# A calculus of the absurd

#### 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

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

Proof: geometry (and hence omitted).

• Theorem 7.3.2

\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}

Proof: 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.

• 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. 5353 These are useful for the integration of $$\cos ^2(x)$$ and $$\sin ^2(x)$$.

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

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:

\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}

• Theorem 7.3.4 We also have a similar identity for $$\sin (x)$$

$$sin(2x) \equiv 2\sin (x)\cos (x). \label {double angle formula for sin}$$

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.

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