Trigonometric Functions

1. We will give (but not prove) Euler’s identity. From Euler’s equation we derive trigonometric functions. We postpone the proof for Euler’s equation until some future post, when we discuss Taylor series.

2. So we have a circle described by the equation
(1)\displaystyle x^{2}+y^{2}=1
where we set the radius to unity (i.e., r=1). Great. What’s its solution in parametric form?

We can use trigonometric functions writing
(2)\displaystyle x(t)=\cos(t) and \displaystyle y(t)=\sin(t).
This is taught to high school kids taking trigonometry. But I never took trigonometry! The only way I learned this stuff was through Euler’s equation. For a graphical idea of what’s going on, consider the following doodle:

Suppose we wrote any pair (x,y)\in\mathbb{R}^{2} as a complex number x+\mathrm{i}y. Then we can work with the complex plane (denoted \mathbb{C}). So what? Well, we have this useful formula:
(3)\displaystyle \exp(\mathrm{i}t)=\mathrm{e}^{\mathrm{i}t}=\cos(t)+\mathrm{i}\sin(t).
We should observe this traces out the unit circle in the complex plane. But we can recover all the trig identities this way!

Remark (Equivalence of \mathbb{R}^{2} and \mathbb{C}). It may seem strange to say “\mathbb{R}^{2} is equivalent to \mathbb{C}“, BUT think about it: our interests lie in the geometry, not the number system.

Complex numbers have two components: the real part and the imaginary part. Ordered pairs have two components too. Thus this component-wise process relating ordered pairs and complex numbers…is kosher.

Moreover, the identification of (x,y) with x+\mathrm{i}y is a one-to-one correspondence. Each distinct complex number gives us an ordered pair, and each ordered pair gives us a distinct complex number. So this process is fairly “sane”.

3. We should first rewrite sine and cosine in terms of \mathrm{e}^{\mathrm{i}t}. Well, we see that
(4)\displaystyle \exp(\mathrm{i}t)+\exp(-\mathrm{i}t)=2\cos(t).
Thus we have
(5a)\displaystyle \cos(t):=\frac{\exp(\mathrm{i}t)+\exp(-\mathrm{i}t)}{2}.
This is cosine.

Sine follows from similar reasoning. We simply have
(5b)\displaystyle \sin(t):=\frac{\exp(\mathrm{i}t)-\exp(-\mathrm{i}t)}{2\mathrm{i}}.
From Equations (5a) and (5b) we can derive everything else in trigonometry.

Exercise 1. What is \sin^{2}(t) in terms of \sin(t) and \cos(t)?

Exercise 2. What is \tan(t) using \exp(\mathrm{i}t)?

Exercise 3. What is \exp(\mathrm{i}\pi)?

Exercise 4. Calculate \sin(a+b) in terms of \sin(a), \cos(a), \sin(b), \cos(b).

Exercise 5. Calculate \cos(a+b) in terms of \sin(a), \cos(a), \sin(b), \cos(b).

Exercise 6. Using Equations (5a) and (5b), what is \sin^{2}(t)+\cos^{2}(t)= ???

Remark. Exercise 6 helps us a lot, because whenever we run into an integral of the form \int\sqrt{1-x^{2}}\,\mathrm{d}x, we can use substitution and rewrite it with trigonometric functions!

Exercise 7 [Math.SE]. Prove \displaystyle\sin^{3}(x)=\frac{3}{4}\sin(x) - \frac{1}{4}\sin(3x)

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Special Functions, Trigonometric Functions. Bookmark the permalink.

2 Responses to Trigonometric Functions

  1. Pingback: Hyperbolic Functions | My Math Blog

  2. Pingback: Taylor Series | My Math Blog

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