Trigonometry formula collection

${\displaystyle {\sqrt[{n}]{x}}}$

This article is a collection of formulas for the topic of trigonometry . Math symbols are used, which are explained in the article List of Math Symbols .

Triangle calculation

The following list contains most of the known formulas from trigonometry in the plane . Most of these relationships use trigonometric functions .

The following terms are used: The triangle has the sides , and , the angles , and at the corners , and . Furthermore, let the radius of the radius , the Inkreisradius and , and the Ankreisradien (namely, the radii of the excircles, the corners , or opposite) of the triangle . The variable stands for half the circumference of the triangle : ${\displaystyle ABC}$ ${\displaystyle a=BC}$ ${\displaystyle b=CA}$ ${\displaystyle c=AB}$ ${\displaystyle \alpha }$ ${\displaystyle \beta }$ ${\displaystyle \gamma }$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle r}$ ${\displaystyle \rho }$ ${\displaystyle \rho _{a}}$ ${\displaystyle \rho _{b}}$ ${\displaystyle \rho _{c}}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle ABC}$ ${\displaystyle s}$ ${\displaystyle ABC}$

${\displaystyle s={\frac {a+b+c}{2}}}$.

Finally, the area of the triangle is with designated. All other terms are explained in the relevant sections in which they appear. ${\displaystyle ABC}$ ${\displaystyle F}$

It should be noted here that the names for the perimeter radius , the Inkreisradius and the three Ankreisradien , , be used. Often notwithstanding, for the same sizes and the names , , , , used. ${\displaystyle r}$ ${\displaystyle \rho }$ ${\displaystyle \rho _{a}}$ ${\displaystyle \rho _{b}}$ ${\displaystyle \rho _{c}}$ ${\displaystyle R}$ ${\displaystyle r}$ ${\displaystyle r_{a}}$ ${\displaystyle r_{b}}$ ${\displaystyle r_{c}}$

Angle sum

${\displaystyle \alpha +\beta +\gamma =180^{\circ }}$

Sine law

Formula 1:

${\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r={\frac {abc}{2F}}}$

Formula 2:

if ${\displaystyle \alpha =90^{\circ }}$

${\displaystyle \sin \beta ={\frac {b}{a}}}$

${\displaystyle \sin \gamma ={\frac {c}{a}}}$

if ${\displaystyle \beta =90^{\circ }}$

${\displaystyle \sin \alpha ={\frac {a}{b}}}$

${\displaystyle \sin \gamma ={\frac {c}{b}}}$

if ${\displaystyle \gamma =90^{\circ }}$

${\displaystyle \sin \alpha ={\frac {a}{c}}}$

${\displaystyle \sin \beta ={\frac {b}{c}}}$

Cosine law

Formula 1:

${\displaystyle a^{2}=b^{2}+c^{2}-2bc\ \cos \alpha }$

${\displaystyle b^{2}=c^{2}+a^{2}-2ca\ \cos \beta }$

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\ \cos \gamma }$

Formula 2:

if ${\displaystyle \alpha =90^{\circ }}$

${\displaystyle \cos \beta ={\frac {c}{a}}}$

${\displaystyle \cos \gamma ={\frac {b}{a}}}$

if ${\displaystyle \beta =90^{\circ }}$

${\displaystyle \cos \alpha ={\frac {c}{b}}}$

${\displaystyle \cos \gamma ={\frac {a}{b}}}$

if ${\displaystyle \gamma =90^{\circ }}$

${\displaystyle a^{2}+b^{2}=c^{2}}$( Pythagorean theorem )

${\displaystyle \cos \alpha ={\frac {b}{c}}}$

${\displaystyle \cos \beta ={\frac {a}{c}}}$

Projection set

${\displaystyle a=b\,\cos \gamma +c\,\cos \beta }$

${\displaystyle b=c\,\cos \alpha +a\,\cos \gamma }$

${\displaystyle c=a\,\cos \beta +b\,\cos \alpha }$

The Mollweide's formulas

${\displaystyle {\frac {b+c}{a}}={\frac {\cos {\frac {\beta -\gamma }{2}}}{\sin {\frac {\alpha }{2}}}},\quad {\frac {c+a}{b}}={\frac {\cos {\frac {\gamma -\alpha }{2}}}{\sin {\frac {\beta }{2}}}},\quad {\frac {a+b}{c}}={\frac {\cos {\frac {\alpha -\beta }{2}}}{\sin {\frac {\gamma }{2}}}}}$

