The Derivatives of the Sine and Cosine Functions

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created September 29, 2025

We shall prove that computing the derivatives of the \( \sin() \) and \( \cos() \) functions is independent of the Pythagorean Theorem and the Pythagorean Identity.

The derivative of \( \sin() \) is computed as follows: $$ \begin{eqnarray*} \frac{d \sin(x)}{dx} &=& \lim_{h\rightarrow 0}\frac{\sin(x+h)-sin(x)}{h} \\ &=& \lim_{h\rightarrow 0} \frac{2\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \\ &=& \left[ \lim_{h\rightarrow 0}\cos\left( \frac{2x+h}{2}\right)\right] \cdot \left[ \lim_{h\rightarrow 0}\frac{\sin(h/2)}{(h/2)} \right] \\ &=& \cos(x)\cdot 1 = \cos(x) \end{eqnarray*} $$ The above used the Sum-to-Product identity for the sine function and the fac \( \lim_{h\rightarrow 0}\frac{\sin(h)}{h}=0 \) proved here.

Because \( \cos(x) = \sin(\pi/2 - x ) \), by the Charin Rule we have \[ \frac{d\cos(x)}{dx} = \frac{d\sin(\pi/2-x)}{dx} =\cos(\pi/2-x)\frac{d(\pi/2-x)}{dx} = -\cos(\pi/2-x) = -\sin(x) \]

Consequently, computing the derivatives of the sine and cosine functions is independent of the Pythagorean Theorem and the Pythagorean Identity.

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