Differentiating Trigonometric Functions

1. Introduction. So we have investigated several useful trig limits, lets consider the bigger problem: differentiating trigonometric functions.

We will use the following properties without proving them:
(1a)\displaystyle \cos(\varphi+\psi)=\cos(\varphi)\cos(\psi)-\sin(\varphi)\sin(\psi)
(1b)\displaystyle \sin(\varphi+\psi)=\cos(\varphi)\sin(\psi)+\sin(\varphi)\cos(\psi)
(1c)\displaystyle \sin(-\varphi)=-\sin(\varphi)
(1d)\displaystyle \cos(-\varphi)=\cos(\varphi)
Where we have \varphi, \psi be arbitrary real numbers.

2. Derivative of Cosine. We see from (1a) that
(2)\displaystyle \begin{array}{rl}\Delta\cos(\varphi)&=\cos(\varphi+\Delta\varphi)-\cos(\varphi)\\  &=\cos(\varphi)\cos(\Delta\varphi)-\sin(\varphi)\sin(\Delta\varphi)-\cos(\varphi)\\  &=\cos(\varphi)\bigl(\cos(\Delta\varphi)-1\bigr)-\sin(\varphi)\sin(\Delta\varphi)\end{array}
We divide both sides by \Delta\varphi obtaining
(3)\displaystyle \frac{\Delta\cos(\varphi)}{\Delta\varphi}=\cos(\varphi)\left(\frac{\cos(\Delta\varphi)-1}{\Delta\varphi}\right)-\sin(\varphi)\frac{\sin(\Delta\varphi)}{\Delta\varphi}.
Taking the limit as \Delta\varphi\to0, and recalling our useful trigonometric limits, we have the definition of the derivative for cosine
(4)\displaystyle \frac{\mathrm{d}\cos(\varphi)}{\mathrm{d}\varphi}=\cos(\varphi)\left(0\right)-\sin(\varphi)\cdot1=-\sin(\varphi).

3. Derivative of Sine. Likewise, using Equation (1b), we find
(5)\displaystyle \begin{array}{rl}  \Delta\sin(\varphi)&=\sin(\varphi+\Delta\varphi)-\sin(\varphi)\\  &=\cos(\varphi)\sin(\Delta\varphi)+\sin(\varphi)\cos(\Delta\varphi)-\sin(\varphi)\\  &=\cos(\varphi)\sin(\Delta\varphi)+\sin(\varphi)\bigl(\cos(\Delta\varphi)-1\bigr)\end{array}.
Dividing through by \Delta\varphi yields
(6)\displaystyle\frac{\Delta\sin(\varphi)}{\Delta\varphi}=\cos(\varphi)\left(\frac{\sin(\Delta\varphi)}{\Delta\varphi}\right)+\sin(\varphi)\left(\frac{\cos(\Delta\varphi)-1}{\Delta\varphi}\right)
Again, taking the limit \Delta\varphi\to 0 yields the derivative
(7)\displaystyle\frac{\mathrm{d}\sin(\varphi)}{\mathrm{d}\varphi}=\lim_{\Delta\varphi\to0}\frac{\Delta\sin(\varphi)}{\Delta\varphi}=\lim_{\Delta\varphi\to0}\cos(\varphi)\left(\frac{\sin(\Delta\varphi)}{\Delta\varphi}\right)+\sin(\varphi)\left(\frac{\cos(\Delta\varphi)-1}{\Delta\varphi}\right)
Using our beloved trig limits produces:
(8)\displaystyle\frac{\mathrm{d}\sin(\varphi)}{\mathrm{d}\varphi}=\cos(\varphi)\left(1\right)+\sin(\varphi)\left(0\right)
Thus we obtain
(9)\displaystyle\frac{\mathrm{d}\sin(\varphi)}{\mathrm{d}\varphi}=\cos(\varphi).

Exercise 1. Prove \displaystyle\frac{\mathrm{d}\tan(x)}{\mathrm{d}x}=\frac{1}{\cos^{2}(x)}.

Exercise 2. Recall the secant function is \sec(x)=1/\cos(x). What is its derivative?

Exercise 3. Recall the cosecant function is \csc(x)=1/\sin(x). What is its derivative?

Exercise 4. What is the derivative of \cos\bigl(\sin(x)\bigr) with respect to x?

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
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3 Responses to Differentiating Trigonometric Functions

  1. Pingback: Fundamental Theorem of Calculus | My Math Blog

  2. Pingback: [Index] Calculus of a Single Variable | My Math Blog

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