WMA12/01 January 2021 Question 6: Trig Identities and Proofs

Question 1

Show that the fraction equation with sin θ and cos θ, where sin θ ≠ ½, can be rearranged into the cubic form 3 sin³ θ + 10 sin² θ − 8 sin θ = 0.

Accepted Answer

Use tan θ = sin θ / cos θ: (3 sin θ cos θ)/(2 sin θ − 1) = 5 sin θ / cos θ Multiply by both denominators: 3 sin θ cos² θ = 5 sin θ(2 sin θ − 1) Use cos² θ = 1 − sin² θ: 3 sin θ(1 − sin² θ) = 10 sin² θ − 5 sin θ 3 sin θ − 3 sin³ θ = 10 sin² θ − 5 sin θ So 3 sin³ θ + 10 sin² θ − 8 sin θ = 0 Answer: Shown: 3 sin θ cos² θ = 5 sin θ(2 sin θ − 1), then cos² θ = 1 − sin² θ gives 3 sin³ θ + 10 sin² θ − 8 sin θ = 0

Question 2

Using part a, solve the same equation with 2x in place of θ for −π/4 < x < π/4, giving answers to 3 decimal places where appropriate.

Accepted Answer

With θ = 2x: 3 sin³ 2x + 10 sin² 2x − 8 sin 2x = 0 sin 2x(3 sin² 2x + 10 sin 2x − 8) = 0 sin 2x(3 sin 2x − 2)(sin 2x + 4) = 0 sin 2x = 0, sin 2x = 2/3 (sin 2x = −4 is impossible) For −π/2 < 2x < π/2: 2x = 0 or 2x = 0.7297… x = 0 or x = 0.365 Answer: x = 0 or x = 0.365

WMA12/01 Jan 2021, Question 6: Trig Identities and Proofs

The question, in short

Part (a) · 4 marks

Part (b) · 4 marks

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