Question

If 0 ≤ x ≤ π and x lies in the IInd quadrant such that $\sin x=\frac{1}{4}$. Find the values of $\cos \frac{x}{2},\sin \frac{x}{2}\mathrm{and}\tan \frac{x}{2}$

Answer 1

Given:

$\sin x=\frac{1}{4}$

$$\begin{gathered} \therefore \sin x=\sqrt{1-{\cos }^{2}x} \\ \Rightarrow {\left(\frac{1}{4}\right)}^2=1-{\cos }^{2}x \\ \Rightarrow \frac{1}{16}-1=-{\cos }^{2}x \\ \Rightarrow \frac{15}{16}={\cos }^{2}x \\ \Rightarrow \cos x=\pm \frac{\sqrt{15}}{4} \end{gathered}$$

Since x lies in the 2nd quadrant, cosx is negative.

Thus,

$\cos x=-\frac{\sqrt{15}}{4}$

Now, using the identity $\cos x=2{\cos }^2\frac{x}{2}-1$, we get

$$\begin{gathered} -\frac{\sqrt{15}}{4}=2{\cos }^2\frac{x}{2}-1 \\ \Rightarrow -\frac{\sqrt{15}}{8}={\cos }^2\frac{x}{2}-\frac{1}{2} \\ \Rightarrow {\cos }^2\frac{x}{2}=\frac{4-\sqrt{15}}{8} \\ \Rightarrow \cos \frac{x}{2}=\pm \frac{4-\sqrt{15}}{8} \end{gathered}$$

Since x lies in the 2nd quadrant and $\frac{x}{2}$ lies in the 1st quadrant, $\cos \frac{x}{2}$ is positive.

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

Again,

$$\begin{gathered} \cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{\sqrt{15}}{4}={\left(\sqrt{{\displaystyle \frac{4-\sqrt{15}}{8}}}\right)}^2-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{\sqrt{15}}{4}={\displaystyle \frac{4-\sqrt{15}}{8}}-{\sin }^2\frac{x}{2} \\ \Rightarrow {\sin }^2\frac{x}{2}=\frac{4+\sqrt{15}}{8} \\ \Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{4+\sqrt{15}}{8}}=\sqrt{\frac{4+\sqrt{15}}{8}} \end{gathered}$$

Now,

$$\begin{gathered} \tan \frac{x}{2}=\frac{\sin {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}} \\ =\frac{\sqrt{{\displaystyle \frac{4+\sqrt{15}}{8}}}}{\sqrt{{\displaystyle \frac{4-\sqrt{15}}{8}}}}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}} \\ =\sqrt{\frac{\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)}{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}} \\ =\frac{4+\sqrt{15}}{4^2-{\left(\sqrt{15}\right)}^2}=\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15} \end{gathered}$$