# Sin X = 1/4, X In Quadrant Ii.Find The Value Of Sin X/2

Given:x =$$\frac{1}{4}$$

sin x $$\frac{1}{4}$$= x is is second quadrant

$$\Rightarrow \frac{\Pi}{2} < x < \Pi$$and cos is negative in second quadrant

Using $$1 – cos x = 2 sin^{2} \frac{x}{2}$$ $$\Rightarrow sin \frac{x}{2} = \pm \sqrt{\frac{1 – cos x}{2}}$$

We get,

$$\frac{x}{2} + \pm \sqrt{\frac{1 – (-\frac{\sqrt{15}}{4})}{2}} = \pm \sqrt{\frac{4 + \sqrt{15}}{8}}$$

As$$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\Pi}{4} < \frac{x}{2} < \frac{\Pi }{2}$$ and sin is positive in first quadrant.

Therefore,$$sin \frac{x}{2} = \frac{\sqrt{8 + 2\sqrt{15}}}{4}$$

Using $$1 + cos x = 2 cos^{2} \frac{x}{2} \Rightarrow cos \frac{x}{2} = \pm \sqrt{\frac{1 + cos x}{2}}$$

We get cos$$\frac{x}{2} = \pm \sqrt{\frac{1 + -(\frac{\sqrt{15}}{4})}{2}}$$ $$\Rightarrow \pm \sqrt{\frac{4 – \sqrt{15}}{8}}$$ $$\Rightarrow \pm \frac{\sqrt{8-2\sqrt{15}}}{4}$$

As $$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\Pi}{4} < \frac{x}{2} < \frac{\Pi }{2}$$ and cos is positive in first quadrant.

Therefore, $$cos \frac{x}{2} = \frac{\sqrt{8-2\sqrt{15}}}{4}$$

Using cos $$x = \frac{1 – tan^{2}\frac{x}{2}}{1 + tan^{2}\frac{x}{2}} \Rightarrow tan \frac{x}{2} = \pm \sqrt{\frac{1-cos x}{1 + cos x}}$$

we get tan $$\frac{x}{2} = \pm \sqrt{\frac{1 – (-\frac{\sqrt{15}}{4})}{1 + (-\frac{\sqrt{15}}{4})}} \Rightarrow = \pm \sqrt{\frac{4 + \sqrt{15}}{4 – \sqrt{15}}}$$

As$$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\Pi}{4} < \frac{x}{2} < \frac{\Pi }{2}$$and tan is positive in first quadrant.

Therefore, tan$$\frac{x}{2} = 4 + \sqrt{15}$$

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