${\displaystyle {\frac {b-c}{a}}={\frac {\sin {\frac {\beta -\gamma }{2}}}{\cos {\frac {\alpha }{2}}}},\quad {\frac {c-a}{b}}={\frac {\sin {\frac {\gamma -\alpha }{2}}}{\cos {\frac {\beta }{2}}}},\quad {\frac {a-b}{c}}={\frac {\sin {\frac {\alpha -\beta }{2}}}{\cos {\frac {\gamma }{2}}}}}$

Tangent theorem

Formula 1:

${\displaystyle {\frac {b+c}{b-c}}={\frac {\tan {\frac {\beta +\gamma }{2}}}{\tan {\frac {\beta -\gamma }{2}}}}={\frac {\cot {\frac {\alpha }{2}}}{\tan {\frac {\beta -\gamma }{2}}}}}$

Analogous formulas apply to and : ${\displaystyle {\frac {a+b}{a-b}}}$ ${\displaystyle {\frac {c+a}{c-a}}}$

${\displaystyle {\frac {a+b}{a-b}}={\frac {\tan {\frac {\alpha +\beta }{2}}}{\tan {\frac {\alpha -\beta }{2}}}}={\frac {\cot {\frac {\gamma }{2}}}{\tan {\frac {\alpha -\beta }{2}}}}}$

${\displaystyle {\frac {c+a}{c-a}}={\frac {\tan {\frac {\gamma +\alpha }{2}}}{\tan {\frac {\gamma -\alpha }{2}}}}={\frac {\cot {\frac {\beta }{2}}}{\tan {\frac {\gamma -\alpha }{2}}}}}$

Because of this , one of these formulas remains valid if both the sides and the associated angles are swapped, for example: ${\displaystyle \tan(-x)=-\tan(x)}$

${\displaystyle {\frac {a+c}{a-c}}={\frac {\tan {\frac {\alpha +\gamma }{2}}}{\tan {\frac {\alpha -\gamma }{2}}}}={\frac {\cot {\frac {\beta }{2}}}{\tan {\frac {\alpha -\gamma }{2}}}}}$

Formula 2:

if ${\displaystyle \alpha =90^{\circ }}$

${\displaystyle \tan \beta ={\frac {b}{c}}}$

${\displaystyle \tan \gamma ={\frac {c}{b}}}$

if ${\displaystyle \beta =90^{\circ }}$

${\displaystyle \tan \alpha ={\frac {a}{c}}}$

${\displaystyle \tan \gamma ={\frac {c}{a}}}$

if ${\displaystyle \gamma =90^{\circ }}$

${\displaystyle \tan \alpha ={\frac {a}{b}}}$

${\displaystyle \tan \beta ={\frac {b}{a}}}$

Formulas with half the circumference

Below is always half the circumference of the triangle , that is . ${\displaystyle s}$ ${\displaystyle ABC}$ ${\displaystyle s={\frac {a+b+c}{2}}}$

${\displaystyle s-a={\frac {b+c-a}{2}}}$

${\displaystyle s-b={\frac {c+a-b}{2}}}$

${\displaystyle s-c={\frac {a+b-c}{2}}}$

${\displaystyle \left(s-b\right)+\left(s-c\right)=a}$

${\displaystyle \left(s-c\right)+\left(s-a\right)=b}$

${\displaystyle \left(s-a\right)+\left(s-b\right)=c}$

${\displaystyle \left(s-a\right)+\left(s-b\right)+\left(s-c\right)=s}$

${\displaystyle \sin {\frac {\alpha }{2}}={\sqrt {\frac {\left(s-b\right)\left(s-c\right)}{bc}}}}$

${\displaystyle \sin {\frac {\beta }{2}}={\sqrt {\frac {\left(s-c\right)\left(s-a\right)}{ca}}}}$

${\displaystyle \sin {\frac {\gamma }{2}}={\sqrt {\frac {\left(s-a\right)\left(s-b\right)}{ab}}}}$

