Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=c^2\Rightarrow a^2+25=676\Rightarrow a=\sqrt{651}\Rightarrow\tan(\frac{5}{\sqrt{651}}).$$ It turned out to be wrong. How to get the right answer?
 A: Let $\theta=\arcsin(5/13)$. Problem: find $t=\tan(\theta/2)$. Now
$$\sin\theta=\frac5{13}=\frac{2t}{1+t^2}.$$
This is a quadratic equation for $t$, etc.
A: I think you need to check you first formula. You can use $\arcsin (\frac {2x}{1+x^2})=2\arctan (x) $ and $\arctan (\frac {2x}{1-x^2})=2\arctan (x) $. Both of which can be proved using $x=\tan (t)$. I think you can continue from here.
A: The formula
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-x^2}{1+x^2}}
$$
is wrong. You were probably thinking to
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}
$$
but there's a better one that doesn't have any sign ambiguity:
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
If $x=\arcsin(5/13)$, then $\cos x>0$, so
$$
\cos x=\sqrt{1-\Bigl(\frac{5}{13}\Bigr)^2}=\frac{12}{13}
$$

As a side note: from
$$
\frac{1}{2}\arcsin\frac{5}{13}=\alpha
$$
you cannot argue that
$$
\sin\alpha=\frac{5}{26}
$$
You cannot bring $1/2$ inside the arcsine.
A: Let $\arcsin(\frac{5}{13})=x$, thus $\sin(x)=\frac{5}{13}$
So, this problem reduced to find value of $\tan(\frac{x}{2})=u$,
According to $\sin(x)=\frac{2u}{u^2+1}$ identity for $u=\tan(\frac{x}{2})$;
$\frac{2u}{u^2+1}=\frac{5}{13}$
$5*(u^2+1)=13*2u$
$5u^2+5=26u$
$5u^2-26u+5=0$
$(u-5)*(5u-1)=0$
Due to $\tan(x)>0$ constraint, $\tan(\frac{x}{2})$ must be equal to $\frac{1}{5}$.
A: A bisector divides opposite sides in the same ratio as adjacent sides. $ 5= 12/5 +13/5$
So by this property the answer is straightaway $1/5$.

EDIT1:

