A calculus of the absurd

14.7 Cool stuff with trigonometry

14.7.1 Proving identities

Example 14.7.1 Show that

\[\cos (a + b) = \cos (a)\cos (b) - \sin (a)\sin (b)\]

Solution:

We can write \(\cos (a + b)\) as the real part of \(e^{(a+b)i}\). This because \(\cos (a+b)\) is equal to the real part of \(\cos (a+b) + i\sin (a+b)\), which is equal to \(e^{(a+b)i}\).

\begin{align*} \cos (a + b) &= \Re (e^{(a + b)i}) \\ &= \Re (e^{ai} e^{bi}) \\ &= \Re ((\cos (a) + i\sin (a))(\cos (b)+i\sin (b))) \\ &= \Re (\cos (a)\cos (b) + i\cos (a)\cos (b) + i\sin (a)\cos (b) + i^2\sin (a)\sin (b)) \\ &= \cos (a)\cos (b) - \sin (a)\sin (b) \end{align*}

Example 14.7.2 Express \(\sin (3x)\) in terms of \(\sin (x)\).

Solution: Firstly, we can write \(\sin (3x)\) as the equation

\begin{equation} \sin (3x) = \Im (\cos (3x) + i\sin (3x)) \end{equation}

We can then apply De Moivre’s theorem¹⁰⁶¹⁰⁶ \(\cos (nx) + i\sin (nx) = (\cos (x) + i\sin (x))^n\) to rewrite the expression in terms of \(\sin (x)\) and \(\cos (x)\)

\begin{equation} \Im (\cos (3x) + i\sin (3x)) = \Im ((\cos (x) + i\sin (x))^3) \end{equation}

We can now expand the binomial obtained, which leads to the result that

\begin{equation} \Im ((\cos (x) + i\sin (x))^3) = \Im ((\cos ^3(x) + 3\cos ^2(x)i\sin (x) + 3\cos (x)i^2\sin ^2(x) + i^3\sin ^3(x))) \end{equation}

Then, we can tidy this up a bit, leading to the expression

\begin{equation} \Im ((\cos ^3(x) + 3\cos ^2(x)\sin (x)i - 3\cos (x)\sin ^2(x) - i\sin ^3(x))) \end{equation}

We are only interested in the imaginary parts of the expansion, so it is therefore equal to just

\begin{equation} 3\cos ^2(x)\sin (x) - \sin ^3(x) \end{equation}

We want \(\sin (3x)\) in terms of \(\sin (x)\), however! There’s a rogue gatecrasher ¹⁰⁷¹⁰⁷ A handy way to remember whether \(\cos (x)\) or \(\sin (x)\) shows a certain property is (as previously mentioned, TODO: mention) that \(\sin (x)\) generally behaves "better" than \(\cos (x)\). in the previous expression - the \(\cos (x)\)! Fortunately we can remove the \(\cos ^2(x)\) without too much difficulty using the Pythagorean identity.

\begin{align} 3(1-\sin ^2(x))\sin (x) - \sin ^3(x) &= 3\sin (x)-3\sin ^3(x) - \sin ^3(x) \\ &= 3\sin (x)-4\sin ^3(x) \end{align}

Thus, we can say that

\begin{equation} \sin (3x) = 3\sin (x)-4\sin ^3(x) \end{equation}