# Trigonometric Identities

## Definitions

Sine and Cosine give the y- and x- coordinates of angles:

$\sin \theta = \frac{opposite}{hypotenuse} = e^{i\theta}\cdot \mathbf{\hat{y}}$ $\cos \theta = \frac{adjacent}{hypotenuse} = e^{i\theta}\cdot \mathbf{\hat{x}}$ $\tan \theta = \frac{\sin\theta}{\cos\theta}$

$$\sec \theta = \frac{1}{\cos \theta}$$, $$\csc \theta = \frac{1}{\sin \theta}$$, $$\cot \theta = \frac{1}{\tan \theta}$$

$e^{i \theta} = \cos \theta + i \sin \theta$ $\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ $\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$

(Clearly it would make more sense for ‘cosine’ to be the simpler name and ‘sine’ to be the derivative one. Oh well.)

Since $$\sin \theta$$ and $$\cos \theta$$ are the x- and y- coordinates of the unit circle $$x^{2} + y^{2} = 1$$, we have:

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

of course $$\sin \theta = \pm \sqrt{1 - \cos^{2} \theta}$$ and vice-versa. Also, $$1 + \tan^{2} \theta = \sec^{2} \theta$$ and $$1 + \cot^{2} \theta = \csc^{2} \theta$$.

## Symmetries and Translation Formulas

$$\sin -\theta = -\sin \theta$$ (sine is odd)

$$\cos -\theta = \cos \theta$$ (cosine is even)

Therefore: $$\tan \theta$$ is odd because it’s a ratio even to odd, etc.

Shifting $$\sin \theta$$ or $$\cos \theta$$ by quarter turns $$\theta \rightarrow \theta + \frac{pi}{2}$$ causes them to rotate through four functions:

$\sin \theta = \cos (\theta - \frac{\pi}{2}) = -\sin (\theta - \pi) = -\cos (\theta- \frac{3\pi}{2})$

Shifting $$\sin -\theta$$ or $$\cos -\theta$$ by quarter turns traverses the list in the opposite direction:

$\sin (-\theta) = -\cos (-\theta - \frac{\pi}{2}) = -\sin (-\theta - \pi) = \cos (-\theta - \frac{3\pi}{2})$

Both of these are just statements that the graphs of $$\sin \theta$$, $$\cos \theta$$, $$\sin -\theta$$, and $$\cos -\theta$$ look the same but are translated differently.

## Angle formulas

$\sin a \pm b = \sin a \cos b \pm \cos a \sin b$ $\cos a \pm b = \cos a \cos b \mp \sin a \sin b$ $\tan a \pm b = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$

Double-angle formulas:

$\sin 2a = 2\sin a \cos a$ $\cos 2a = \cos^{2} a - \sin^{2} a$ $\tan 2a = \frac{2 \tan a}{1 - \tan^{2} a}$

General multiples are most easily handled through

$(\cos n\theta + i\sin n\theta) = e^{in\theta} = (\cos \theta + i \sin \theta)^{n}$

Half-angle:

$\sin \frac{a}{2} = \sqrt{\frac{1}{2}(1-\cos a)}$

(though the square root can be negative if $$\sin \frac{a}{2}$$ is negative, so if $$\frac{a}{2}$$ is in $$(\pi, 2\pi)$$, so only if $$a$$ is allowed to be greater than $$2\pi$$ without wrapping around.)

$\cos \frac{a}{2} = \pm\sqrt{\frac{1}{2}(1+\cos a)}$

(the sign is negative if $$\cos \frac{a}{2}$$ is negative, so, if $$\frac{a}{2}$$ is in $$(\frac{\pi}{2}, \frac{3\pi}{2})$$).

$\tan \frac{a}{2} = \frac{\sin a}{1 + \cos a}$ $\tan \frac{a+b}{2} = \frac{\sin a + \sin b}{\cos a + \cos b}$

Product-to-sum:

$2 \cos a \cos b = \cos (a-b) + \cos (a+b)$ $2 \sin a \sin b = \cos (a-b) - \cos (a+b)$ $2 \sin a \cos b = \sin (a+b) + \sin (a-b)$

Sum-to-product:

$\sin 2a \pm \sin 2b = 2 \sin (a \pm b) \cos (a \mp b)$ $\cos 2a + \cos 2b = 2 \cos (a + b) \cos (a - b)$ $\cos 2a - \cos 2b = -2 \sin (a+b) \sin (a-b)$

Powers:

