Expressing $\sin\theta$ and $\cos\theta$ in terms of $\tan(\theta/2)$ This is the question:

Let $t = \tan(\theta/2)$. Express the following functions in terms of $t$:
  
  
*
  
*$\tan \theta$
  
*$\sin\theta$
  
*$\cos\theta$
  

I know that for part (1),
$$\tan\theta = \frac{2t}{1-t^2}$$
How do I get parts (2) and (3)?
If $\tan\theta = \frac{2t}{1-t^2}$ then I would multiply by $\cos \theta$ to get $$\sin \theta = 2t\frac{\cos \theta}{1-t^2}$$
but that doesn't look right.

$$z_{\frac{\theta}2} = \cos\frac{\theta}2 + i \sin \frac{\theta}2 = (1+t^2)^{-\frac12}(1+it)
$$
by deMoivre's theorem
$$
z_{\theta} = z_{\frac{\theta}2}^2=(1+t^2)^{-1}\left((1-t^2)+2it\right) = \cos\theta+i\sin\theta
$$
hence, equating real parts:
$$
\cos \theta = \frac{1-t^2}{1+t^2}
$$
and imaginary parts:
$$
\sin \theta = \frac{2t}{1+t^2}
$$
then taking the ratio:
$$
\tan \theta = \frac{2t}{1-t^2}
$$
