1.40: Past-paper question 8

4PM1/1/November/2024 — Question 8 · 11 marks

(i) (a) Using a formula given on page 2, show that
tan2A=2tanA1−tan2A\mathrm{tan} 2A = \frac{2 \mathrm{tan} A}{1 - \mathrm{tan}^2 A}
(2)
(b) Hence, solve the equation
tanA∘−tan2A∘=0for0≤A≤180\mathrm{tan} A^\circ - \mathrm{tan} 2A^\circ = 0 \quad \mathrm{for} 0 \leq A \leq 180
(5)
(ii) Using a formula given on page 2, solve, giving your solutions as exact values
cos(x−π6)=sinxfor−π≤x≤2π\mathrm{cos}\left(x - \frac{\pi}{6}\right) = \mathrm{sin} x \quad \mathrm{for} -\pi \leq x \leq 2\pi
(4)