TITLE: How do I prove: $\cos (\theta + 90^\circ) \equiv - \sin \theta $
QUESTION [5 upvotes]: How do I go about proving this?
I know one method is:
$\eqalign{
   \cos (90^\circ + \theta ) &\equiv \cos90^\circ \cos\theta  - \sin90^\circ \sin\theta   \cr 
  &  \equiv (0)(\cos\theta ) - (1)(\sin \theta )  \cr 
  &  \equiv  - \sin \theta  \cr} $
I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.
Thanks alot.

REPLY [6 votes]: I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.

The picture represents the trigonometric unit circle. The angle $\theta=\angle AOP$ is represented in the first quadrant, but it is a generic orientate angle. $|AP|=\sin\theta$. $\angle AOQ=\theta+90^\circ$. The right triangle $[O,A,P]$ is similar to the right triangle $[O,B,Q]$, because $\angle BQO=\angle AOP=\theta$. Since $|OP|=|OQ|=1$, both triangles are congruent and $$\cos(\theta+90^\circ)=-|OB|=-|AP|=-\sin\theta.$$