The Structure Behind Trigonometric Functions

This entry is dedicated to my friend Bud in Milwaukee.

Sine, cosine, tangent, etc. are actually abbreviations for combinations of complex exponential functions:

\text{sin}(\theta) = \frac{1}{2i} (e^{i\theta} - e^{-i\theta})

\text{cos}(\theta) = \frac{1}{2} (e^{i\theta} + e^{-i\theta})

\text{tan}(\theta) = \frac{ e^{i\theta} - e^{-i\theta} }{ i (e^{i\theta} + e^{-i\theta}) }

It’s just easier to write out something like “cos”, instead of the complex exponential form. However, it is handy to know the complex exponential forms to derive trig identities on the fly. So how exactly are trig functions made of complex exponential functions?

In a previous entry, I briefly discuss how the exponential function relates angle to Cartesian coordinates. When you put an imaginary number in the exponential function it results in a complex number (a number with a real and an imaginary component). If you treat the real component like an x coordinate, and the imaginary component like a y coordinate, then any complex number can represent a point  on a 2-D plane.

To visualize what’s actually going on in terms of geometry, see the figure below. Imagine a point (pictured in red) that can travel along a unit circle (a circle with radius 1). You can track the point’s angular position \varphi along the circle, and you can also track its position in terms of x and y. There is a relationship between the two ways to track position, and this relationship can be expressed as:
e^{i\varphi} = x + iy
euler3
At this point, you may see that x = \text{cos}(\varphi) and y = \text{sin}(\varphi).
By definition, cosine is adjacent over hypotenuse h
\text{cos}(\varphi) = x/h
and sine is opposite over hypotenuse
\text{sin}(\varphi) = y/h
and since radius = h = 1, you get:
x = \text{cos}(\varphi) and y = \text{sin}(\varphi)

So e^{i\varphi} = x + iy is really:
e^{i\varphi} = \text{cos}(\varphi) + i \text{sin}(\varphi)

From this equation, you can get the equations listed at the beginning by adding or subtracting complex conjugates. For example:
Take the complex conjugate (which just means to change all the i to -i).
e^{-i\varphi} = \text{cos}(\varphi) - i \text{sin}(\varphi)
Now add it to the original to get:
e^{i\varphi} + e^{-i\varphi} = \text{cos}(\varphi) + i \text{sin}(\varphi) + \text{cos}(\varphi) - i \text{sin}(\varphi)
e^{i\varphi} + e^{-i\varphi} = 2 \text{cos}(\varphi)
Now divide by 2 and you have the identity of cosine shown at the beginning!
\text{cos}(\varphi) = \frac{1}{2} (e^{i\varphi} + e^{-i\varphi})

So how do we know that something like e^{i\varphi} breaks up into a complex number like x + iy? One way is through Taylor expansions. I won’t get into how Taylor expansions are derived, since it requires calculus. The only thing you need to know here is that a function can be expressed as an infinitely long polynomial. In the case of the exponential function:
e^{z} = 1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + ...
or to be complete, we can write it as a summation:
e^{z} = \sum\limits_{n=0}^{\infty} \frac{1}{n!} z^n

So if we replace the variable z with i \varphi we get:
e^{i\varphi} = 1 + i\varphi + \frac{1}{2}(i\varphi)^2 + \frac{1}{6}(i\varphi)^3 + \frac{1}{24}(i\varphi)^4 + ...
The terms in the polynomial with odd exponents will be imaginary, and the terms with even exponents will be real. Recall that:
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
So then we get:
e^{i\varphi} = 1 + i\varphi - \frac{1}{2}\varphi^2 - \frac{1}{6}i\varphi^3 + \frac{1}{24}\varphi^4 + ...
We can then group all the real terms together and all the imaginary terms together:
e^{i\varphi} = \left(1 - \frac{1}{2}\varphi^2 + \frac{1}{24}\varphi^4 -... \right) + i \left( \varphi - \frac{1}{6} \varphi^3 + ... \right)
or to be complete:
e^{i\varphi} = \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \varphi^{2n} + i \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \varphi^{2n+1}
So you can see that e^{i\varphi} indeed breaks up into a complex number. By definition, this is equivalent to:
e^{i\varphi} = \text{cos}(\varphi) + i \text{sin}(\varphi)

It is also worth noting that this gives us the Taylor series for sine and cosine:
\text{sin}(\varphi) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \varphi^{2n+1}

\text{cos}(\varphi) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \varphi^{2n}

For an example, let’s derive a trig identity using the complex exponential forms.
Let’s show that:
\text{sin}(\theta) \text{cos}(\theta) = \frac{1}{2} \text{sin}(2 \theta)

First write the complex exponential forms:
\text{sin}(\theta) = \frac{1}{2i} (e^{i\theta} - e^{-i\theta})

\text{cos}(\theta) = \frac{1}{2} (e^{i\theta} + e^{-i\theta})
Then multiply them:
\frac{1}{2i} (e^{i\theta} - e^{-i\theta}) \cdot \frac{1}{2} (e^{i\theta} + e^{-i\theta})
= \frac{1}{2} \cdot \frac{1}{2i} (e^{2 i \theta} + 1 - 1 - e^{-2 i \theta})
= \frac{1}{2} \cdot \frac{1}{2i} (e^{2 i \theta} - e^{-2 i \theta})
This is the same exponential form as the sine function with a couple adjustments (a factor of 1/2 out front and a factor of 2 inside the function):
= \frac{1}{2} \text{sin} (2 \theta)

Look up some other trig identities and try to derive them using complex exponentials.

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