Given: sin x = \(\frac{\sqrt{5}}{3}\) and \(\frac{\pi}{2}\) < x < π i.e, x lies in the Quadrant II .
To Find:
(i) sin\(\frac{x}{2}\)
(ii) cos\(\frac{x}{2}\)
(iii) tan\(\frac{x}{2}\)
Now, since sin x = \(\frac{\sqrt{5}}{3}\)
We know that cos x = \(\pm \sqrt{1-sin^2\text{x}}\)
since cos x is negative in II quadrant, hence cos x = \(-\frac{2}{3}\)
(i) sin\(\frac{x}{2}\)
Formula used:
Since sinx is positive in II quadrant, hence sin \(\frac{x}{2}=\sqrt{\frac{5}{6}}{}\)
(ii) cos\(\frac{x}{2}\)
Formula used:
since cosx is negative in II quadrant, hence cos\(\frac{x}{2}=-\frac{1}{\sqrt{6}}\)
(iii) tan\(\frac{x}{2}\)
Formula used:
Here, tan x is negative in II quadrant.