Answer
a)-c) See proof
Work Step by Step
a)
If $t$ = $\tan(\frac{x}{2})$
then $\frac{x}{2}$ = $\tan^{-1}t$
$\cos\left(\frac{x}{2}\right)$ = $\frac{1}{\sqrt {1+t^2}}$
$\sin\left(\frac{x}{2}\right)$ = $\frac{t}{\sqrt {1+t^2}}$
b)
$\cos x$ = $\cos\left(2\cdot\frac{x}{2}\right)$ = $2\cos^2\left(\frac{x}{2}\right)-1$ = $2\left(\frac{1}{\sqrt {1+t^2}}\right)^2-1$ = $\frac{1-t^2}{1+t^2}$
$\sin x =\sin\left(2\cdot\dfrac{x}{2}\right)= 2\sin\frac{x}{2}\cos \frac{x}{2}$
$ = 2\cdot\frac{t}{\sqrt{1+t^2}}\cdot\frac{1}{\sqrt{1+t^2}}=\frac{2t}{1+t^2}$
c)
$\frac{x}{2}$ = $\tan^{-1}t$
$x$ = $2\tan^{-1}t$
$dx$ = $\frac{2}{1+t^2}dt$