Why is $\sin(x) = \sin(180^{\circ}-x)$ I cannot seem to understand why this is true. Same for $\cos(x) = -\cos(180^{\circ}-x)$ and $\tan(x) = -\tan(180^{\circ}-x)$. Without the use of the compound angle formulas. 

Thanks
Hint  Rebember that 
$
\tan( 180^o \pm x  ) = \frac{\sin( 180^o \pm x)}{\cos(180^o\pm x)}.
$
Now draw the  trigonometric cycle and observes the measures of  sine and cosine.
Because, if you take $\cos(180°-x)$ you are effectively getting the opposite of $\cos(x)$. Due to $180°$ being halfway over the circle diameter. So if you do take the opposite of that $-\cos(180°-x)$ you get $x$ back, because you switch the sign.

Easy to see with this image: