A calculus of the absurd

7.3.4 Proving trigonometric identities

  • Example 7.3.1 Prove that

    \begin{equation} \csc ^2(\theta )(\tan ^2(\theta ) - \sin ^2(\theta )) = \tan ^2(\theta ) \end{equation}

This is a standard application of the general method of proving identities (as in Section 6.2); we pick one side (and my usual heuristic is to always start with the more complex side) and work toward the other side.

In this case we start with

\begin{align} \csc ^2(\theta )(\tan ^2(\theta ) - \sin ^2(\theta )) = \csc ^2(\theta ) \tan ^2(\theta ) - \csc ^2\sin ^2(\theta ) \end{align}

This seems like a reasonable thing to do (we must often form an “ansatz” - that is, an educated guess and try to apply this; this was my first one, but always remember the story of the mouse - as in Section 2.4 - we can always try something else if the first idea does not work), and we can proceed by unrolling some definitions

\begin{align} \csc ^2(\theta )(\tan ^2(\theta ) - \sin ^2(\theta )) &= \csc ^2(\theta ) \tan ^2(\theta ) - \csc ^2\sin ^2(\theta ) \\ &= \frac {1}{\sin ^2(\theta )} \frac {\sin ^2(\theta )}{\cos ^2(\theta )} - \frac {1}{\sin ^2(\theta )} \sin ^2(\theta ) \\ &= \frac {1}{\cos ^2(\theta )} - 1 & \sin ^2(\theta ) \ne 0 \\ &= \frac {1 - \cos ^2(\theta )}{\cos ^2(\theta )} \\ &= \frac {\sin ^2(\theta )}{\cos ^2(\theta )} & \text {By the Pythagorean identity} \end{align}