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A geometric proof of trigonometric derivatives

May 19, 2009 5 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.

geo_proof1

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.  geo_proof2The 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.