# Trigonometry#

Trigonometry studies relationships involving lengths and angles of triangles.

## Sin, Cos, Tan#

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

with angle $$\alpha$$

Explanation:

In the unit circle, the sinus and cosinus at a given angle form an orthogonal triangle with the edges $$a,b,c$$. The length of the hypotenuse $$c$$ corresponds to the radius $$r$$, which equals 1 (unit circle). Thus, we can apply the law of Pythagoras $$a^2 + b^2 = c^2$$ with $$a = \sin(\alpha)$$, $$b = \cos(\alpha)$$, and $$c = 1$$.

$$\(\sin(\alpha) = \frac{b}{r}$$\) with angle $$\alpha$$, radius $$r$$

Symmetry: $$\sin(\alpha) = - \sin(α)$$

$$\(\cos(\alpha) = \frac{a}{r}$$\) with angle $$\alpha$$, radius $$r$$

Symmetry: $$\cos(\alpha) = \cos(-\alpha)$$

$$\(\tan(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)} = \frac{b}{a}$$\) with angle $$\alpha$$

Properties Equation
Symmetry $$\sin(-x)=-\sin(x)$$ $$\cos (-x) = \cos (x)$$
Complex $$e^{\i x}=\cos(x)+\i\sin(x)$$ $$e^{-\i x}=\sin(x)-\i\cos(x)$$
$$x$$
$$\scriptstyle{ \alpha }$$
$$0$$
$$\scriptstyle{0^\circ}$$
$$\pi / 6$$
$$\scriptstyle{30^\circ}$$
$$\pi / 4$$
$$\scriptstyle{45^\circ}$$
$$\pi / 3$$
$$\scriptstyle{60^\circ}$$
$$\frac{1}{2}\pi$$
$$\scriptstyle{90^\circ}$$
$$\pi$$
$$\scriptstyle{180^\circ}$$
$$1\frac{1}{2}\pi$$
$$\scriptstyle{270^\circ}$$
$$2 \pi$$
$$\scriptstyle{360^\circ}$$
$$\sin$$ $$0$$ $$\frac{1}{2}$$ $$\frac{1}{\sqrt{2}}$$ $$\frac{\sqrt 3}{2}$$ $$1$$ $$0$$ $$-1$$ $$0$$
$$\cos$$ $$1$$ $$\frac{\sqrt 3}{2}$$ $$\frac{1}{\sqrt 2}$$ $$\frac{1}{2}$$ $$0$$ $$-1$$ $$0$$ $$1$$
$$\tan$$ $$0$$ $$\frac{\sqrt{3}}{3}$$ $$1$$ $$\sqrt{3}$$ $$\pm \infty$$ $$0$$ $$\mp \infty$$ $$0$$
$$\cos (x - \frac{\pi}{2}) = \sin x$$ $$\int x \cos(x) \diff x = \cos(x) + x \sin(x)$$
$$\sin (x + \frac{\pi}{2}) = \cos x$$ $$\int x \sin(x) \diff x = \sin(x) - x \cos(x)$$
$$\sin 2x = 2 \sin x \cos x$$ $$\int \sin^2(x) \diff x = \frac12 \bigl(x - \sin(x)\cos(x) \bigr)$$
$$\cos 2x = 2\cos^2 x - 1$$ $$\int \cos^2(x) \diff x = \frac12 \bigl(x + \sin(x)\cos(x) \bigr)$$
$$\sin(x) = \tan(x)\cos(x)$$ $$\int \cos(x)\sin(x) = -\frac12 \cos^2(x)$$

## Hyperboles sinh, cosh, tanh#

$$\cosh (x) + \sinh (x) = e^{x}$$
$$\cosh^2 (x) - \sinh^2 (x) = 1$$

$\sinh(x) = \frac{1}{2}(e^x -e^{-x}) = - \i \, \sin(\i x)$
$\cosh(x) = \frac{1}{2}(e^x +e^{-x}) = \cos(\i x)$
$\tanh(x) =\frac {\sinh x}{\cosh x} = {\frac {\mathrm {e} ^{x}-\mathrm {e} ^{-x}}{\mathrm {e} ^{x}+\mathrm {e} ^{-x}}}={\frac {\mathrm {e} ^{2x}-1}{\mathrm {e} ^{2x}+1}}=1-{\frac {2}{\mathrm {e} ^{2x}+1}}$
$\coth(x) ={\frac {\cosh x}{\sinh x}}={\frac {\mathrm {e} ^{x}+\mathrm {e} ^{-x}}{\mathrm {e} ^{x}-\mathrm {e} ^{-x}}}={\frac {\mathrm {e} ^{2x}+1}{\mathrm {e} ^{2x}-1}}=1+{\frac {2}{\mathrm {e} ^{2x}-1}}$

### Inverse#

$\mathrm{arcsinh}(x) := \ln\left(x+\sqrt{x^2+1}\right)$
$\mathrm{arccosh}(x) := \ln\left(x+\sqrt{x^2-1}\right)$
$\mathrm{artanh}(x) = \frac{1}{2} \ln {\frac {1+x}{1-x}}$
$\mathrm{arcoth}(x) = \frac{1}{2} \ln {\frac {x+1}{x-1}}$

## Cardinal Sinus#

$\mathrm{si}(x) = \frac{\sin(x)}{x}$

normalized: $$\(\sinc(x) = \frac{\sin(\pi x)}{\pi x}$$\)