# 7.3 Trigonometric identities

## 7.3.1 Identities for $\sin$ and $\cos$

None of this document is prose (it’s all my ramblings on mathematics, and an exposition of all the things I failed to grasp explained in probably what is far too much verbosity) but if you thought the rest of it was bad (in terms of being dry) then this section is probably worse.

Here is a list of a bunch of trigonometric identities which are useful.

###### Theorem 7.3.1

Let $\theta$ be a real number, then

$\sin^{2}(\theta)+\cos^{2}(\theta)=1$ (7.6)

Proof: geometry (and hence omitted).

###### Theorem 7.3.2
 $\displaystyle\cos(\alpha\mp\beta)\equiv\cos(\alpha)\cos(\beta)\mp\sin(\alpha)% \sin(\beta)$ (7.7) $\displaystyle\sin(\alpha\pm\beta)\equiv\cos(\alpha)\sin(\beta)\pm\sin(\alpha)% \cos(\beta)$ (7.8)

Proof: The \sayeasy 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.

###### Theorem 7.3.3

A special case of these are the "double angle formulae" which are what we get if we set $\alpha,\beta=x$ in Equations 7.7 and 7.8. 55 5 These are useful for the integration of $\cos^{2}(x)$ and $\sin^{2}(x)$.

$cos(2x)\equiv\cos^{2}(x)-\sin^{2}(x)$ (7.9)

Proof: This also serves as an easy way to remember the identities (or, to recall them if needed). This is derived from Equation 7.7 by replacing $\alpha$ and $\beta$ with $x$, and then simplifying a bit:

 $\displaystyle\cos(x+x)$ $\displaystyle\equiv\cos(x)\cos(x)-\sin(x)\sin(x)$ (7.10) $\displaystyle\cos(2x)$ $\displaystyle\equiv\cos^{2}(x)-\sin^{2}(x)$ (7.11)
###### Theorem 7.3.4

We also have a similar identity for $\sin(x)$

$sin(2x)\equiv 2\sin(x)\cos(x).$ (7.12)

This is derived for 7.8 in a similar way to how the double-angle formula for cos is derived: replace $\alpha$ and $\beta$ with $x$, and simplify.

 $\displaystyle\sin(x+x)$ $\displaystyle\equiv\cos(x)\sin(x)+\sin(x)\cos(x)$ (7.13) $\displaystyle\sin(2x)$ $\displaystyle\equiv 2\cos(x)\sin(x)$ (7.14)

## 7.3.3 Secant, cosecant, cotangent and friends

These are a pain to remember. All of these functions are defined as the reciprocal of a trig function (I remember someone saying that they’re not exactly defined like this because, mumble, mumble division by zero, mumble, mumble), but the basic idea is this:

###### Definition 7.3.1

We define

 $\displaystyle\csc(\theta)=\frac{1}{\sin(\theta)}$ (7.15) $\displaystyle\sec(\theta)=\frac{1}{\cos(\theta)}$ (7.16) $\displaystyle\cot(\theta)=\frac{1}{\tan(\theta)}$ (7.17)

Here’s a helpful trick for remembering this

1. 1.

cosecant $\mapsto$ $\sin$

2. 2.

secant $\mapsto$ $\cos$

3. 3.

cotangent $\mapsto$ $\tan$

## 7.3.4 Proving trigonometric identities

###### Example 7.3.1

Prove that

$\csc^{2}(\theta)(\tan^{2}(\theta)-\sin^{2}(\theta))=\tan^{2}(\theta)$ (7.18)

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.

 $\displaystyle\csc^{2}(\theta)(\tan^{2}(\theta)-\sin^{2}(\theta))=\csc^{2}(% \theta)\tan^{2}(\theta)-\csc^{2}\sin^{2}(\theta)$ (7.19)
 $\displaystyle\csc^{2}(\theta)(\tan^{2}(\theta)-\sin^{2}(\theta))$ $\displaystyle=\csc^{2}(\theta)\tan^{2}(\theta)-\csc^{2}\sin^{2}(\theta)$ (7.20) $\displaystyle=\frac{1}{\sin^{2}(\theta)}\frac{\sin^{2}(\theta)}{\cos^{2}(% \theta)}-\frac{1}{\sin^{2}(\theta)}\sin^{2}(\theta)$ (7.21) $\displaystyle=\frac{1}{\cos^{2}(\theta)}-1$ $\displaystyle\sin^{2}(\theta)\neq 0$ (7.22) $\displaystyle=\frac{1-\cos^{2}(\theta)}{\cos^{2}(\theta)}$ (7.23) $\displaystyle=\frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}$ By the Pythagorean identity (7.24)