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Trigonometric Identities and Equations — Answers

Answer sheet with worked solutions for the Year 12 trigonometry worksheet.

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Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Trigonometry

Trigonometric Identities and Equations — Answers

Part A: Identities and exact ratios

Pythagorean identity: sin2θ + cos2θ = 1 for every angle θ. Quotient identity: tan θ = (when cos θ ≠ 0). Rearranged forms are useful: sin2θ = 1 − cos2θ and cos2θ = 1 − sin2θ.

When you take a square root, use the quadrant of θ to choose the sign. Sine is positive in quadrants 1 and 2, cosine in quadrants 1 and 4, and tangent in quadrants 1 and 3.

Worked example. sin θ = with θ in the second quadrant. cos2θ = 1 − = , and cosine is negative there, so cos θ = − and tan θ = ÷ (−) = −[frac 3/4].

1.If cos θ = and θ is acute, find the exact value of sin θ.

Answer:

  1. sin2θ = 1 − cos2θ = 1 − =
  2. θ is acute, so sin θ > 0: sin θ = [frac 12/13]
2.If sin θ = − and π < θ < , then cos θ = (answer space) and tan θ = (answer space).

Answer: −,

cos2θ = 1 − = , so cos θ = ±. In the third quadrant cosine is negative, so cos θ = −. Then tan θ = = (−) ÷ (−) = (tangent is positive in the third quadrant).

3.If sin x = 0.6 and x is obtuse, then cos x = (answer space).

Answer: −0.8

cos2x = 1 − 0.36 = 0.64, so cos x = ±0.8. An obtuse angle is in the second quadrant, where cosine is negative: cos x = −0.8.

4.For sin θ ≠ 0, the expression simplifies to
  1. Asin θ
  2. Bcos θ
  3. Ctan θ
  4. D1

Answer: sin θ

1 − cos2θ = sin2θ, and = sin θ.

5.Expand and simplify (sin θ + cos θ)2 − 2 sin θ cos θ.

Answer: 1

  1. (sin θ + cos θ)2 = sin2θ + 2 sin θ cos θ + cos2θ
  2. Subtract 2 sin θ cos θ: sin2θ + cos2θ = 1
6.The identity sin2θ + cos2θ = 1 is true only for acute angles.

Answer: False

It comes from the unit circle, where the point (cos θ, sin θ) is always 1 unit from the origin, so it holds for every angle θ.

Part B: Solving trigonometric equations

Method: make the trigonometric ratio the subject, find the reference angle (the acute angle with that ratio, ignoring the sign), then use the signs in each quadrant to place every solution in the domain.

For an equation in 2x (or 3x, …), first change the domain: if 0 ≤ x ≤ 2π then 0 ≤ 2x ≤ 4π. Find every value of 2x in that domain, then divide by 2.

Worked example. Solve tan x = −1 for 0 ≤ x ≤ 2π. The reference angle is , and tangent is negative in quadrants 2 and 4, so x = π − = [frac 3π/4] or x = 2π − = [frac 7π/4].

7.Solve √2 cos x = 1 for 0 ≤ x ≤ 2π.

Answer: ,

  1. cos x = , so the reference angle is .
  2. Cosine is positive in quadrants 1 and 4: x = or x = 2π − =
  3. x = [frac π/4], [frac 7π/4]
8.Solve tan x = √3 for 0° ≤ x ≤ 360°.

Answer: 60°, 240°

  1. Reference angle: tan 60° = √3.
  2. Tangent is positive in quadrants 1 and 3: x = 60° or 180° + 60° = 240°
  3. x = 60°, 240°
9.Solve sin x = −0.4 for 0° ≤ x ≤ 360°, giving answers correct to 1 decimal place.

Answer: 203.6°, 336.4°

  1. Reference angle: sin−1 0.4 = 23.578…°
  2. Sine is negative in quadrants 3 and 4.
  3. x = 180° + 23.578…° = 203.6° or x = 360° − 23.578…° = 336.4°
10.How many solutions does sin 2x = have for 0 ≤ x ≤ 2π?
  1. A2
  2. B4
  3. C6
  4. D1

Answer: 4

If 0 ≤ x ≤ 2π then 0 ≤ 2x ≤ 4π, which is two full cycles of sine. Each cycle has two solutions, so there are 4: x = , , , .

