I was wondering why do we need to reflect the graph in vertical axis in the trigonometric identity:
$\cos(\frac{\pi}{2} -\theta) = \sin(\theta)$.
It seems that if we only translate the graph of $\cos(\theta)$ by $\frac{\pi}{2}$ it would take the same values as the graph of $\sin(\theta)$, so why do we reflect the graph by changing sign of $\theta$ to negative.
Hint:$$\cos(-x)=\cos(x)$$Cosine is even and that is one of the properties of an even function.
Note that $$\cos(\frac{\pi}2+\theta)\ne\sin(\theta)$$
It isn't that $\cos(\theta - \pi/2) = \sin(\theta)$ is more correct or less correct than $\cos(\pi/2 - \theta) = \sin(\theta)$: both are true. However, the second one also expresses the familiar concept of "complementary angles": in any right triangle, the sine of one of the two (non-right) angles is always equal to the cosine of the other angle. This follows from elementary opposite/adjacent/hypotenuse considerations, and thus makes it a more natural way to remember the identity.
Because of this naturality, the principle also generalizes much more readily than a simple shift would: $$\cos(\pi/2 - \theta) = \sin(\theta)$$ $$\sin(\pi/2 - \theta) = \cos(\theta)$$ $$\cot(\pi/2 - \theta) = \tan(\theta)$$ $$\tan(\pi/2 - \theta) = \cot(\theta)$$ $$\csc(\pi/2 - \theta) = \sec(\theta)$$ $$\sec(\pi/2 - \theta) = \csc(\theta)$$
If you tried to express these in terms of $\theta \pm \pi/2$ you'd have to fiddle with the signs in each case to get it right.
I don't understand why we change sign of $\theta$, it seems that $cos(\frac{\pi}{2}+\theta) = sin(\theta)$ should be correct.
– Pawel Dec 29 '15 at 13:17