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Prove formulas for the nth derivative of sin and cos

by RoRi

Let

    \[ f(x) = \cos x, \qquad g(x) = \sin x. \]

Prove that

    \[ f^{(n)} (x) = \cos \left( x + \frac{1}{2} n \pi \right), \qquad g^{(n)}(x) = \sin \left( x + \frac{1}{2}n \pi \right).\]

Proof. The proof is by induction. For n = 1 we have

    \[ f'(x) = -\sin x = \cos \left ( x+ \frac{1}{2} \pi \right), \qquad g'(x) = \cos x = \sin \left( x + \frac{1}{2} \pi \right). \]

These equalities follow from the co-relations of sine and cosine (Theorem 2.3 part (d) on page 96 of Apostol). Thus, the formulas are true for the case n = 1. Assume then that they are true for some n = k \in \mathbb{Z}_{>0}. For f(x) we then have

    \begin{align*} f^{(k+1)}(x) &= \left( f^{(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) &(\text{Co-relations}) \\ &= \cos \left( x + \frac{1}{2} (k+1) \pi \right). \end{align*}

Similarly, for g(x) we have

    \begin{align*} g^{(k+1)}(x) &= \left( g^{(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) &(\text{Co-relations})\\ &= \sin \left( x + \frac{1}{2} (k+1) \pi \right). \end{align*}

Therefore, the theorem follows by induction for all positive integers. \qquad \blacksquare

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