Trigonometric Functions — Practice

Question 1

Convert 5π/6 radians into degree measure.

Accepted Answer

Degree measure = (180/π) × Radian measure = (180/π) × 5π/6 = 150°. Answer: 5π/6 radians = 150°.

Question 2

Find the length of an arc that subtends an angle of 60° at the centre of a circle of radius 21 cm (use π = 22/7).

Accepted Answer

θ = 60° = π/3 radian. l = rθ = 21 × (22/7)/3 = 21 × 22/21 = 22 cm. Answer: The arc length is 22 cm.

Question 3

sin x = 3/5, x lies in the second quadrant. Find cos x and tan x.

Accepted Answer

cos²x = 1 − sin²x = 1 − 9/25 = 16/25, so cos x = ±4/5. In the second quadrant cosine is negative, so cos x = −4/5. Then tan x = sin x/cos x = (3/5)/(−4/5) = −3/4. Answer: cos x = −4/5 and tan x = −3/4.

Question 4

Evaluate cos(1035°).

Accepted Answer

1035° = 2 × 360° + 315°, so cos(1035°) = cos(315°) using periodicity. 315° = 360° − 45°, so cos(315°) = cos(45°) = 1/√2 using the 2π−x allied-angle rule. Answer: cos(1035°) = 1/√2.

Question 5

Find the value of cos 75°.

Accepted Answer

Write 75° = 45° + 30°. cos 75° = cos45°cos30° − sin45°sin30° = (1/√2)(√3/2) − (1/√2)(1/2) = (√3−1)/(2√2). Answer: cos 75° = (√3−1)/(2√2).

Question 6

Prove that sin(x+y)/sin(x-y) = (tanx+tany)/(tanx-tany), assuming none of the denominators vanish.

Accepted Answer

LHS numerator: sinx cosy + cosx siny. LHS denominator: sinx cosy − cosx siny. Dividing every term by cosx cosy: numerator becomes tanx+tany and denominator becomes tanx−tany. Answer: So sin(x+y)/sin(x-y) = (tanx+tany)/(tanx-tany), matching the RHS.

Question 7

If tan x = 3/4 and x lies in the first quadrant, find sin 2x, cos 2x, and tan 2x.

Accepted Answer

From tanx=3/4 with x in QI: sinx=3/5, cosx=4/5. sin2x = 2sinx cosx = 2(3/5)(4/5) = 24/25. cos2x = cos²x−sin²x = 16/25−9/25 = 7/25. tan2x = sin2x/cos2x = 24/7. Answer: sin2x=24/25, cos2x=7/25, tan2x=24/7.

Question 8

If sin x = 1/3, find sin 3x.

Accepted Answer

sin3x = 3sinx − 4sin³x = 3(1/3) − 4(1/3)³ = 1 − 4/27 = 23/27. Answer: sin 3x = 23/27.

Question 9

Prove that (sin5x + sin3x)/(cos5x + cos3x) = tan4x.

Accepted Answer

Numerator: sin5x+sin3x = 2 sin4x cosx (using sinA+sinB = 2 sin((A+B)/2) cos((A-B)/2)). Denominator: cos5x+cos3x = 2 cos4x cosx. Dividing, the shared 2cosx factor cancels, leaving sin4x/cos4x = tan4x. Answer: So (sin5x+sin3x)/(cos5x+cos3x) = tan4x.

Question 10

Express 2 cos5x sin3x as a sum, then evaluate it at x = π/12.

Accepted Answer

2cosA sinB = sin(A+B) − sin(A−B), so 2cos5x sin3x = sin8x − sin2x. At x=π/12: sin(8π/12) − sin(2π/12) = sin(2π/3) − sin(π/6) = √3/2 − 1/2. Answer: The value is (√3−1)/2.

Practice — Trigonometric Functions

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