1.52: Past-paper question 11

4PM1/1/November/2025 — Question 11 · 10 marks

(a) Using a formula on page 2, show that
1−cos2A1+cos2A=tan2A\frac{1 - \mathrm{cos} 2A}{1 + \mathrm{cos} 2A} = \mathrm{tan}^2 A
(3)
(b) Hence, or otherwise, solve in degrees to one decimal place
3−3cos4x1+cos4x+5sin2xcos2x=2for−90∘<x<90∘\frac{3 - 3 \mathrm{cos} 4x}{1 + \mathrm{cos} 4x} + \frac{5 \mathrm{sin} 2x}{\mathrm{cos} 2x} = 2 \quad \mathrm{for} -90^\circ < x < 90^\circ
(7)