A calculus of the absurd

7.2 Trigonometric identities

This section goes through a bunch of trigonometric identities. The first one is the "Pythagorean identity" which is that

\[sin^2(\theta )+cos^2(\theta ) \equiv 1\]

It looks a bit like Pythagoras’ theorem! 5252 Why this is true was explored in the previous section.

The next identities are the "addition formulae" which state that

\begin{align} & \cos (\alpha \mp \beta ) \equiv \cos (\alpha )\cos (\beta ) \mp \sin (\alpha )\sin (\beta ) \label {angle addition formula for cos} \\ & \sin (\alpha \pm \beta ) \equiv \cos (\alpha )\sin (\beta ) \pm \sin (\alpha )\cos (\beta ) \label {angle addition formula for sin} \end{align}

These are in the formula booklet. 5353 The "easy" way to prove this is using complex numbers. I was going to point out that they can be proven by using triangles, however, as previously mentioned geometry is not my thing. For a geometric proof see https://www.youtube.com/watch?v=2SlvKnlVx7U. In the "complex numbers" section of this document there’s a proof of this identity which uses the properties of complex numbers.

A special case of these are the "double angle formulae" which are what we get if we set \(\alpha ,\beta =x\) in Equations 7.4 and 7.5. 5454 These are useful for the integration of \(\cos ^2(x)\) and \(\sin ^2(x)\).

\begin{equation} cos(2x) \equiv \cos ^2(x) - \sin ^2(x) \label {double angle formula for cos} \end{equation}

This is derived from Equation 7.4 by replacing \(\alpha \) and \(\beta \) with \(x\), and then simplifying a bit:

\begin{align} \cos (x + x) &\equiv \cos (x)\cos (x) - \sin (x)\sin (x) \\ \cos (2x) &\equiv \cos ^2(x) - \sin ^2(x) \end{align}

\begin{equation} sin(2x) \equiv 2\sin (x)\cos (x) \label {double angle formula for sin} \end{equation}

This is derived for 7.5 in a similar way to how the double-angle formula for cos is derived: replace \(\alpha \) and \(\beta \) with \(x\), and simplify.

\begin{align} \sin (x + x) &\equiv \cos (x)\sin (x) + \sin (x)\cos (x) \\ \sin (2x) &\equiv 2\cos (x)\sin (x) \end{align}

What about \(tan(\theta )\)? Don’t memorise identities for \(tan(\theta )\) because it’s equal to \(\frac {sin(\theta )}{cos(\theta )}\). Just use the identities for \(sin(\theta )\) and \(cos(\theta )\)!