
cos B = 12/13 (positive since B is acute)

Step 3: Apply the compound angle formula sin(A + B) = sin A cos B + cos A sin B.
sin(A + B) = (3/5)(12/13) + (4/5)(5/13)
           = 36/65 + 20/65
           = 56/65

The answer is 56/65.

Question 4

Find the principal value of sin⁻¹(sin 5π/6).

Accepted Answer

We need to find the principal value of sin⁻¹(sin 5π/6).

Step 1: Recall that the range of the principal value of sin⁻¹ is [-π/2, π/2].

Step 2: Evaluate sin(5π/6).
5π/6 is in the second quadrant (between π/2 and π).
sin(5π/6) = sin(π - π/6) = sin(π/6) = 1/2

Step 3: Find sin⁻¹(1/2) within the principal range [-π/2, π/2].
sin⁻¹(1/2) = π/6 (since π/6 is in [-π/2, π/2] and sin(π/6) = 1/2)

The answer is π/6.

Question 5

Simplify: (sin 2θ) / (1 + cos 2θ)

Accepted Answer

We need to simplify (sin 2θ) / (1 + cos 2θ).

Step 1: Apply the double angle formulas:
- sin 2θ = 2 sin θ cos θ
- cos 2θ = 2cos²θ - 1, which gives 1 + cos 2θ = 1 + (2cos²θ - 1) = 2cos²θ

Alternatively, cos 2θ = cos²θ - sin²θ, so 1 + cos 2θ = 1 + cos²θ - sin²θ = cos²θ + cos²θ = 2cos²θ

Step 2: Substitute into the original expression:
(sin 2θ) / (1 + cos 2θ) = (2 sin θ cos θ) / (2cos²θ)

Step 3: Simplify (assuming cos θ ≠ 0):
= (2 sin θ cos θ) / (2cos²θ) = sin θ / cos θ = tan θ

The answer is tan θ.

Question 6

Simplify: sin²(π/8) + sin²(3π/8)

Accepted Answer

We need to simplify sin²(π/8) + sin²(3π/8).

Step 1: Observe that 3π/8 = π/2 - π/8.
Therefore, sin(3π/8) = sin(π/2 - π/8) = cos(π/8) (using the co-function identity sin(π/2 - x) = cos x)

Step 2: Rewrite the expression:
sin²(π/8) + sin²(3π/8) = sin²(π/8) + cos²(π/8)

Step 3: Apply the Pythagorean identity sin²θ + cos²θ = 1:
sin²(π/8) + cos²(π/8) = 1

The answer is 1.

Question 7

If sin θ = 3/5 and θ lies in the second quadrant, find the value of cos θ.

Accepted Answer

We use the fundamental trigonometric identity sin²θ + cos²θ = 1.

Given: sin θ = 3/5

Step 1: Square both sides of the given equation.
sin²θ = (3/5)² = 9/25

Step 2: Apply the Pythagorean identity.
cos²θ = 1 - sin²θ = 1 - 9/25 = 16/25

Step 3: Take the square root.
cos θ = ±4/5

Step 4: Determine the sign using the quadrant.
Since θ lies in the second quadrant, cosine is negative.
Therefore, cos θ = -4/5

Question 8

Simplify: sin(60° + θ) cos(60° - θ) + cos(60° + θ) sin(60° - θ)

Accepted Answer

We recognize this expression as the sine addition formula in reverse.

Recall: sin(A + B) = sin A cos B + cos A sin B

Step 1: Identify the pattern.
The given expression has the form: sin(60° + θ) cos(60° - θ) + cos(60° + θ) sin(60° - θ)

Step 2: Apply the sine addition formula with A = (60° + θ) and B = (60° - θ).
sin[(60° + θ) + (60° - θ)] = sin(60° + θ) cos(60° - θ) + cos(60° + θ) sin(60° - θ)

Step 3: Simplify the argument.
(60° + θ) + (60° - θ) = 120°

Step 4: Evaluate.
sin(120°) = sin(180° - 60°) = sin(60°) = √3/2

Question 9

Find the principal value of sin⁻¹(sin(7π/6)).

Accepted Answer

We must find the principal value, which requires understanding the range of the inverse sine function.

Step 1: Recall the range of sin⁻¹.
The principal value of sin⁻¹ has range [-π/2, π/2].

Step 2: Evaluate sin(7π/6).
7π/6 = π + π/6 (third quadrant)
sin(7π/6) = -sin(π/6) = -1/2

Step 3: Find sin⁻¹(-1/2).
We need an angle in [-π/2, π/2] whose sine is -1/2.
sin(-π/6) = -1/2 and -π/6 ∈ [-π/2, π/2]

Step 4: Conclusion.
sin⁻¹(sin(7π/6)) = sin⁻¹(-1/2) = -π/6

Question 10

If tan θ = 5/12 and θ is acute, find sin(2θ).

