Question

ok i really need help on this one!

Answer 1

\[\cos \left( \theta \right)= \cos^2\left( \frac \theta 2 \right)-\sin^2\left( \frac \theta 2 \right)\]\[=1 - 2\cdot \sin^2\left( \frac \theta 2 \right)\]
\[\implies 2\sin^2\left( \frac \theta 2 \right) = -\cos(\theta)+1\]\[\large \implies \sin\left( \frac \theta 2 \right) = \pm \sqrt{-\cos(\theta)+1 \over 2}\]

Answer 2

how would i put that with the letterS?

Answer 3

Did you follow how I got the formula ?

Answer 4

not really :(

Answer 5

Have you learned the the identity: \[\sin^2(x) + \cos^2(x) =1\]
and the formulas for \[\sin(\alpha+\beta)\ \&\ \cos(\alpha+\beta)\]

Answer 6

no

Answer 7

All right let's try this another way ... I drew the angle bisector for theta

Answer 8

Not sure if I can do it this way...
If the length of angle bisector is Y ... \[\sin \left( \frac \theta 2 \right)= \frac XY\]

We already know:\[\large Y^2= A^2+X^2\]