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Trigonometry#

Trigonometry studies relationships involving lengths and angles of triangles.

Unit Circle#

image/svg+xml sin(α) = 0.500 cos(α) = 0.866 tan(α) = 0.577 α = 30°

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

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