Answer
The correct use of the Pythagorean Identity:
$sin^2\theta+cos^2\theta=1$
$1-cos^2\theta=sin^2\theta$
If $\theta=180°$:
$1-cos~180°=1-(-1)=2\ne sin~180°=0$
Work Step by Step
$sin^2\theta+cos^2\theta=1$
$sin^2\theta=1-cos^2\theta$
$sin^2\theta=(1-cos~\theta)(1+cos~\theta)$
$1-cos~\theta=\frac{sin^2\theta}{1+cos~\theta}$
or just square both sides:
$(1-cos~\theta)^2\ne sin^2\theta$
$1-2cos~\theta+cos^2~\theta\ne sin^2\theta$
