Given $\tan A=\frac{-1}{2}$ and $\sin B=\frac{1}{3}$,where angle A and angle B are in the same quadrant, find the value of $\sin \frac{B}{2}$. Can anyone give me some hints on this?


Since $\sin B>0$ and $\tan A<0$, you know that $\pi/2<B<\pi$. Therefore $\cos B<0$. Now use the fact that $\pi/4<B/2<\pi/2$, so $\sin(B/2)>0$ and $$ \sin\frac{B}{2}= \sqrt{\frac{1-\cos B}{2}} $$ Since $$ \cos B=-\sqrt{1-\sin^2B}=-\sqrt{1-\frac{1}{9}}=-\frac{2}{3}\sqrt{2} $$ you have all the ingredients for doing the computation

$\displaystyle\sin\frac{B}{2}=\sqrt{\frac{3+2\sqrt{2}}{6}}=\sqrt{\frac{2+2\sqrt{2}+1}{6}}=\frac{\sqrt{2}+1}{\sqrt{6}}\approx 0.9856$

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    $\begingroup$ It's better show the way(formula) than showing the destination :) $\endgroup$ – lab bhattacharjee Feb 21 '16 at 16:15

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