The derivative of sin(x)

We know that we can use the limit definition

    \[ \frac{d}{dx}\sin(x)=\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h} \]

to compute the derivative of \sin(x) “from first principles”. And this limit can be evaluated analytically in several ways. We could for example use the addition law for sine and write the limit as

    \[ \lim_{h\to 0}\left(\frac{\cos(h)-1}{h}\sin(x)+\frac{\sin(h)}{h}\cos(x)\right) \]

and then compute the individual limits. This is where that fine old mathematical chestnut

    \[ \lim_{h\to 0}\frac{\sin(h)}{h}=1 \]


Or we could use the sine difference formula

    \[ \sin(X)-\sin(Y)=2\cos\left(\frac{X+Y}{2}\right)\sin\left(\frac{X-Y}{2}\right) \]

and so write the limit as

    \[ \lim_{h\to 0}\frac{2\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)}{h} \]

Neither of these approaches is wrong, but as teaching tools they leave a lot to be desired. They are very abstract, and several removes from the geometric notion of the derivative as a rate of change, or of the gradient of a tangent. Beginning students, especially those with weak backgrounds, are likely to be bemused.

So here is a geometric approach. I discovered that it is in fact very well known (well, something so simple was bound to be), but in fact I’d never come across this until today. So here it is.

Start with \sin(x) and \sin(x+h) on a diagram:

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Then \sin(x+h)-\sin(x) is the vertical side of the small blue triangle in this next diagram:

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Since h is very small, the hypotenuse of this triangle will be very close to the arc length of the circle, which is also h, since we are working in radians. Thus the hypotenuse can be considered to be h.

Now let’s look at a few angles. First the outer angles in the triangle created by h will be equal, since the triangle is isosceles (the two long sides are both radii of the circle):

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Both these angles will be equal to

    \[ \frac{\pi-h}{2}=\frac{\pi}{2}-\frac{h}{2}. \]

Since the upper green angle in the first diagram is

    \[ \frac{\pi}{2}-(x+h) \]

the upper angle in the blue triangle will be

    \[ \frac{\pi}{2}-\frac{h}{2}-\left(\frac{\pi}{2}-(x+h)\right) \]

which simplifies to

    \[ x+\frac{h}{2}. \]

Just to make this quite clear, here’s the top point:

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We have

    \[ A=\frac{\pi}{2}-(x+h),\qquad A+B=\frac{\pi}{2}-\frac{h}{2} \]

and so

    \[ B=x+\frac{h}{2} \]

as we found above.

So, the blue triangle has hypotenuse h, vertical side \sin(x+h)-\sin(x) and upper angle x+\frac{h}{2}:

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This means that, for small h:

    \[ \sin(x+h)-\sin(x)=h\cos\left(x+\frac{h}{2}\right) \]

or alternatively that

    \[ \frac{\sin(x+h)-\sin(x)}{h}=\cos\left(x+\frac{h}{2}\right) \]

Taking the limit of both sides as h tends to zero:

    \[ \frac{d}{dx}\sin(x)=\cos(x). \]



Note that the base of the blue triangle, checking with the first diagram, is

    \[ \cos(x)-\cos(x+h). \]

This means that

    \[ \frac{\cos(x)-\cos(x+h)}{h}=\sin\left(x+\frac{h}{2}\right) \]


    \[ \frac{\cos(x+h)-\cos(x)}{h}=-\sin\left(x+\frac{h}{2}\right) \]

from which, by taking limits of both sides as h tends to zero, we obtain

    \[ \frac{d}{dx}\cos(x)=-\sin(x). \]

3 thoughts on “The derivative of sin(x)

  1. Excellent! I once tried to find a geometric explanation while in the shower, but it didn’t seem trivial . This is nice and makes sense.
    Note, there seems to be a typo on your line for lim h-> 0 sin h/h

  2. Excellent geometric proof! I like to dabble in these myself, as I’m just learning about Calculus in school at the moment. Great to see someone dissecting these, very inspiring.

  3. I had what I thought was a nice realization the last time I taught the derivatives of trig functions, and it goes something like this:

    1. The algebra we go through in the limit calculations for the derivative of either sin(x) or cos(x), using the angle-addition formula and rearranging, accomplishes one thing: reducing these limits to the computation of the special limits sin(h)/h or (cos(h)-1)/h as h goes to 0.

    2. Those two limits are just exactly the limits you would compute to evaluate the particular values sin'(0) or cos'(0).

    3. If we know anything about the cosine function, it’s that it has a maximum at 0. Thus cos'(0) must be zero, without doing any calculation.

    4. Thus we can complete the differentiation of the cosine function entirely without even the difficulty of going through the geometric arguments involving triangles inscribed in circles.

    Kind of neat, I thought! Also it was nice to realize, and point out, the common mathematical theme of reducing a general case to an easier-to-solve particular case.

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