${\displaystyle \cos {\frac {\alpha }{2}}={\sqrt {\frac {s\left(s-a\right)}{bc}}}}$

${\displaystyle \cos {\frac {\beta }{2}}={\sqrt {\frac {s\left(s-b\right)}{ca}}}}$

${\displaystyle \cos {\frac {\gamma }{2}}={\sqrt {\frac {s\left(s-c\right)}{ab}}}}$

${\displaystyle \tan {\frac {\alpha }{2}}={\sqrt {\frac {\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}}}$

${\displaystyle \tan {\frac {\beta }{2}}={\sqrt {\frac {\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}}}$

${\displaystyle \tan {\frac {\gamma }{2}}={\sqrt {\frac {\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}}}}$

${\displaystyle s=4r\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}}$

${\displaystyle s-a=4r\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}}$

Area and radius

The area of ​​the triangle is denoted here with (not, as is common today, with , in order to avoid confusion with the triangle corner): ${\displaystyle F}$ ${\displaystyle A}$ ${\displaystyle A}$

Heron's formula:

${\displaystyle F={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\frac {1}{4}}{\sqrt {\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}}$

${\displaystyle F={\frac {1}{4}}{\sqrt {2\left(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)}}}$

Further area formulas:

${\displaystyle F={\frac {1}{2}}bc\sin \alpha ={\frac {1}{2}}ca\sin \beta ={\frac {1}{2}}ab\sin \gamma }$

${\displaystyle F={\frac {1}{2}}ah_{a}={\frac {1}{2}}bh_{b}={\frac {1}{2}}ch_{c}}$Wherein , and the lengths of from , or outgoing heights of the triangle are. ${\displaystyle h_{a}}$ ${\displaystyle h_{b}}$ ${\displaystyle h_{c}}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle ABC}$

${\displaystyle F=2r^{2}\sin \,\alpha \,\sin \,\beta \,\sin \,\gamma }$

${\displaystyle F={\frac {abc}{4r}}}$

${\displaystyle F=\rho s=\rho _{a}\left(s-a\right)=\rho _{b}\left(s-b\right)=\rho _{c}\left(s-c\right)}$

${\displaystyle F={\sqrt {\rho \rho _{a}\rho _{b}\rho _{c}}}}$

${\displaystyle F=4\rho r\cos \,{\frac {\alpha }{2}}\,\cos \,{\frac {\beta }{2}}\,\cos \,{\frac {\gamma }{2}}}$

${\displaystyle F=s^{2}\tan \,{\frac {\alpha }{2}}\,\tan \,{\frac {\beta }{2}}\,\tan \,{\frac {\gamma }{2}}}$

${\displaystyle F=\rho ^{2}{\sqrt {\dfrac {h_{a}\,h_{b}\,h_{c}}{(h_{a}-2\rho )(h_{b}-2\rho )(h_{c}-2\rho )}}}}$, With ${\displaystyle {\dfrac {1}{\rho }}={\dfrac {1}{h_{a}}}+{\dfrac {1}{h_{b}}}+{\dfrac {1}{h_{c}}}}$

${\displaystyle F={\sqrt {\dfrac {r\,h_{a}\,h_{b}\,h_{c}}{2}}}}$

${\displaystyle F={\dfrac {\,h_{a}\,h_{b}\,h_{c}}{2\rho \,{(\sin \alpha +\sin \beta +\sin \gamma )}}}}$

Extended sine law:

${\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r={\frac {abc}{2F}}}$

${\displaystyle a=2r\,\sin \alpha }$

${\displaystyle b=2r\,\sin \beta }$

${\displaystyle c=2r\,\sin \gamma }$

${\displaystyle r={\frac {abc}{4F}}}$

Inside and circle radii

This section lists formulas in which the incircle radius , the circle radius , and the triangle appear. ${\displaystyle \rho }$ ${\displaystyle \rho _{a}}$ ${\displaystyle \rho _{b}}$ ${\displaystyle \rho _{c}}$ ${\displaystyle ABC}$

${\displaystyle \rho =\left(s-a\right)\tan {\frac {\alpha }{2}}=\left(s-b\right)\tan {\frac {\beta }{2}}=\left(s-c\right)\tan {\frac {\gamma }{2}}}$

