# Proof: Proof: nth derivative of sin and cos

Let $$f(x) = \sin x$$ and $$g(x)=\cos x$$. The nth derivative of sine and co-sine is then given by

$f^{(n)}(x) = \sin\left(x+\frac{1}{2}n\pi\right),\; g^{(n)}(x) = \cos\left(x+\frac{1}{2}n\pi\right)$

## Proof by induction

For $$n=1$$ we know

$f'(x) = \cos x = \sin\left(x+\frac{1}{2}\pi\right),\; g'(x) = -\sin x = \cos\left(x+\frac{1}{2}\pi\right) \text{✓}$

Now assume that they are true for $$n=k\in\mathbb{Z}_{\geq 1}$$:

$\begin{array}{rl} f^{(k+1)} &= \left(f^{(k)}(x)\right)'\\ &= \left(\sin\left(x+\frac{1}{2}k\pi\right)\right)'\\ &= \cos\left(x+\frac{1}{2}k\pi\right)\\ &= \sin\left(\left(x+\frac{1}{2}k\pi\right)+\frac{1}{2}\pi\right)\\ &= \sin\left(x+\frac{1}{2}(k+1)\pi\right) \end{array}$

$\begin{array}{rl} g^{(k+1)} &= \left(g^{(k)}(x)\right)'\\ &= \left(\cos\left(x+\frac{1}{2}k\pi\right)\right)'\\ &= -\sin\left(x+\frac{1}{2}k\pi\right)\\ &= \cos\left(\left(x+\frac{1}{2}k\pi\right)+\frac{1}{2}\pi\right)\\ &= \cos\left(x+\frac{1}{2}(k+1)\pi\right) \end{array}$

By induction the claim holds for all positive integers. ■