Question

Let sin(t)= -4/5
pi 15 years ago

Answer 1

\[sin\frac{t}{2}=\pm \sqrt{\frac{1-cos(t)}{\sqrt2}}\]
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