${\displaystyle \rho =4r\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}=s\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}}$

${\displaystyle \rho =r\left(\cos \alpha +\cos \beta +\cos \gamma -1\right)}$

${\displaystyle \rho ={\frac {F}{s}}={\frac {abc}{4rs}}}$

${\displaystyle \rho ={\sqrt {\frac {\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}}={\frac {1}{2}}{\sqrt {\frac {\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}{a+b+c}}}}$

${\displaystyle \rho ={\frac {a}{\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}}}={\frac {b}{\cot {\frac {\gamma }{2}}+\cot {\frac {\alpha }{2}}}}={\frac {c}{\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}}}}$

${\displaystyle a\cdot b+b\cdot c+c\cdot a=s^{2}+\rho ^{2}+4\cdot \rho \cdot r}$

Major inequality ; Equality only occurs when triangle is equilateral. ${\displaystyle 2\rho \leq r}$ ${\displaystyle ABC}$

${\displaystyle \rho _{a}=s\tan {\frac {\alpha }{2}}=\left(s-b\right)\cot {\frac {\gamma }{2}}=\left(s-c\right)\cot {\frac {\beta }{2}}}$

${\displaystyle \rho _{a}=4r\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}=\left(s-a\right)\tan {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}}$

${\displaystyle \rho _{a}=r\left(-\cos \alpha +\cos \beta +\cos \gamma +1\right)}$

${\displaystyle \rho _{a}={\frac {F}{s-a}}={\frac {abc}{4r\left(s-a\right)}}}$

${\displaystyle \rho _{a}={\sqrt {\frac {s\left(s-b\right)\left(s-c\right)}{s-a}}}={\frac {1}{2}}{\sqrt {\frac {\left(a+b+c\right)\left(c+a-b\right)\left(a+b-c\right)}{b+c-a}}}}$

The circles are equal: Each formula for applies analogously to and . ${\displaystyle \rho _{a}}$ ${\displaystyle \rho _{b}}$ ${\displaystyle \rho _{c}}$

${\displaystyle {\frac {1}{\rho }}={\frac {1}{\rho _{a}}}+{\frac {1}{\rho _{b}}}+{\frac {1}{\rho _{c}}}}$

Heights

The lengths of from , or outgoing heights of the triangle are with , and referred to. ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle ABC}$ ${\displaystyle h_{a}}$ ${\displaystyle h_{b}}$ ${\displaystyle h_{c}}$

${\displaystyle h_{a}=b\sin \gamma =c\sin \beta ={\frac {2F}{a}}=2r\sin \beta \sin \gamma =2r\left(\cos \alpha +\cos \beta \cos \gamma \right)}$

${\displaystyle h_{b}=c\sin \alpha =a\sin \gamma ={\frac {2F}{b}}=2r\sin \gamma \sin \alpha =2r\left(\cos \beta +\cos \alpha \cos \gamma \right)}$

${\displaystyle h_{c}=a\sin \beta =b\sin \alpha ={\frac {2F}{c}}=2r\sin \alpha \sin \beta =2r\left(\cos \gamma +\cos \alpha \cos \beta \right)}$

${\displaystyle h_{a}={\frac {a}{\cot \beta +\cot \gamma }};\;\;\;\;\;h_{b}={\frac {b}{\cot \gamma +\cot \alpha }};\;\;\;\;\;h_{c}={\frac {c}{\cot \alpha +\cot \beta }}}$

${\displaystyle F={\frac {1}{2}}ah_{a}={\frac {1}{2}}bh_{b}={\frac {1}{2}}ch_{c}}$

${\displaystyle {\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}}={\frac {1}{\rho }}={\frac {1}{\rho _{a}}}+{\frac {1}{\rho _{b}}}+{\frac {1}{\rho _{c}}}}$

If the triangle has a right angle at (is therefore ), then applies ${\displaystyle ABC}$ ${\displaystyle C}$ ${\displaystyle \gamma =90^{\circ }}$

