How does integrating trigonometric functions work?

Question

I tried searching Google but all I get is what integrating each trigonometric function equals to. Nothing tells me why.

I can sort of see why:

${d\over dx} \sin(x) = \cos(x)$

And...

${d\over dx} \cos(x) = -\sin(x)$

Purely by looking at the two graphs. But I want to know the maths behind it. How does integrating both functions work? Why the results?

I've also seen from Wolfram Alpha that trigonometric functions can be written as series, but is there any article I can read that explains how these series are derived? I can't seem to find any.

UPDATE: Wow. Thank you everyone for sharing your knowledge. You have no idea how much you've helped :).

Consider googling for Proofs of the derivative of cosx and sinx as it might help make clear why the integrations work the way they do. — patrickh, Dec 29 '15 at 02:10
@math635 you need the notion of derivative to define series expansion... — gt6989b, Dec 29 '15 at 02:10
@math635 - I've seen the series expansions for the trigonometric functions, but can't find articles on how they're derived. — Max Echendu, Dec 29 '15 at 02:11
@patrickh - I will do...I think my problem is, I'm just not good with Google haha. I never know what to type :( — Max Echendu, Dec 29 '15 at 02:13
Your title only mentions integrating, but the question itself expresses some doubt about differentiating. The answers to this question might help: "Intuitive understanding of the derivatives of $\sin x$ and $\cos x$". — Blue, Dec 29 '15 at 02:37
A textbook on the Calculus should have explantations of these formulas. Formulas for derivatives of trig functions. Taylor series. Other types of series. But this is too broad to include in an answer here. — GEdgar, Dec 29 '15 at 02:50

gt6989b · Accepted Answer · 2015-12-29T13:09:31.307

$$ \begin{split} \frac{d [\sin x]}{dx} &= \lim_{h \to 0^+} \frac{\sin(x+h) - \sin x}{h} \\ &= \lim_{h \to 0^+} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ &= \lim_{h \to 0^+} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h} \\ &= \sin x \left(\lim_{h \to 0^+} \frac{\cos h - 1}{h} \right) + \cos x \left( \lim_{h \to 0^+} \frac{\sin h}{h} \right)\\ &= \sin x \cdot 0 + \cos x \cdot 1 \\ &= \cos x. \end{split} $$

UPDATE To finish this off you need a proof that $$\lim_{h \to 0^+} \frac{\cos h - 1}{h} = 0 \quad \text{and} \quad \lim_{h \to 0^+} \frac{\sin h}{h} = 1.$$

Here is a creative way to show the first limit assuming the second limit. Note that $$\lim_{h \to 0^+} \frac{\cos h -1}{h} \times \frac{\cos h +1}{h} = \lim_{h \to 0^+} \frac{\cos^2 h -1}{h} = \lim_{h \to 0^+} \frac{-\sin^2 h}{h^2} = -1.$$ Since $$\lim_{h \to 0^+} \frac{\cos h +1}{h} = +\infty$$ we must conclude that $$\lim_{h \to 0^+} \frac{\cos h -1}{h} = 0.$$

Now the only thing left is to show the second limit, which is a geometric proof to establish the inequality $$\cos h \le \frac{\sin h}{h} \le 1,$$ and then using the Squeezing Theorem. You can see the details, for example, here.

This is REALLY clear... more so than I would have hoped :) thank you. — Max Echendu, Dec 29 '15 at 02:21
Question... is (cos(h)-1)\h) undefined when h=0?... and isn't (sin(h)/h) the same as (cos(h)-1)\h) when h=0? — Max Echendu, Dec 29 '15 at 02:31
You have to show that $(\cos h -1)/h \to 0$. The easiest way is with the half-angle cos formula combined with $\sin h/h$ \to 1$. — marty cohen, Dec 29 '15 at 02:46

score 3 · Answer 2 · 2015-12-29T02:46:05.683

3

I'll assume you're aware that $\frac d{dx} e^{bx} = be^{bx}$ where $b \in \Bbb R$. It turns out that this also holds when $b$ is a complex number. Using this and the definition of the trig functions in terms of the exponential function it's easy to see that

$$\begin{align}\frac d{dx}\cos(x) &= \frac d{dx}\left(\frac{e^{ix}+e^{-ix}}{2}\right) \\ &= \frac 12\left(\frac d{dx}e^{ix}+\frac d{dx}e^{-ix}\right) \\ &= \frac 12\left(ie^{ix}-ie^{-ix}\right) \\ &= \frac 12\left(\frac iiie^{ix}-\frac iiie^{-ix}\right) \\ &= -\frac{e^{ix}-e^{-ix}}{2i}\ \\ &= -\sin(x)\end{align}$$

edited Dec 29 '15 at 02:46

answered Dec 29 '15 at 02:38

Oh now this looks good. Thank you so much :). Do you know any article I can look at regarding the series expansions of sine and cosine? – Max Echendu Dec 29 '15 at 02:44
1

Any book on infinite series should cover that. – Dec 29 '15 at 02:47
Alright thank you for your help :). Much appreciated. – Max Echendu Dec 29 '15 at 02:48
Usually Euler's formula requires using the Taylor expansion of the sine and cosine functions to prove, which first requires knowing their derivatives, so isn't this cyclic? Can one show Euler's formula without knowing the trigonometric derivatives? – Brevan Ellefsen Dec 29 '15 at 03:04
I'm simply defining $\sin$ and $\cos$ in terms of $\exp$. Of course that's a bit of a cheat, but OP will learn more about this later anyway. – Dec 29 '15 at 03:09

score 1 · Answer 3 · answered Dec 29 '15 at 02:56

The most important limit formula for trig functions is $\lim_{h \to 0} \frac{\sin h}{h} = 1$. You can convince yourself of this by looking at a triangle with angle $h$ in a circle of radius $1$. The best way to show this analytically is probably to use the power series, but the diagram of the small angle in the circle is where it all comes from.

Once you have this, the formulas for the derivatives of the trig functions readily follow.

Another result is that $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0 $. But this follows from the $\sin$ limit combined with $\cos h - 1 =-2\sin^2(h/2) $ so $\lim_{h \to 0} \frac{\cos h - 1}{h} =\lim_{h \to 0} \frac{-2\sin^2(h/2)}{h} =\lim_{h \to 0} \sin(h/2)\frac{-\sin(h/2)}{h/2} = 0 $ since the right term goes to $-1$ and the left term goes to zero.

That helped clear things up. Thank you :). – Max Echendu Dec 29 '15 at 03:02 — Max Echendu, Dec 29 '15 at 03:02

How does integrating trigonometric functions work?

3 Answers3