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. ■