Question

If $\tan x=-\frac{3}{4},\frac{3\pi }{2}<x<2\pi ,thenfind\left(i\right)\sin 2x\left(ii\right)\cos 2x\left(iii\right)\tan 2x$

Answer 1

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

$=\frac{2\times \frac{-3}{4}}{1+{\left(\frac{-3}{4}\right)}^2}$

$=\frac{\frac{-3}{2}}{\frac{16+9}{16}}$

$=\frac{-3}{2}\times \frac{16}{25}=\frac{-24}{25}$

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

$=\frac{1-{\left(\frac{-3}{4}\right)}^2}{1+{\left(\frac{-3}{4}\right)}^2}$

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

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

$\left(iii\right)\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{-24}{25}}{\frac{7}{25}}=\frac{-24}{7}$