cos(pi-o)+ sin(pi/2 +0)=0 verifying the identity

Question

anonymous · Answer

theta?

anonymous · Answer

$$cos(\pi-\theta)+sin({\pi\over2}+\theta)=0~?$$

whpalmer4 · Answer

Think about what happens to the value of $\cos \theta$ if you add $\pi$ to $\theta$.  It flops over to the diagonally opposite spot on the unit circle, because the period of the sin and cos functions is $2\pi$, so adding $\pi$ takes you halfway around the circle from where you started.  Also, $$\cos \theta = \cos -\theta$$which means the left hand term is simply the cos function shifted in phase by $\pi$ radians or $180^\circ$.

Now, what does adding $\pi/2$ to the argument of the sin function do?  Does the graph of $\sin(\dfrac{\pi}{2}+\theta)$ resemble the graph of any other basic trig function?