9709 P31 - Jun 2021 - Q4
1725
(a) Prove that \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^2 \theta\).
(b) Hence find the exact value of \(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1 - \cos 2\theta}{1 + \cos 2\theta} \, d\theta\).
Solution
(a) Use the double angle formula for cosine: \(\cos 2\theta = 1 - 2\sin^2 \theta\).
Substitute into the expression: \(\frac{1 - (1 - 2\sin^2 \theta)}{1 + (1 - 2\sin^2 \theta)} = \frac{2\sin^2 \theta}{2\cos^2 \theta} = \tan^2 \theta\).
(b) Express \(\tan^2 \theta\) in terms of \(\sec^2 \theta\): \(\tan^2 \theta = \sec^2 \theta - 1\).
Integrate: \(\int (\sec^2 \theta - 1) \, d\theta = \tan \theta - \theta\).
Evaluate from \(\theta = \frac{1}{6}\pi\) to \(\theta = \frac{1}{3}\pi\):
\(\left[ \tan \theta - \theta \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} = \left( \tan \frac{1}{3}\pi - \frac{1}{3}\pi \right) - \left( \tan \frac{1}{6}\pi - \frac{1}{6}\pi \right)\).
Calculate: \(\tan \frac{1}{3}\pi = \sqrt{3}\), \(\tan \frac{1}{6}\pi = \frac{1}{\sqrt{3}}\).
Result: \(\left( \sqrt{3} - \frac{1}{3}\pi \right) - \left( \frac{1}{\sqrt{3}} - \frac{1}{6}\pi \right) = \frac{2}{3}\sqrt{3} - \frac{1}{6}\pi\).
