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## A geometric proof of trigonometric derivatives

May 19, 2009 9 comments

As we already know, the first derivatives of $\sin(x)$ and $\cos(x)$ are $\cos(x)$ and $-\sin(x)$ respectively. The question is how do we prove it?

Since high school, the only proof that I knew is to use two of the trigonometric identities which are:

• $\sin \alpha - \sin \beta = 2 \sin (\frac{\alpha-\beta}{2}) \cos (\frac{\alpha+\beta}{2})$, and
• $\cos \alpha - \cos \beta = -2 \sin (\frac{\alpha-\beta}{2}) \sin (\frac{\alpha+\beta}{2})$.

Hence,

$\frac{d}{dx}\sin x = \displaystyle{\lim_{\Delta x \rightarrow 0} \frac{\sin(x+\Delta x)-\sin(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2 \sin(\frac{\Delta x}{2})\cos(\frac{2x+\Delta x}{2})=\cos x}$.

The derivative of $\cos x$ can also be derived using the second identity.

However, there is another proof, which is (I think) more elegant. It is using a geometric interpretation of $\sin x$ and $\cos x$. Let us draw a unit circle in $\mathbb{R}^2$.

Thus we have $y=\sin \theta$ and $x=\cos \theta$. We are looking for $\frac{dy}{d \theta}$ and$\frac{dx}{d \theta}$, however we are not going to discuss the first derivative of $\cos \theta$ as it can be done in the same way as $\sin \theta$.

$\displaystyle{\frac{d \sin \theta}{d \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\sin(\theta + \Delta \theta)-\sin(\theta)}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{y+\Delta y - y}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\Delta y}{\Delta \theta}}.$

We are now going to estimate $\Delta y$, see the second figure.  The following relationship applies:

$\sin \phi = \frac{\Delta y}{r}$. or $\Delta y = r \sin \phi$.

Thus,

$\displaystyle{\frac{d \sin \theta}{d \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\Delta y}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \sin \phi \frac{r}{\Delta \theta}}$,

where $r$ is the chord PQ and $\Delta \theta$ is the arc PQ. As $\Delta \theta$ goes to zero, the ratio of the arc PQ and the chord PQ tends to 1 and the chord PQ tends to become a tangent line of the unit circle at $(x,y)$, therefore $\sin \phi = \sin (\theta + \pi/2)$. Assuming that the limit of multiplication is the same as the multiplication of limit, we have the following:

$\frac{d \sin \theta}{d \theta}=\sin(\theta + \pi/2) = \cos(\theta)$.

As mentioned before, the first derivative of $\cos \theta$ can be done the same way.

Also, I should thank Pak Yudi Soeharyadi as I got the idea of this geometric proof from him.