Derivative of trigonometric functions

From Applied Science

When we have trigonometric functions, the derivatives are all related to the trigonometric identities. Let's plot sine, cosine and tangent on the same space:

Both the sine and the tangent change the sign of their respective rates of change when the function is zero. The point where the sine is maximum is also the point where the cosine is zero. The opposite is also true, where the cosine is maximum the sine is zero. The point where the cosine and the sine intercept each other is the angle [math]\displaystyle{ \pi / 4 }[/math] in the unit circle. The point where the tangent and the cosine intercept each other is the angle [math]\displaystyle{ 3 \pi / 4 }[/math].

Sine and cosine are identical functions with the exception that their respective roots differ by an angle of [math]\displaystyle{ \pi / 2 }[/math]. We have that [math]\displaystyle{ \sin(x + \pi / 2) = \cos(x) }[/math] and [math]\displaystyle{ cos(x - \pi / 2) = \sin(x) }[/math]. The fact that both functions can match each other is what we need to prove that cosine is the derivative of sine.

[math]\displaystyle{ \lim_{h \ \to \ 0} \sin(x) = \frac{\sin(x + h) - \sin(x)}{h} }[/math]

There is more than one trigonometric identity that can help here. I'm going to follow the identity that expresses a difference of sines as a product of sines and cosines. The idea is that sine and cosine "disappear" at certain angles, because we end up multiplying by zero or one depending on the angle:

[math]\displaystyle{ = \frac{2 \sin \left(\frac{h}{2}\right) \cos \left(\frac{2x + h}{2}\right)}{h} }[/math] (the limit of the product is the product of the limits)

[math]\displaystyle{ = 2 \lim_{h \ \to \ 0} \frac{\sin(h/2)}{h/2} \lim_{h \ \to \ 0} \cos \left(\frac{2x + h}{2}\right) \frac{1}{2} }[/math] (we can move a constant factor out of the limit and multiply the limit by the inverse of that constant to keep the same result)

[math]\displaystyle{ = 2 \frac{1}{2} \cos(x) }[/math] (the fundamental trigonometric limit is equal to one)

Note: one may have asked about the derivative of the product not being the product of the derivatives. What happened in the beginning is that we did begin with a derivative, but the sine was rewritten as a product of sine x cosine. We were not attempting to differentiate the product. It was a substitution, nothing more.

For the proof of [math]\displaystyle{ \cos'(x) = -\sin(x) }[/math] the idea is very much the same. We begin with a limit and have to use some trigonometric identity to reach the result.

For the proof of [math]\displaystyle{ \tan'(x) = \sec^2(x) }[/math] we can use the fact that [math]\displaystyle{ \tan(x) = \frac{\sin(x)}{\cos(x)} }[/math] and proceed from there.

All the proofs are nothing more than pure algebraic manipulations. All relating to the unit circle.

Links for the proofs

Retrieved from "https://www.0kdesign.org:443/appliedscience/index.php?title=Derivative_of_trigonometric_functions&oldid=2088"