Why is the derivative of sine function is only cosine function?

Question

Is there any other way to prove it rather than proving it using the formula by definition ?

Answer 1

I'll present an "algebraic" proof of why the derivative of $\sin x$ in indeed "only $\cos x$." There are of course other, more geometric explanations that are more insightful as to why this intuitively makes sense.

By the definition of a derivative, \begin{equation*} \frac{\mathrm{d}}{\mathrm{d} x} \sin (x) = \lim\limits_{h \to 0} \frac{\sin (x+h) - \sin x}{h}. \end{equation*} Applying the trigonometric identity $\sin (a \pm b) = \sin a \cos b \pm \sin b \cos a$ gives us \begin{align*} \lim\limits_{h \to 0} \frac{\sin (x+h) - \sin x}{h} &= \lim\limits_{h \to 0} \frac{\sin x \cos h + \sin h \cos x -\sin x}{h} \\ &= \lim\limits_{h \to 0} \frac{\sin x (\cos h - 1) + \sin h \cos x}{h} \\ &= \lim\limits_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + \lim\limits_{h \to 0} \frac{\sin h \cos x}{h}. \end{align*} Note that $\sin x$ and $\cos x$ are constant when we evaluate the derivative, so \begin{equation*} \lim\limits_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + \lim\limits_{h \to 0} \frac{\sin h \cos x}{h} = \sin x \left(\lim\limits_{h \to 0} \frac{(\cos h - 1)}{h}\right) + \cos x \left(\lim\limits_{h \to 0} \frac{\sin h}{h}\right). \end{equation*} We can see that $\lim\limits_{h \to 0} \frac{(\cos h - 1)}{h}$ converges to $0$ and $\lim\limits_{h \to 0} \frac{\sin h}{h}$ converges to $1$. Applying these results to the previous equations gives us \begin{equation*} \frac{\mathrm{d}}{\mathrm{d} x} \sin (x) = (\sin x)(0) + (\cos x)(1) = \cos x. \end{equation*}

Note that similar logic can be used to derive the derivative of $\cos x$. These two results can then be used to derive the derivatives of other trigonometric functions such as $\tan x$ or $\csc x$.