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.

here is a way to see how and why you get the expressions for $\sin t, \cos t$ and $\tan t$ using the unit circle. it is easier to write $m = \tan(t/2).$  midpoint of the line segment(chord) connecting the points $ (x,y)=(\cos t, \sin t)$ and $(1, 0 )$ which is $((x+1)/2, y/2)$ on the the radius with a slope $m.$
now we have two equations for $x$ and $y.$
they are $$ x^2 + y^2 = 1, y = m(x+1)$$ solving the quadratic equation for $x$ we get $x = {1 - m^2 \over 1 + m^2}, -1.$ the answer $x = -1$ corresponds to $t = \pi, m = \infty$ and $y = {2m \over 1 + m^2}, 0.$ finally putting both cases we get $$ \cos t = {1 - m^2 \over 1 + m^2}, \sin t = {2m \over 1 + m^2}.$$ 
