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Trigonometry 

Sinus Cosinus Tangens Arctan

Definition

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

$\cos(\alpha) = \frac{a}{r}$

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

Hyperboles sinh, cosh, tanh

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

$\sinh(x) = \frac{1}{2}(e^x -e^{-x}) = - \text{i}\, \sin(\text{i}x)$

$\cosh(x) = \frac{1}{2}(e^x +e^{-x}) = \cos(\text{i}x)$

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

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

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