Question

Prove that:

(i) If $cosx=-\frac{3}{5}$ and x lies in the IIIrd quadrant, find the values of $cos\frac{x}{2},sin\frac{x}{2}$ and sin 2x.

(ii) If $cosx=-\frac{3}{5}$ and x lies in the IInd quadrant, find the values of sin 2x and sin $\frac{x}{2}$

Answer 1

(i) Since $cosx=-\frac{3}{5}=\frac{b}{h}$

$\Rightarrow$ b = 3, y = 5

$\Rightarrow$ p = 4

Now, x lies on third quad.

$\therefore$ sin 2x = 2 sin x. cos x

$=2.\left(\frac{-4}{5}\right).\left(\frac{-3}{5}\right)=\frac{24}{25}$

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

Which means $\frac{x}{2}$ lies in second quadrant.

So, $cos\frac{x}{2}=\sqrt{\frac{1+cosx}{2}}$

$[\because 1+cos2\theta =2co{s}^2\theta ]=\sqrt{\frac{1-\frac{3}{5}}{2}}=\frac{-1}{\sqrt{5}}$

[-ve sign because of second quad. where cos D is -ve]

Also,

$sin\frac{x}{2}=\frac{sinx}{2cos\frac{x}{2}}$

[$\because$ sin 2A = 2 sin A cos A]

$=\left(\frac{\frac{-4}{5}}{2\left(\frac{-1}{\sqrt{5}}\right)}\right)=\frac{2}{\sqrt{5}}$

(ii) $\because$ x lies $I{I}^{nd}$ quadrant.

$\Rightarrow \frac{\pi }{2}<x<\pi \Rightarrow \pi <2x<2\pi \Rightarrow \text{2x lies in}{1}^{st}quad.$

Also, $cosx=\frac{-3}{5}=\frac{b}{h}$

$\Rightarrow$ b = 3

h = 5

$\Rightarrow$ p = 4

So, $sinx=\frac{p}{h}=\frac{4}{5}$

$\therefore$ sin 2x = 2 sin x cos x

$=2.\frac{4}{5}.\left(\frac{-3}{5}\right)=\frac{-24}{25}sin\frac{x}{2}=\frac{sinx}{2cos\frac{x}{2}}or\sqrt{\frac{1-cosx}{2}}=\sqrt{\frac{1-(1-\frac{3}{5})}{2}}=\frac{2}{\sqrt{5}}$