这是一份umass麻省大学 MATH 103作业代写的成功案例
$\operatorname{proj}{o x} O P=\operatorname{proj}{o x} O A+\operatorname{proj}_{o x} A P .$
By the first projection theorem, this becomes:
$$
O P \cos (\alpha+\beta)=O A \cos \alpha+A P \cos \left(90^{\circ}+\alpha\right) .
$$
Or, since
$$
\cos \left(90^{\circ}+\alpha\right)=-\sin \alpha,
$$
we have:
$O P \cos (\alpha+\beta)=O A \cos \alpha-A P \sin \alpha$
Dividing by $O P$, we have:
$$
\cos (\alpha+\beta)=\cos \alpha\left(\frac{O A}{O P}\right)-\sin \alpha\left(\frac{A P}{O P}\right)
$$
Or, since
$$
\frac{O A}{O P}=\cos \beta
$$
and
$$
\frac{A P}{O P}=\sin \beta
$$
MATH104 COURSE NOTES ：
Hence we have:
$$
\cos \alpha=1-2 \sin ^{2} \frac{\alpha}{2}
$$
or:
$$
2 \sin ^{2} \frac{\alpha}{2}=1-\cos \alpha
$$
or:
$$
\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} .
$$
Or, finally,
$$
\sin \frac{\alpha}{2}=\pm \sqrt{\frac{1-\cos \alpha}{2}}
$$