Integrals and derivatives of trigonometric functions can be tricky. You may have been told in your math class that $\dfrac{d}{dx} \sin(x) = \cos(x)$, but not given a good explanation for it. In this article, we will show you the derivatives and integrals of all the trigonometric functions while also proving that they are true.
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 $\sin x$: $\dfrac{d}{dx} \sin x = \displaystyle \lim_{h\to 0} \frac{\sin(x+h) – \sin(x)}{h} = \frac{\sin x \cos h-\cos x \sin h-\sin x}{h} = \cos x(\lim_{h\to0} \frac{\sin h}{h}) + \sin x(\lim_{h\to0} \frac{1-\cos h}{h}) = \cos x$.
We can do the same for $\cos x$: $\dfrac{d}{dx} \cos x = \displaystyle \lim_{h\to 0} \frac{\cos(x+h) – \cos(x)}{h} = \frac{\cos x \cos h-\sin x \sin h-\cos x}{h} = \cos x(\lim_{h\to0} \frac{\cos h-1}{h}) – \sin x(\lim_{h\to0} \frac{\sin h}{h}) = -\sin x$.
Integrals of \(\sin(x)\) and \(\cos(x)\)
Once we have the derivatives, the integrals will be quite easy to do. For the integral of $\sin x$, we simply take integral of both sides of the following: $\dfrac{d}{dx} \cos x = -\sin x \to \cos x + C =\displaystyle \int -\sin x\,dx \to -\cos x + C =\displaystyle \int \sin x\,dx$. It is the same for $\cos x$: $\dfrac{d}{dx} \sin x = \cos x \to \sin x + C = \displaystyle\int \cos x\,dx$.
Conclusion
You now know how to take the derivatives and integrals of $\sin x$ and $\cos x$, and, even better, you know the underlying logic, so if you ever forget the derivatives or integrals, you will remember how to find them again.
