# How to show $\cos(3\pi/2 - a) = -\sin(a)$?

$$\cos(3\pi/2 - a) = -\sin(a)$$

According to an answer to one of the questions in my book that's true, but come up with that? $$\cos(\pi/2 - a) = \sin(a),$$ but this is $3\pi/2$. Do you just ignore if $\pi$ is multiplied by anything or?

• That's because $\cos{(x\pm y)}=\cos{x}\cos{y}\mp\sin{x}\sin{y}$ – CIJ Mar 27 '15 at 5:34
• Does it help to picture a diameter drawn through the unit circle? If it crosses the circle at $(x,y)$, it should also cross at $(-x,-y)$ and the 2 radii are separated by 180 degrees. – Mike Mar 27 '15 at 5:44
Hint $cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ can you get it now.