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.

Since $\tan(\theta) = \dfrac{2t}{1-t^2}$, draw a picture using the fact that $\tan(\theta) = \frac{\mbox{opposite}}{\mbox{adjacent}}$ in a right triangle.

From here, find the hypotenuse using the Pythagoren Theorem. Then, use the definition for $\sin(\theta)$ and $\cos(\theta)$ in a right triangle. 
