$$\tan\left(\frac{1}{2}\arccos \frac{2}{5}\right)$$ How do we evaluate this by using trigonometric identity?
The options are $\dfrac{\sqrt{24}}{1}$, $\dfrac{\sqrt{21}}{7}$, $\dfrac{\sqrt{70}}{10}$, $\dfrac{2\sqrt{5}}{5}$, $\dfrac{3\sqrt{10}}{10}$.
I know that
$$\arccos \frac {2}{5} \neq \cos \frac {2}{5}$$
However, I don't have any idea about how to.
On
By the bisector theorem, it is $\frac{1}{7}\sqrt{21}=\sqrt{\frac{3}{7}}$:
More generally, for any $r\in[0,1]$ we have $$\tan\left(\frac{1}{2}\arccos r\right)=\sqrt{\frac{1-r}{1+r}}. $$
On
$$\tan\left(\frac{1}{2}\arccos \frac{2}{5}\right)= \tan(x) $$ Where $$ x=\frac{1}{2}\arccos \frac{2}{5}$$ Thus $$2x = \arccos\frac25$$ Which implies $$\cos(2x)=\frac25$$ Note that $$\cos(2x)= 1-2\sin^2(x)=\frac25 \implies \sin^2(x)=\frac3{10}$$
Thus $\cos^2(x)=\frac7{10}$ and $\tan^2(x) = \frac37.$
That implies that $$\tan(x)= \sqrt {\frac37}=.6546536707.$$
We have $$\tan\frac A2=\pm\sqrt{\frac{1-\cos A}{1+\cos A}}$$ so $$\tan\left(\frac12\arccos\frac25\right)=\pm\sqrt{\frac{1-2/5}{1+2/5}}=\pm\sqrt{\frac37}$$
Note that as $0<\arccos\dfrac25<\dfrac\pi2$ the positive root is taken so $$\boxed{\tan\left(\frac12\arccos\frac25\right)=\sqrt{\frac37}}$$