Trigonometric Identities Worksheet Answer Key – Precalculus

Problem 1 Reciprocal and Quotient Values

tan θ = (5/13) ÷ (12/13) = 5/12. csc θ = 1 ÷ (5/13) = 13/5. sec θ = 13/12. cot θ = 12/5.

Problem 2 Simplify with Pythagorean Identities

(a) 5(1 − sin² x) = 5cos² x. (b) 3(sec² x − 1) = 3tan² x. (c) 1. (d) cos x · sin x/cos x = sin x.

Problem 3 Find Sine and Tangent from Cosine

sin² θ = 1 − 25/169 = 144/169. In Quadrant IV sine is negative: sin θ = −12/13.

tan θ = (−12/13) ÷ (5/13) = −12/5.

Problem 4 Check a Point on the Unit Circle

(8/17)² + (15/17)² = 64/289 + 225/289 = 289/289 = 1, so the point is on the circle.

cos θ = 8/17, sin θ = 15/17, tan θ = 15/8, sec θ = 17/8.

Problem 5 Identity or Not?

(a) (1/2)² + (√3/2)² = 1/4 + 3/4 = 1.

(b) sin(π/3) = √3/2 ≈ 0.866, but sin(π/6) + sin(π/6) = 1. Since 0.866 ≠ 1, it is not an identity.

Problem 6 Exact Value with a Difference Formula

cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2).

= (√6 + √2)/4 ≈ 0.966.

Problem 7 Verify a Quotient Identity

Left side: 1/cos x − cos x = (1 − cos² x)/cos x = sin² x/cos x = sin x · (sin x/cos x) = sin x tan x. ∎

Problem 8 Verify with Reciprocals

Left side: 1/sin x − sin x = (1 − sin² x)/sin x = cos² x/sin x = cos x · (cos x/sin x) = cos x cot x. ∎

Problem 9 Tangent of a Sum

tan(π/3 + π/4) = (√3 + 1) ÷ (1 − √3 · 1) = (√3 + 1)/(1 − √3).

Multiply top and bottom by (1 + √3): (√3 + 1)²/(1 − 3) = (4 + 2√3)/(−2) = −2 − √3 ≈ −3.732.

Problem 10 Sum of Two Angles from Given Values

cos A = 15/17 and sin B = 4/5.

sin(A + B) = (8/17)(−3/5) + (15/17)(4/5) = (−24 + 60)/85 = 36/85.

cos(A + B) = (15/17)(−3/5) − (8/17)(4/5) = (−45 − 32)/85 = −77/85. Check: 36² + 77² = 85² ✓.

Problem 11 Double Angle from Cosine

sin θ = 24/25 (positive in Quadrant II).

sin 2θ = 2(24/25)(−7/25) = −336/625.

cos 2θ = 49/625 − 576/625 = −527/625.

Problem 12 Double Angle from Tangent

The reference triangle is 3-4-5, with sin θ = −3/5 and cos θ = 4/5.

sin 2θ = 2(−3/5)(4/5) = −24/25; cos 2θ = 16/25 − 9/25 = 7/25.

tan 2θ = −24/7. Check with the formula: 2(−3/4) ÷ (1 − 9/16) = −24/7 ✓.

Problem 13 Verify a Double-Angle Quotient

Use 1 − cos 2x = 1 − (1 − 2sin² x) = 2sin² x and sin 2x = 2 sin x cos x.

The fraction is 2sin² x/(2 sin x cos x) = sin x/cos x = tan x. ∎

Problem 14 Simplify to a Constant

Expand: sin x cos x · sin x/cos x + sin x cos x · cos x/sin x = sin² x + cos² x = 1.

Problem 15 A Longer Verification

Common denominator sin x(1 + cos x): the numerator is sin² x + (1 + cos x)² = sin² x + 1 + 2cos x + cos² x = 2 + 2cos x = 2(1 + cos x).

The fraction becomes 2(1 + cos x)/(sin x(1 + cos x)) = 2/sin x = 2 csc x. ∎

Problem 16 Angle of Elevation of a Treetop

tan(α + β) = (1/4 + 3/5) ÷ (1 − (1/4)(3/5)) = (17/20) ÷ (17/20) = 1.

Since tan 45° = 1 and the angle is acute, the treetop’s elevation is 45°.

Problem 17 Range of a Launched Ball

cos θ = 0.8, so sin 2θ = 2(0.6)(0.8) = 0.96. The factor 60²/32 = 112.5, so R = 112.5 × 0.96 = 108 feet.

At 45°, sin 90° = 1 and R = 112.5 feet, only 4.5 feet farther.

Problem 18 Interfering Sound Waves

sin(x + π/6) = sin x cos(π/6) + cos x sin(π/6) and sin(x − π/6) = sin x cos(π/6) − cos x sin(π/6).

Adding, the cos x terms cancel: 2 sin x cos(π/6) = 2 sin x (√3/2) = √3 sin x.

Problem 19 Solve cos 2x = cos x

2cos² x − 1 = cos x, so 2cos² x − cos x − 1 = 0 and (2cos x + 1)(cos x − 1) = 0.

cos x = 1 gives x = 0; cos x = −1/2 gives 2π/3 and 4π/3. x = 0, 2π/3, 4π/3.

Problem 20 Solve sin 2x = sin x

2 sin x cos x − sin x = 0, so sin x(2cos x − 1) = 0.

sin x = 0 gives x = 0, π; cos x = 1/2 gives π/3, 5π/3. x = 0, π/3, π, 5π/3.

Dividing by sin x would lose the solutions 0 and π, where sin x = 0.