1.27: Circular measure and sectors, trigonometric identities and trigonometric equations

4PM1/1/November/2023 — Question 10 · 17 marks

(a) Using formulae on page 2, show that

(i) \mathrm{sin} 2A = 2 \mathrm{sin} A \mathrm{cos} A

(ii) \mathrm{cos} 2A = 2 \mathrm{cos}^2 A - 1

(3)

f(\theta) = \frac{2 \mathrm{tan} \theta}{1 + \mathrm{tan}^2 \theta}

(b) Show that

f(\theta) = \mathrm{sin} 2\theta

(4)

-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}

, the equation

5 \mathrm{tan}\left(x + \frac{\pi}{6}\right) = \left[1 + \mathrm{tan}^2\left(x + \frac{\pi}{6}\right)\right] \left[1 - 2 \mathrm{cos}^2\left(x + \frac{\pi}{6}\right)\right]

(6)

(d) Using calculus, find the exact value of

\int_0^{\frac{\pi}{2}} \left( \frac{4 \mathrm{tan} \theta}{1 + \mathrm{tan}^2 \theta} - \mathrm{cos} 5\theta + 2 \right) \mathrm{d}\theta

(4)