We always learn the trigonometric functions from the geometry and use them to solve geometry problem. It it quite difficult to memorize some trigonometric identities. We re-define the sine and cosine function from the view of the solution of the ordinary differential equations in this article. It is similar with the Bessel functions.
The Helmholtz equation in frequency domain in cartesian coordinates is
\(\nabla^2 E(x)+k^2 E(x)=0\)
It is a 2nd order linear differential equation, there must be two linearly independent solutions. The plane wave solutions are \(e^{-ikx}\) and \(e^{ikx}\), which reprent the wave along x-axis and -x respectly. Their linear combinations are also the solutions of the Helmholtz equation. According to the Euler’s equation, we get
\(\cos kx=\frac{e^{ikx}+e^{-ikx}}{2}\)
\(\sin kx=\frac{e^{ikx}-e^{-ikx}}{2i}\)
It is clear that they are linear independent. They both represent the standing wave. It provides a powerful connection between analysis and trigonometry. Because the plane wave are much easier to manipulate than the standing wave. One technique to manipulate the trigonometry functions is simply to decompose the standing wave in to the plane wave. After the manipulations, the simplified result is still real-valued. For example:
Prove \(\cos (x-y)=\cos x \cos y+\sin x \sin y\)
\(\begin{eqnarray}\cos x \cos y+\sin x \sin y&=&\frac{e^{ix}+e^{-ix}}{2}\frac{e^{iy}+e^{-iy}}{2}+\frac{e^{ix}-e^{-ix}}{2i}\frac{e^{iy}-e^{-iy}}{2i}\\
&=& \frac{e^{i(x+y)}+e^{i(x-y)}+e^{-i(x-y)}+e^{-i(x+y)}}{4}-\frac{e^{i(x+y)}-e^{i(x-y)}-e^{-i(x-y)}+e^{-i(x+y)}}{4}\\
&=&\frac{e^{i(x-y)}+e^{-i(x-y)}}{2}\\
&=&\cos (x-y)
\end{eqnarray}\)
Simplify \(\cos^2 x\)
\(\begin{eqnarray}\cos^2 x &=& (\frac{e^{ix}+e^{-ix}}{2})^2 \\
&=&\frac{e^{2ix}+e^{-2ix}+2}{4}\\
&=&\frac{1}{2}(\frac{e^{2ix}+e^{-2ix}}{2}+1)\\
&=&\frac{\cos 2x +1}{2}\end{eqnarray}\)
Universal trigonometric subsitution
There are several trigonometric functions in a express, we have to specify them with only one variable. The universal substitution is useful. If we set \(\tan \frac{x}{2}=t\), then we get
\[\begin{eqnarray}
\sin x &=& \frac{2 t}{1+t^2} \\
\cos x &=& \frac{1-t^2}{1+t^2} \\
\tan x &=& \frac{2 t}{1-t^2}
\end{eqnarray}\]
Prove:
\[\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\frac{e^{i \frac {x}{2}}-e^{-i \frac {x}{2}}}{2 i}}{\frac{e^{i \frac {x}{2}}+e^{-i \frac {x}{2}}}{2}}=\frac{e^{i \frac {x}{2}}-e^{-i \frac {x}{2}}}{e^{i \frac {x}{2}}+e^{-i \frac {x}{2}}}\frac{1}{i}=t\]
\[\tan^2 \frac{x}{2}=(\frac{e^{i \frac {x}{2}}-e^{-i \frac {x}{2}}}{e^{i \frac {x}{2}}+e^{-i \frac {x}{2}}}\frac{1}{i})^2=-\frac{e^{ix}+e^{-ix}-2}{e^{ix}+e^{-ix}+2}=-\frac{2\cos x-2}{2 \cos x +2}=-\frac{\cos x-1}{\cos x +1}=t^2\]
then
\[\cos x=\frac{1-t^2}{1+t^2}\]
\[\sin x=\sqrt{1-\cos^2 x}=\sqrt{1-(\frac{1-t^2}{1+t^2})^2}=\frac{\sqrt{(1+t^2)^2-(1-t^2)^2}}{1+t^2}=\frac{2 t}{1+t^2}\]