Accepted Answer

We use the double angle formula for sine and derive sin θ and cos θ from the given tangent value.

Step 1: Construct a right triangle from tan θ = 5/12.
If tan θ = opposite/adjacent = 5/12, then opposite = 5, adjacent = 12.
Hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13

Step 2: Find sin θ and cos θ.
sin θ = 5/13
cos θ = 12/13

Step 3: Apply the double angle formula for sine.
sin(2θ) = 2 sin θ cos θ

Step 4: Substitute and calculate.
sin(2θ) = 2 × (5/13) × (12/13) = 2 × 60/169 = 120/169

Question 11

Simplify: cos²(π/8) - sin²(π/8)

Accepted Answer

We recognize this as the double angle formula for cosine.

Recall: cos(2α) = cos²α - sin²α

Step 1: Identify the pattern.
The expression cos²(π/8) - sin²(π/8) matches the form cos²α - sin²α with α = π/8.

Step 2: Apply the double angle formula.
cos(2 × π/8) = cos²(π/8) - sin²(π/8)

Step 3: Simplify the argument.
2 × π/8 = π/4

Step 4: Evaluate.
cos(π/4) = √2/2

Question 12

Simplify: sin(60° + θ) + sin(60° − θ)

Accepted Answer

We use the sum-to-product identity or expand using the compound angle formula.

Method: Expand using sin(A ± B) = sin A cos B ± cos A sin B

Step 1: Expand sin(60° + θ).
sin(60° + θ) = sin 60° cos θ + cos 60° sin θ = (√3/2) cos θ + (1/2) sin θ

Step 2: Expand sin(60° − θ).
sin(60° − θ) = sin 60° cos θ − cos 60° sin θ = (√3/2) cos θ − (1/2) sin θ

Step 3: Add the two expressions.
sin(60° + θ) + sin(60° − θ) = [(√3/2) cos θ + (1/2) sin θ] + [(√3/2) cos θ − (1/2) sin θ]
                              = √3 cos θ

Therefore, the answer is √3 cos θ.

Question 13

If tan θ = 5/12 and θ is acute, find sin 2θ.

Accepted Answer

We use the double angle formula sin 2θ = 2 sin θ cos θ, and derive sin θ and cos θ from tan θ.

Step 1: Construct a right triangle.
If tan θ = 5/12, then opposite = 5, adjacent = 12.
Hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13

Step 2: Find sin θ and cos θ.
sin θ = opposite/hypotenuse = 5/13
cos θ = adjacent/hypotenuse = 12/13

Step 3: Apply the double angle formula.
sin 2θ = 2 sin θ cos θ = 2 · (5/13) · (12/13) = 2 · 60/169 = 120/169

Therefore, sin 2θ = 120/169.

Question 14

If cos(A) + cos(B) = 1/2 and sin(A) + sin(B) = 1/3, then cos(A − B) is equal to:

Accepted Answer

We use sum-to-product identities and the Pythagorean theorem.

Given:
• cos(A) + cos(B) = 1/2  ... (1)
• sin(A) + sin(B) = 1/3  ... (2)

Square both equations:
• [cos(A) + cos(B)]² = 1/4
  cos²(A) + 2cos(A)cos(B) + cos²(B) = 1/4  ... (3)

• [sin(A) + sin(B)]² = 1/9
  sin²(A) + 2sin(A)sin(B) + sin²(B) = 1/9  ... (4)

Add equations (3) and (4):
[cos²(A) + sin²(A)] + [cos²(B) + sin²(B)] + 2[cos(A)cos(B) + sin(A)sin(B)] = 1/4 + 1/9

1 + 1 + 2cos(A − B) = 1/4 + 1/9

2 + 2cos(A − B) = 9/36 + 4/36 = 13/36

2cos(A − B) = 13/36 − 2 = 13/36 − 72/36 = −59/36

cos(A − B) = −59/72.

Question 15

If sin(A + B) = 3/5 and cos(A − B) = 4/5, where A + B and A − B are acute angles, then the value of sin(2A) is:

Accepted Answer

We use the compound angle formulas and product-to-sum identities.

Given: sin(A + B) = 3/5 and cos(A − B) = 4/5, both acute angles.

From sin(A + B) = 3/5 (acute), we get cos(A + B) = √(1 − 9/25) = 4/5.
From cos(A − B) = 4/5 (acute), we get sin(A − B) = √(1 − 16/25) = 3/5.

