Home > Justincase > A geometric proof of trigonometric derivatives

## A geometric proof of trigonometric derivatives

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.

Categories: Justincase
1. March 14, 2013 at 12:34 pm

Nice to get to know your all posts and surfing inside!

• March 15, 2013 at 1:07 am

Thank you.. It is also nice to read yours. Quite motivational.

-salam hangat dari Tangerang-

2. May 3, 2013 at 3:28 am

I do accept as true with all the ideas you’ve presented in your post. They are very convincing and will certainly work. Nonetheless, the posts are too quick for newbies. May just you please extend them a bit from subsequent time? Thanks for the post.

3. May 2, 2017 at 1:15 am

I think your second-listed identity should read $\cos\alpha-\cos\beta = -2 \sin\left(\frac{\alpha-\beta}{2}\right) \sin\left(\frac{\alpha+\beta}{2}\right)$

• May 2, 2017 at 1:34 am

Thanks for the comment. I edited it.