$\sin \frac{x}{2}, \cos \frac{x}{2}$ અને $\tan \frac{x}{2}$ ની કિંમતો શોધો.: $\tan x=\frac{-4}{3}, x$ એ બીજા ચરણમાં છે.
Here, $x$ is in quadrant $II$.
i.e., $\frac{\pi}{2} < x < \pi$
$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$
There, $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$
are lies in first quadrant. It is given that $\tan x=-\frac{4}{3}$
$\sec ^{2} x=1+\tan ^{2} x=1+\left(\frac{-4}{3}\right)^{2}=1+\frac{16}{9}=\frac{25}{9}$
$\therefore \cos ^{2} x=\frac{9}{25}$
$\Rightarrow \cos x=\pm \frac{3}{5}$
As $x$ is in quadrant $II$, $\cos x$ is negative.
$\cos x=\frac{-3}{5}$
Now, $\cos x=2 \cos ^{2} \frac{x}{2}-1$
$\Rightarrow \frac{-3}{5}=2 \cos ^{2} \frac{x}{2}-1$
$\Rightarrow 2 \cos ^{2} \frac{x}{2}=1-\frac{3}{5}$
$\Rightarrow 2 \cos ^{2} \frac{x}{2}=\frac{2}{5}$
$\Rightarrow \cos ^{2} \frac{x}{2}=\frac{1}{5}$
$ \Rightarrow \cos \frac{x}{2} = \frac{1}{{\sqrt 5 }}\quad \left[ {\because \cos \frac{x}{2}\,is\,positve} \right]$
$\therefore \cos \frac{x}{2}=\frac{\sqrt{5}}{5}$
$\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}=1$
$\Rightarrow \sin ^{2} \frac{x}{2}+\left(\frac{1}{\sqrt{5}}\right)^{2}=1$
$\Rightarrow \sin ^{2} \frac{x}{2}=1-\frac{1}{5}=\frac{4}{5}$
$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{2}{\sqrt{5}} \quad\left[\because \sin \frac{x}{2} \text { is positive }\right]$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{1}{\sqrt{5}}\right)}=2$
Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5},$ and $2.$
રેડિયન માપ શોધો : $-47^{\circ} 30^{\prime}$
સાબિત કરો કે : $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
$2({\sin ^6}\theta + {\cos ^6}\theta ) - 3({\sin ^4}\theta + {\cos ^4}\theta ) + 1 =$
$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ =
જો $x + \frac{1}{x} = 2\cos \alpha $, તો ${x^n} + \frac{1}{{{x^n}}} = $