Now, sin(2A) = sin[(A + B) + (A − B)].
Using sin(X + Y) = sin X cos Y + cos X sin Y:
sin(2A) = sin(A + B)·cos(A − B) + cos(A + B)·sin(A − B)
        = (3/5)·(4/5) + (4/5)·(3/5)
        = 12/25 + 12/25
        = 24/25.

Verification: This is consistent with 2A being acute (since sin(2A) < 1).

Question 16

If sin(A + B) = 1 and sin(A − B) = 1/2, where A, B ∈ [0, π/2], then the value of tan(A)/tan(B) is:

Accepted Answer

We use the compound angle formulas:
sin(A + B) = sin A cos B + cos A sin B = 1
sin(A − B) = sin A cos B − cos A sin B = 1/2

Adding these two equations:
2 sin A cos B = 3/2
⟹ sin A cos B = 3/4

Subtracting the second from the first:
2 cos A sin B = 1/2
⟹ cos A sin B = 1/4

Dividing the first result by the second:
(sin A cos B)/(cos A sin B) = (3/4)/(1/4) = 3

Therefore:
tan A / tan B = (sin A / cos A) / (sin B / cos B) = (sin A cos B)/(cos A sin B) = 3

Question 17

The value of cos⁻¹(cos(7π/6)) is:

Accepted Answer

We need to find cos⁻¹(cos(7π/6)).

First, compute cos(7π/6):
7π/6 = π + π/6
cos(7π/6) = cos(π + π/6) = −cos(π/6) = −√3/2

Now we need cos⁻¹(−√3/2).

The range of cos⁻¹ is [0, π]. We seek the unique angle θ ∈ [0, π] such that cos(θ) = −√3/2.

Since −√3/2 is negative, θ must be in (π/2, π).
cos(5π/6) = −cos(π/6) = −√3/2 ✓

Since 5π/6 ∈ [0, π], we have:
cos⁻¹(cos(7π/6)) = 5π/6

Question 18

If tan(θ/2) = t, then sin(θ) − cos(θ) equals:

Accepted Answer

We use the Weierstrass substitution formulas. Let tan(θ/2) = t.

Then:
sin(θ) = 2t/(1 + t²)
cos(θ) = (1 − t²)/(1 + t²)

Therefore:
sin(θ) − cos(θ) = 2t/(1 + t²) − (1 − t²)/(1 + t²)
                 = [2t − (1 − t²)]/(1 + t²)
                 = [2t − 1 + t²]/(1 + t²)
                 = [t² + 2t − 1]/(1 + t²)

Question 19

If sin⁻¹(x) + sin⁻¹(y) = π/2, then the value of x² + y² is:

Accepted Answer

Given: sin⁻¹(x) + sin⁻¹(y) = π/2

Let sin⁻¹(x) = α and sin⁻¹(y) = β.
Then α + β = π/2, which means β = π/2 − α.

From the definitions:
sin(α) = x
sin(β) = y

Since β = π/2 − α:
sin(β) = sin(π/2 − α) = cos(α)

Therefore:
y = cos(α)

Now, we know that sin(α) = x and cos(α) = y.
Using the Pythagorean identity sin²(α) + cos²(α) = 1:

x² + y² = sin²(α) + cos²(α) = 1

Question 20

The value of sin(π/8) · sin(3π/8) · sin(5π/8) · sin(7π/8) is:

Accepted Answer

We use the symmetry of sine and product-to-sum identities.

First, note the symmetries:
sin(5π/8) = sin(π − 5π/8) = sin(3π/8)
sin(7π/8) = sin(π − 7π/8) = sin(π/8)

Therefore:
sin(π/8) · sin(3π/8) · sin(5π/8) · sin(7π/8) = sin(π/8) · sin(3π/8) · sin(3π/8) · sin(π/8)
                                                 = [sin(π/8) · sin(3π/8)]²

Now, note that 3π/8 = π/2 − π/8, so sin(3π/8) = cos(π/8).

Therefore:
[sin(π/8) · sin(3π/8)]² = [sin(π/8) · cos(π/8)]²

Using the double angle formula: sin(2α) = 2sin(α)cos(α), we have sin(α)cos(α) = sin(2α)/2.

With α = π/8:
sin(π/8) · cos(π/8) = sin(π/4)/2 = (√2/2)/2 = √2/4

Therefore:
[sin(π/8) · cos(π/8)]² = (√2/4)² = 2/16 = 1/8

NDA Trig Functions & Identities

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Key Points to Remember

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60-Second Revision — Trig Functions & Identities