If cos theta = frac{-3}{5} and pi

Question

(C) Given $\cos 	heta = \frac{-3}{5}$ and $\pi < 	heta < \frac{3\pi}{2}$.Since $\pi < 	heta < \frac{3\pi}{2}$,we have $\frac{\pi}{2} < \frac{	heta

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) Given $\cos 	heta = \frac{-3}{5}$ and $\pi Since $\pi In the $2^{nd}$ quadrant,$\sin \frac{	heta}{2} > 0$,$\cos \frac{	heta}{2} Using the half-angle formulas:$\cos 	heta = 1 - 2 \sin^2 \frac{	heta}{2}$ $\Rightarrow \frac{-3}{5} = 1 - 2 \sin^2 \frac{	heta}{2}$ $\Rightarrow 2 \sin^2 \frac{	heta}{2} = \frac{8}{5}$ $\Rightarrow \sin^2 \frac{	heta}{2} = \frac{4}{5}$.Since $\sin \frac{	heta}{2} > 0$,we get $\sin \frac{	heta}{2} = \frac{2}{\sqrt{5}}$.$\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$ $\Rightarrow \frac{-3}{5} = 2 \cos^2 \frac{	heta}{2} - 1$ $\Rightarrow 2 \cos^2 \frac{	heta}{2} = \frac{2}{5}$ $\Rightarrow \cos^2 \frac{	heta}{2} = \frac{1}{5}$.Since $\cos \frac{	heta}{2} Then $	an \frac{	heta}{2} = \frac{\sin \frac{	heta}{2}}{\cos \frac{	heta}{2}} = \frac{2/\sqrt{5}}{-1/\sqrt{5}} = -2$.Substituting these values into the expression:$	an \frac{	heta}{2} + \sin \frac{	heta}{2} + 2 \cos \frac{	heta}{2} = -2 + \frac{2}{\sqrt{5}} + 2 \left( -\frac{1}{\sqrt{5}} ight) = -2 + \frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}} = -2$.

If $\cos \theta = \frac{-3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$,then $\tan \frac{\theta}{2} + \sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2} =$

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