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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