Differentials from the unit circle
We've seen how to work out differentials graphs and limits. Now let's see how the geometry of the unit circle gives us the differential of \(\sin\theta\) directly.
Take your time with each video. Pause, rewind, and think before moving on!
Dr Brian Brooks
Mathematics InSight
Here's a completely different way of thinking about the differential of sine. In a way it's much more direct, but it's not explored that often. Start with this video clip: it's the usual unit circle, but with an extra angle \(h\) added in:
What is the length of the pink vertical line segment in terms of \(\theta\) and \(h\)?
The radius of the circle is \(1\) so the pink segment length is \(\sin(\theta+h)\).
What is the length of the vertical blue segment?
The blue vertical segment length is the pink vertical minus the yellow vertical; that is \(\sin(\theta+h)-\sin\theta\).
What is the length of the pink arc?
Since the angle is in radians and the radius of the circle is \(1\), the measure of the angle is the arc length, so the arc length is \(h\).
What is the yellow shaded angle?
It's \(\theta\).
What is the blue shaded angle?
The angle between tangent and radius is \(90^\circ\), so the blue angle is \(90^\circ-\theta\).
Now watch what happens to the light green angle and to \(\alpha\) as \(h\) gets increasingly small
What does happen to the light green angle as \(h\) gets increasingly small? What does this mean for \(\alpha\)?
The light green angle gets increasingly small, which means that \(\alpha\) gets increasingly close to \(\theta\).
This time, watch what happens to the blue hypotenuse as \(h\) gets increasingly small
What does happen as \(h\) gets increasingly small?
The blue hypotenuse gets increasingly close to the arc length, \(h\).
What happens to \(\cos\alpha\) as \(h\) gets increasingly small?
What is \[\frac{\mathrm{d}}{\mathrm{d}\theta}\sin\theta\;?\]
\[\begin{align*} \frac{\mathrm{d}}{\mathrm{d}\theta}\cos\alpha&=\lim_{h\to 0}\frac{\sin(\theta+h)-\sin\theta}{\text{hypotenuse}}\\[12pt] &=\lim_{h\to 0}\to \frac{\sin(\theta+h)-\sin\theta}{h}\\[12pt] &=\lim_{h\to 0}\cos\alpha\\[12pt] &=\cos\theta \end{align*}\].
Well done!
You've seen how the geometry of the unit circle directly gives us
\[\frac{\mathrm{d}}{\mathrm{d}\theta}\sin\theta = \cos\theta\]without needing to expand \(\sin(\theta+h)\) using the addition formula.
Dr Brian Brooks
Mathematics InSight