Trigonometric Functions — MATHS STD 11 Science — Question

2. $\sin4\beta$

Solution:

it is given thet $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}.$

Now,

$\tan\beta=\frac{2\tan\beta}{1-\tan^2\beta}$

$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}$

$=\frac{3}{4}$

$\therefore\tan(\alpha+2\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}$

$=\frac{\frac{25}{28}}{\frac{25}{28}}$

$=1$

$\tan(\alpha+2\beta)=1=\tan\frac{\pi}{4}$

$\Rightarrow\alpha+2\beta=\frac{\pi}{4}$

$\Rightarrow\alpha=\frac{\pi}{4}-2\beta$

$\Rightarrow2\alpha=\frac{\pi}{2}-4\beta$

$\Rightarrow\cos2\alpha=\cos\big(\frac{\pi}{2}-4\beta\big)=\sin4\beta$

$\therefore\cos2\alpha=\sin4\beta$

Hence, the correct answer is option B.