## anonymous 5 years ago Let sin(t)= -4/5 pi<t<3pi/5 Find sin(t/2)

$sin\frac{t}{2}=\pm \sqrt{\frac{1-cos(t)}{\sqrt2}}$ so you need $cos(t)$ since $sin(t)=-\frac{4}{5}$ you know by pythagoras that $cos(t)=\pm \frac{3}{5}$ and since you are in quadrant III cosine is negative, so you get $\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{8}{10}}$ square root should be over the whole thing