# $\sin x$ shifts by $\frac{\pi}{2}$ are equivalent to its derivatives?

An interesting thing in trigonometry that I noticed is that $$\sin(x + \pi/2) = \cos x \\ \sin(x +\pi) = \cos(x + \pi/2) = -\sin x \\ \sin (x + 3\pi/2) = -\sin(x+ \pi/2) = -\cos x$$

However, upon learning calculus (basic derivatives), I noticed that this pattern is exhibited in the derivatives of $\sin x$ as well: $$\frac{d}{dx}[\sin x] = \cos x \\ \frac{d}{dx}[\cos x] = -\sin x \\ \frac{d}{dx}[-\sin x] = -\cos x$$

What is the reason for this? How is the shift of $\sin x$ by $\pi/2$ related to the derivatives of $\sin x$? Of course this would also imply that a shift of $-\pi/2$ would give the integral of the function from which it is shifted.

The derivative of $t \mapsto e^{it}$ is $t \mapsto i \ e^{it}$ and multiplying a complex number by $i$ is equivalent to a $\frac{\pi}{2}$-rotation.

EDIT: To see what happens about derivative of circular functions, you have to consider the complex plane.

$e^{it} = \cos(t) + i \sin(t)\$ hence$\ \frac{d}{dt}e^{it} =\frac{d}{dt} \cos(t) + i \frac{d}{dt}\sin(t) \ \ \ \ (i)$

Otherwise $\frac{d}{dt}e^{it} = i\ e^{it}$ and, since $i = e^{i \frac{\pi}{2}}$

$$i\ e^{it} = e^{i(t+\frac{\pi}{2})} = \cos(t+\frac{\pi}{2}) + i \sin(t+\frac{\pi}{2}) \ \ \ \ (ii)$$

So, by identifying real and imaginary parts between $(i)$ and $(ii)$, both $\sin$ and $\cos$ have a phase change of $\frac{\pi}{2}$ due to derivation.

• I'm afraid I don't understand your answer, or how it relates to my question. Could you please elaborate? Commented Feb 4, 2018 at 21:44
• How come the rotation acquired by multiplying by $i$ causes a phase shift in the real plane? Commented Feb 5, 2018 at 20:36