11.Solve cos 2x = − for 0 ≤ x ≤ 2π.

Answer: , , ,

  1. Let u = 2x. Since 0 ≤ x ≤ 2π, the new domain is 0 ≤ u ≤ 4π.
  2. cos u = −: reference angle ; cosine is negative in quadrants 2 and 3.
  3. u = , , , (add 2π to the first two)
  4. x = = [frac π/3], [frac 2π/3], [frac 4π/3], [frac 5π/3]
12.The equation 3 sin x = 4 has no solutions.

Answer: True

It gives sin x = , but −1 ≤ sin x ≤ 1 for every x, so there is no solution.

Part C: Equations that need an identity

If an equation mixes sin x and cos x, use an identity so that only one ratio remains. Replace cos2x by 1 − sin2x (or sin2x by 1 − cos2x), or divide by cos x to make tan x.

A quadratic in sin x can be factorised like an ordinary quadratic: let s = sin x, solve for s, then solve each sin x = s. Reject any value outside −1 ≤ s ≤ 1.

Worked example. Solve 2 cos2x = 1 + cos x for 0° ≤ x ≤ 360°. Rearrange: 2 cos2x − cos x − 1 = 0, so (2 cos x + 1)(cos x − 1) = 0. cos x = 1 gives x = 0°, 360°; cos x = − gives x = 120°, 240°.

13.Solve 2 sin2x − sin x − 1 = 0 for 0 ≤ x ≤ 2π.

Answer: , ,

  1. This is a quadratic in sin x: (2 sin x + 1)(sin x − 1) = 0
  2. sin x = 1 gives x = .
  3. sin x = −: reference angle , quadrants 3 and 4, so x = or .
  4. x = [frac π/2], [frac 7π/6], [frac 11π/6]
14.Solve 2 cos2x + 3 sin x − 3 = 0 for 0 ≤ x ≤ 2π.

Answer: , ,

  1. Replace cos2x with 1 − sin2x: 2 − 2 sin2x + 3 sin x − 3 = 0
  2. Rearrange: 2 sin2x − 3 sin x + 1 = 0, so (2 sin x − 1)(sin x − 1) = 0
  3. sin x = gives x = , ; sin x = 1 gives x = .
  4. x = [frac π/6], [frac π/2], [frac 5π/6]
15.Solve sin x = √3 cos x for 0 ≤ x ≤ 2π.

Answer: ,

  1. cos x = 0 is not a solution (then sin x = ±1 ≠ 0), so divide both sides by cos x: tan x = √3
  2. Reference angle ; tangent is positive in quadrants 1 and 3.
  3. x = [frac π/3], [frac 4π/3]

Part D: Extended problem

A Ferris wheel turns at a steady rate. The height h (in m) of a seat above the ground t minutes after it passes the lowest point is h = 11 − 9 cos().

(a) Find the height of the seat at t = 0, and find the greatest height and when it first occurs.

(b) Find the height after 5 minutes.

(c) Find, correct to 2 decimal places, the two times during the first rotation (0 ≤ t ≤ 20) when the seat is 15.5 m above the ground.

(d) For how long in each rotation is the seat higher than 15.5 m?

(e) Find the height of the seat after 3 minutes, correct to 1 decimal place.

16.Answer parts (a) to (e), showing full working.[9 marks]

Model answer: (a) h(0) = 11 − 9 cos 0 = 2 m. The greatest height is 11 + 9 = 20 m, when cos() = −1, that is = π, so t = 10 minutes. (b) h(5) = 11 − 9 cos() = 11 − 0 = 11 m. (c) 11 − 9 cos() = 15.5 gives cos() = −. For 0 ≤ t ≤ 20 the angle runs from 0 to 2π, so = or , giving t = ≈ 6.67 min and t = ≈ 13.33 min. (d) − = minutes ≈ 6 minutes 40 seconds in each rotation. (e) h(3) = 11 − 9 cos(0.3π) = 11 − 9 × 0.5878 = 5.7 m (1 d.p.).

Success criteria:

  • (a) 2 m at the start; greatest height 20 m at t = 10 min (2 marks)
  • (b) 11 m (1 mark)
  • (c) forms cos() = − and finds both times in the first rotation (3 marks)
  • (d) min ≈ 6 min 40 s (1 mark)
  • (e) 5.7 m, with the calculator in radian mode (2 marks)
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