$\sin \frac{x}{2}, \cos \frac{x}{2}$ અને $\tan \frac{x}{2}$ ની કિંમતો શોધો.: $\cos x=-\frac{1}{3}, x$ એ બીજા ચરણમાં છે.

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Here, $x$ is in quadrant $III$.

i.e., $\pi < x < \frac{3 \pi}{2}$

$\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$

Therefore, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative, where $\sin \frac{x}{2}$ as is positive.

It is given that $\cos x=-\frac{1}{3}$

$\cos x=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}$

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

$\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \quad\left[\because \sin \frac{x}{2} \text { is positive }\right]$

$\therefore \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

Now $\cos x=2 \cos ^{2} \frac{x}{2}-1$

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

$\Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \quad\left[\because \cos \frac{x}{2} \text { is negative }\right]$

$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(\frac{-1}{\sqrt{3}}\right)}=-\sqrt{2}$

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are  $\frac{\sqrt{6}}{3}, \frac{-\sqrt{3}}{3},$ and $-\sqrt{2}.$

Similar Questions

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જો $x$ ની વાસ્તવિક કિમત માટે  $\cos \theta = x + \frac{1}{x},$ તો  . . .. 

જો ${\sin ^2}\theta = \frac{{{x^2} + {y^2} + 1}}{{2x}}$, તો $x$ એ ફરજિયાત  . . . હોવો જોઈએ. 

જો કાટકોણ ત્રિકોણમાં કર્ણની લંબાઈ કર્ણની સામેના શિરોબિંદુથી લંબ અંતર કરતાં $2 \sqrt 2$ ગણું હોય તો બીજા લઘુકોણોના માપ મેળવો . 

સાબિત કરો કે : $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$