$\sin^{2} a = \frac{1}{2}(1-\cos 2a)$ $\cos^{2} a = \frac{1}{2}(1+\cos 2a)$ $\sin^{2} a \cos^{2} a = \frac{1}{8}(1 - \cos 4a)$

## Geometric identities

For a triangle with angles $$a, b, c$$ and side lengths $$A, B, C$$:

$\frac{\sin a}{A} = \frac{\sin b}{B} = \frac{\sin c}{C}$

(Law of Sines)

$C^{2} = A^{2} + B^{2} + 2AB \cos c$

(Law of Cosines = general case of Pythagorean Theorem)

If $$a + b + c = \pi$$, then

$\sin 2a + \sin 2b + \sin 2c = 4 \sin a \sin b \sin c$ $\tan a + \tan b + \tan c = \tan a \tan b \tan c$

If $$a + b + c + d = \pi$$, then: (Ptolemy’s Theorem)

$\sin(a+b) + \sin(b+c) = \sin(a+c) \sin(b + d)$

On spheres with radius R, the Pythagorean Theorem becomes

$\cos \frac{C}{R} = \cos \frac{A}{R} \cos \frac{B}{R}$

(as $$R \rightarrow \infty$$, $$\cos \frac{x}{R} \rightarrow 1 - \frac{1}{2} (\frac{x}{R})^{2}$$, so this becomes $$1 - \frac{C^{2}}{2R^{2}} = 1 - \frac{A^{2}}{2R^{2}} - \frac{B^{2}}{2R^{2}} - o(R^{4}) \rightarrow C^{2} = A^{2} + B^{2}$$

And the Law of Cosines becomes:

$\cos \frac{C}{R} = \cos \frac{A}{R} \cos \frac{A}{R} + \sin \frac{A}{R} \sin \frac{B}{R} \cos c$

or

$\cos c = -\cos a \cos b + \sin a \sin b \cos \frac{C}{R}$

## Inverse functions

It’s really antiquated to call these $$\arcsin$$ instead of $$\sin^{-1}$$, in my opinion, but whatever.

Domain and Range:

Inverse trig functions require some fanangling with their domains to keep things behaving like proper inverses. $$\arcsin x$$ takes values in $$(-1, 1)$$ (the range of $$\sin$$) and typically returns $$(-\frac{\pi}{2}, \frac{\pi}{2})$$, though that’s a convention. $$\arccos x$$ has the same range and typically returns values in $$(0, \pi)$$.

$$\arctan x$$ gives real results for all real inputs (since the range of tangent is all reals) and typically has range $$(-\frac{\pi}{2}, \frac{\pi}{2})$$

There’s also $$\text{atan2 } x$$, the oriented / two-argument / sign-preserved arctan function, which I don’t feel like going into here. It is usually what you actually want in code.

### Symmetries

These are, unsurprisingly, inverted forms of the regular trigonometric symmetry identities above.

$\arcsin (-x) = - \arcsin x$ $\arccos (-x) = \pi - \arccos x$ $\arcsin x = \frac{\pi}{2} - \arccos x$ $\arccos x = \arcsin \sqrt{1- x^{2}}, x \in [0, 1]$ $\arccos x = \frac{1}{2} \arccos (2x^{2}-1), x \in [0, 1]$ $\arcsin x = \frac{1}{2} \arccos (1 - 2x^{2}), x \in [0,1]$ $\arctan x = \arcsin \frac{x}{\sqrt{1 + x^{2}}}$ $\arcsin x = 2\arctan \frac{x}{1 + \sqrt{1- x^{2}}}$ $\arccos x = 2\arctan \frac{\sqrt{1 - x^{2}}}{1 + x}, x \in (-1, 1]$

### Other Formulas

Inversions:

Relationships between trig and inverse trig functions can be thought of as relationships between parts of a triangle. For example, $$\arcsin x$$ is “the angle that gives $$\frac{opposite}{hypotenuse}$$ ratio x”. Then $$\cos \arcsin x$$ must be the $$\frac{adjacent}{hypotenuse}$$ ratio such a triangle. If $$\frac{o}{h}=x$$, then let $$h=1$$, so $$o = x$$ and $$a = \sqrt{1-x^{2}}$$.

