Thursday, October 4, 2012

$\cos \theta=2$

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Question: Write down one solution to each of the following:

(a) $\cos \theta=0.5$
(b) $\cos \theta=1$
(c) $\cos \theta=-0.17$
(d) $\cos \theta=2$

Note that there is no ° in the question, so the answer should be in radian.

A high school student could do (a) and (b) without a calculator. Calculator will give a solution of (c) with $\pi/2<\theta<\pi$. For (d), school algebra tells us that there is no solution, as the value of cosine lies in between -1 and 1 -- however, it is not the case if we consider complex number.

There are three definitions of cosine:

1. As the ratio of the adjacent to the hypotenuse in a right angled triangle.  Under this definition, $\theta$ can only be $0<\theta<\pi/2$ and we have $0<\cos \theta<1$.

2. As the $x$-coordinate of point with polar coordinate $(1, \theta)$. Under this definition, $\theta$ can be any real number and it is still true that $-1\le \cos \theta\le 1$.

3. By the formula discovered by Leonhard Euler (1707-1783):
$$\cos \theta=1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6!}\theta^6+\cdots$$
or equivalently
$$\cos \theta=\frac{e^{i\theta}+e^{-i\theta}}{2}.$$
Nowadays, the study of wave behaviour in physics and modern axiomatic approach in mathematics, cosine is no more an "angle" thing, and therefore Euler's formula serves as a better definition.

By using Euler's formula, we can have the cosine of a complex number.
$$\begin{eqnarray*}
\cos\theta&=&2\\
\frac{e^{i\theta}+e^{-i\theta}}{2}&=&2\\
e^{i\theta}+e^{-i\theta}&=&4\\
1+(e^{-i\theta})^2&=&4e^{-i\theta}\\
(e^{-i\theta})^2-4e^{-i\theta}+1&=&0
\end{eqnarray*}$$
One solution of the quadratic equation is $e^{-i\theta}=2+2\sqrt{3}$ and hence one solution of (d) is $\theta=i\log_e(2+\sqrt{3})$.