Prove $\sin^{(k)}{x}=\sin{\left(x+\frac{k\pi}{2}\right)}$

Question

That means that the k-the derivative of $\sin{x}$ can be expressed this way. This seems to be a basic result, but I cannot find any reference to it and cannot see how to prove it on my own. Any hint?

See also : http://math.stackexchange.com/questions/1752455/100-th-derivative-of-the-function-fx-ex-cosx/1752462#1752462 — lab bhattacharjee, Mar 25 '17 at 19:00

score 1 · Accepted Answer · answered Mar 25 '17 at 19:00

1

Hint. From De Moivre's formula, $$ e^{ix}=\cos x+i\sin x, $$ one may differentiate $k$ times to get $$ i^k \cdot e^{ix}=\cos^{(k)} x+i\sin^{(k)} x, $$ then take the imaginary part, for $x \in \mathbb{R}$, noticing that $$ i=e^{i \pi/2}. $$

answered Mar 25 '17 at 19:00

Olivier Oloa

120,989

but what to do with the cos? Don't see it... – Averroes2 Mar 25 '17 at 19:03
You also get a formula for $\cos^{(k)} x$... – Olivier Oloa Mar 25 '17 at 19:04
Observe that $i^k \cdot e^{ix}=e^{i(x+k\pi/2)}$. – Olivier Oloa Mar 25 '17 at 19:06

score 0 · Answer 2 · answered Mar 25 '17 at 19:08

0

$$\sin'(x) = \cos(x) = \sin(x+\frac\pi 2) $$ Now use induction on $k$ to find the higher derivatives.

answered Mar 25 '17 at 19:08

hmakholm left over Monica

286,031

Prove $\sin^{(k)}{x}=\sin{\left(x+\frac{k\pi}{2}\right)}$

2 Answers2