Understanding integrals and derivatives of trigonometric functions can be tough. For example, You may have been told that $\dfrac{d}{dx} \sin(x) = \cos(x)$, but not given a good explanation for it. In this article, we will show you with proof how to find the derivatives and integrals of $\cos x$ and $\sin x$ functions.
Derivatives of \(\sin(x)\) and \(\cos(x)\)
To find the derivatives of $\sin x$ and $\cos x$, we need two identities: $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$, and $\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x} = 0$.
Now, we can use the definition of a derivative on $\sin x$:
\begin{align*}\dfrac{d}{dx} \sin x &= \displaystyle \lim_{h\to 0} \frac{\sin(x+h) – \sin(x)}{h}\\ &= \frac{\sin x \cos h-\cos x \sin h-\sin x}{h}\\ &= \cos x\bigg(\lim_{h\to0} \frac{\sin h}{h}\bigg) + \sin x\bigg(\lim_{h\to0} \frac{1-\cos h}{h}\bigg)\\ &= \cos x \end{align*}
We can do the same for $\cos x$:
\begin{align*}\dfrac{d}{dx} \cos x &= \displaystyle \lim_{h\to 0} \frac{\cos(x+h) – \cos(x)}{h}\\ &= \frac{\cos x \cos h-\sin x \sin h-\cos x}{h}\\ &= \cos x\bigg(\lim_{h\to0} \frac{\cos h-1}{h}\bigg) – \sin x\bigg(\lim_{h\to0} \frac{\sin h}{h}\bigg)\\ &= -\sin x\end{align*}
Integrals of \(\sin(x)\) and \(\cos(x)\)
At this point, we can integrate the two functions, since we know their derivatives. For example, to find the integral of $\sin x$, we take integrate both sides of
\begin{align*}\dfrac{d}{dx} \cos x& = -\sin x\\ \cos x + C& = \int -\sin x\,dx\\ -\cos x + C& =\int \sin x\,dx\end{align*}
Similarly, $\cos x$
\begin{align*} \dfrac{d}{dx} \sin x &= \cos x\\\sin x + C &= \int \cos x\,dx\end{align*}
Conclusion
You now know the proofs for both integrals and derivatives of $\sin x$ and $\cos x$. We hope that now you feel more comfortable using this knowledge in your math classes.
