Question

If $sinx=-\frac{1}{2},\frac{3\pi }{2}<x<2\pi ,$ find the values of $sin\frac{x}{2},cos\frac{x}{2}$ and $tan\frac{x}{2}.$

Or

If $tan\left(\pi cos\theta \right)=cot\left(\pi sin\theta \right),$ prove that $cos(\theta -\frac{\pi }{4})=\pm \frac{1}{2\sqrt{2}}.$

Answer 1

It is given that, x lies in IVth quadrant in which cos x is positive.

$\therefore sinx=-\frac{1}{2}$

$\Rightarrow cosx=+\sqrt{1-si{n}^{2}x}$

[in IVth quadrant, cos x is positive]

= $\sqrt{1-{(-\frac{1}{2})}^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

Now, $\frac{3\pi }{2}<x<2\pi$

$\Rightarrow \frac{3\pi }{4}<\frac{x}{2}<\pi$ [divide both sides by 2]

$\Rightarrow \frac{x}{2}$ lies in IInd quadrant.

$\Rightarrow sin\frac{x}{2}>0,cos\frac{x}{2}<0$ and $tan\frac{x}{2}<0...\left(i\right)$

Now, $sin\frac{x}{2}=\sqrt{\frac{1-cosx}{2}}$

$\Rightarrow sin\frac{x}{2}=\sqrt{\frac{1-\sqrt{3}/2}{2}}=\sqrt{\frac{2-\sqrt{3}}{{}^{.}4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$

$\Rightarrow cos\frac{x}{2}=-\sqrt{\frac{1+cosx}{2}}$ [from Eq.(i)]

$\Rightarrow cos\frac{x}{2}=-\sqrt{\frac{1+\sqrt{3}/2}{2}}=-\sqrt{\frac{2+\sqrt{3}}{4}}$

= $-\frac{\sqrt{2+\sqrt{3}}}{2}$

and $tan\frac{x}{2}=-\sqrt{\frac{1-cosx}{1+cosx}}$ [from Eq.(i)]

= $-\sqrt{\frac{1-\sqrt{3}/2}{1+\sqrt{3}/2}}=-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$

Or

We have, $tan\left(\pi cos\theta \right)=cot\left(\pi sin\theta \right)$

$\Rightarrow \frac{sin\left(\pi cos\theta \right)}{cos\left(\pi cos\theta \right)}=\frac{cos\left(\pi sin\theta \right)}{sin\left(\pi sin\theta \right)}$

$\Rightarrow sin\left(\pi cos\theta \right).sin\left(\pi sin\theta \right)$

= $cos\left(\pi cos\theta \right).cos\left(\pi sin\theta \right)$

$\Rightarrow cos\left(\pi cos\theta \right).cos\left(\pi sin\theta \right)$

$-sin\left(\pi cos\theta \right).sin\left(\pi sin\theta \right)=0$

$\Rightarrow cos(\pi cos\theta +\pi sin\theta )=0$

$[\because cos(A+B)=cosAcosB-sinAsinB]$

$\Rightarrow \pi cos\theta +\pi sin\theta =\pm \frac{\pi }{2}[\because cos(\pm \frac{\pi }{2})=0]$

$\Rightarrow cos\theta +sin\theta =\pm \frac{1}{2}$

On multiplying both sides by $\frac{1}{\sqrt{2}},$ we get

$\frac{1}{\sqrt{2}}cos\theta +\frac{1}{\sqrt{2}}sin\theta =\pm \frac{1}{2\sqrt{2}}$

$\Rightarrow cos\frac{\pi }{4}cos\theta +sin\frac{\pi }{4}sin\theta =\pm \frac{1}{2\sqrt{2}}$

$\therefore cos(\theta -\frac{\pi }{4})=\pm \frac{1}{2\sqrt{2}}$

$[\because cos(A-B)=cosAcosB+sinAsinB]$

Hence proved.