Question

If $\tan x=-\frac{3}{4},$ $\frac{3}{2}\pi <x<2\pi$ then find the
i) $\sin 2x$
ii)$\cos 2x$
iii)$\tan 2x$

Answer 1

$\tan x=-\frac{3}{4},$ for $x\in (\frac{3\pi }{2},2\pi )$

$\left(i\right)\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}$

$=\frac{2\times -\frac{3}{4}}{1+\frac{9}{16}}=\frac{-\frac{3}{2}}{\frac{25}{16}}=-\frac{3}{2}\times \frac{16}{25}=-\frac{24}{25}$

$\left(ii\right)\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16-9}{16+9}=\frac{7}{25}$

$\left(iii\right)\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$

$=\frac{2\times -\frac{3}{4}}{1-\frac{9}{16}}=\frac{2\times -\frac{3}{4}}{\frac{16-9}{16}}=-\frac{3}{2}\times \frac{16}{7}=-\frac{24}{7}$