$\sin \arccos x = \cos \arcsin x = \sqrt{1-x^{2}}$ $\sin \arctan x = \cos \arctan x = \frac{x}{\sqrt{1+x^{2}}}$ $\tan \arcsin x = \frac{x}{\sqrt{1-x^{2}}}$ $\tan \arccos x = \frac{\sqrt{1-x^{2}}}{x}$

Logarithm definitions:

$\arcsin x = -i \ln (ix + \sqrt{1-x^{2}})$ $\arccos x = i \ln (x - i\sqrt{1-x^{2}})$ $\arctan x = \frac{i}{2} \ln \frac{i+x}{i-x}$

## Calculus

Limits:

$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ $\lim_{x\rightarrow 0} \frac{1 - \cos x}{x} = 0$

### Derivatives:

$\sin' = \cos, \cos' = -\sin$ $\tan' = \sec^{2}, \cot' = -\csc^{2}$

(by the product / division rule)

$\sec' = \tan \sec, \csc' = -\csc \cot$

Inverse function derivatives (defined except where denominators would be 0):

$(\arcsin z)' = \frac{1}{\sqrt{1-x^{2}}} = -(\arccos z)'$ $(\arctan z)' = \frac{1}{1+z^{2}} = -(\text{arccot } z)'$

Inverse functions can also be defined as integrals:

$\arcsin x = \int_{0}^{x} \frac{1}{\sqrt{1-z^{2}}} dz$ $\arccos x = \int_{x}^{1} \frac{1}{\sqrt{1-z^{2}}} dz$ $\arctan x = \int_{0}^{x} \frac{1}{1+z^{2}} dz$

### Non-trivial integrals:

$\int \arcsin x dx = x \arcsin x + \sqrt{1-x^{2}} + C$ $\int \arccos x dx = x \arccos x - \sqrt{1-x^{2}} + C$ $\int \arctan x dx = x \arctan x - \frac{1}{2} \ln (1+x^{2}) + C$ $\int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}(\frac{x}{a}) + C$ $\int \frac{dx}{a^{2} + x^{2}} = \frac{1}{a}\tan^{-1}(\frac{x}{a}) + C$ $\int \frac{dx}{x\sqrt{a^{2} - x^{2}}} = \frac{1}{a}\sec^{-1}\vert \frac{x}{a} \vert + C$

### Taylor Series:

$\sin x = x - \frac{x^{3}}{3!} \ldots = \sum \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}$ $\cos x = 1 - \frac{x^{2}}{2!} \ldots = \sum \frac{(-1)^{n}}{(2n)!}x^{2n}$

Clearly the sum of series for $$\cos x + i \sin x$$ gives the series for $$e^{ix} = \sum \frac{(ix)^{n}}{n!}$$

The series for other trig functions are awkward and use special functions. Some of the inverse ones are okay, though:

$\arcsin x = \sum \frac{(2n)!}{4^{n}(n!)^{2}(2n+1)} x^{2n+1}, \vert x \vert \leq 1$ $\arccos x = \frac{\pi}{2} - \arcsin x, \vert x \vert \leq 1$ $\arctan x = \sum \frac{(-1)^{n}}{2n+1)} x^{2n+1}, \vert x \vert \leq 1, x \neq \pm i$

Miscellaneous series:

$\sin x = x \prod_{n=1}^{\infty} (1 - \frac{x^{2}}{\pi^{2} n^{2}})$ $\cos x = \prod_{n=1}^{\infty} (1- \frac{x^{2}}{\pi^{2}(n-\frac{1}{2})^{2}})$ $\sin x = x\prod \cos \frac{x}{2^{n}}$

## Hyperbolic functions

Hyperbolic trig functions are defined on hyperbolic angles the same way regular trig functions are defined on circular angles, with the unit hyperbola as the graph of the function $$x^{2} - y^{2} =1$$.

A ‘hyperbolic angle’ $$\phi$$ rotates from 0 to $$\infty$$ as the point on the unit hyperbola approaches the asymptote $$x=y$$, and towards $$-\infty$$ as it approaches $$x = -y$$. As with trig functions, $$\cosh x$$ gives the x-coordinate and $$\sinh y$$ gives the y coordinate.

Anyway, it’s a chore to store a lot of intuition about hyperbolic functions, or to write down a lot of identities. The important ones are:

$\tanh \phi = \frac{\sinh \phi}{\cosh \phi}$ $\cosh^{2} \phi - \sinh^{2} \phi = 1$ $e^{\phi} = \cosh \phi + \sinh \phi$ $\cosh \phi = \frac{1}{2}(e^{\phi} + e^{-\phi})$ $\sinh \phi = \frac{1}{2}(e^{\phi} - e^{-\phi})$ $\cosh \phi = \cos ix = \sum \frac{x^{2n}}{(2n)!}$ $\sinh \phi = -i \sin ix = \sum \frac{x^{2n+1}}{(2n+1)!}$ $\tanh \phi = -i \tan i\phi$

And you can just grab the rest here.