${\displaystyle h_{c}={\frac {ab}{c}}}$

${\displaystyle h_{a}=b}$

${\displaystyle h_{b}=a}$

Bisector

The lengths of from , or outgoing medians of the triangle is , and called. ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle ABC}$ ${\displaystyle s_{a}}$ ${\displaystyle s_{b}}$ ${\displaystyle s_{c}}$

${\displaystyle s_{a}={\frac {1}{2}}{\sqrt {2b^{2}+2c^{2}-a^{2}}}={\frac {1}{2}}{\sqrt {b^{2}+c^{2}+2bc\cos \alpha }}={\sqrt {{\frac {a^{2}}{4}}+bc\cos \alpha }}}$

${\displaystyle s_{b}={\frac {1}{2}}{\sqrt {2c^{2}+2a^{2}-b^{2}}}={\frac {1}{2}}{\sqrt {c^{2}+a^{2}+2ca\cos \beta }}={\sqrt {{\frac {b^{2}}{4}}+ca\cos \beta }}}$

${\displaystyle s_{c}={\frac {1}{2}}{\sqrt {2a^{2}+2b^{2}-c^{2}}}={\frac {1}{2}}{\sqrt {a^{2}+b^{2}+2ab\cos \gamma }}={\sqrt {{\frac {c^{2}}{4}}+ab\cos \gamma }}}$

${\displaystyle s_{a}^{2}+s_{b}^{2}+s_{c}^{2}={\frac {3}{4}}\left(a^{2}+b^{2}+c^{2}\right)}$

Bisector

We denote by , and the lengths of the of , or outgoing bisecting the triangle . ${\displaystyle w_{\alpha }}$ ${\displaystyle w_{\beta }}$ ${\displaystyle w_{\gamma }}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle ABC}$

${\displaystyle w_{\alpha }={\frac {2bc\cos {\frac {\alpha }{2}}}{b+c}}={\frac {2F}{a\cos {\frac {\beta -\gamma }{2}}}}={\frac {\sqrt {bc(b+c-a)(a+b+c)}}{b+c}}}$

${\displaystyle w_{\beta }={\frac {2ca\cos {\frac {\beta }{2}}}{c+a}}={\frac {2F}{b\cos {\frac {\gamma -\alpha }{2}}}}={\frac {\sqrt {ca(c+a-b)(a+b+c)}}{c+a}}}$

${\displaystyle w_{\gamma }={\frac {2ab\cos {\frac {\gamma }{2}}}{a+b}}={\frac {2F}{c\cos {\frac {\alpha -\beta }{2}}}}={\frac {\sqrt {ab(a+b-c)(a+b+c)}}{a+b}}}$

General trigonometry in the plane

The trigonometric functions on the unit circle :

${\displaystyle {\overline {CP}}=\sin b}$	${\displaystyle {\overline {SP}}=\cos b}$
${\displaystyle {\overline {DT}}=\tan b}$	${\displaystyle {\overline {EK}}=\cot b}$
${\displaystyle {\overline {OT}}=\operatorname {sec} \,b}$	${\displaystyle {\overline {OK}}=\operatorname {csc} \,b}$

periodicity

${\displaystyle \sin x\quad =\quad \sin(x+2n\pi );\quad n\in \mathbb {Z} }$

${\displaystyle \cos x\quad =\quad \cos(x+2n\pi );\quad n\in \mathbb {Z} }$

${\displaystyle \tan x\quad =\quad \tan(x+n\pi );\quad n\in \mathbb {Z} }$

${\displaystyle \cot x\quad =\quad \cot(x+n\pi );\quad n\in \mathbb {Z} }$

Mutual representation

The trigonometric functions can be converted into one another or represented mutually. The following relationships apply:

${\displaystyle \tan x={\frac {\sin x}{\cos x}}}$

${\displaystyle \sin ^{2}x+\cos ^{2}x=1}$      (" Trigonometric Pythagoras ")

${\displaystyle 1+\tan ^{2}x={\frac {1}{\cos ^{2}x}}=\sec ^{2}x}$

${\displaystyle 1+\cot ^{2}x={\frac {1}{\sin ^{2}x}}=\csc ^{2}x}$

(See also the section on phase shifts .)

Using these equations, the three functions that occur can be represented by one of the other two:

${\displaystyle \sin x\;=\;{\sqrt {1-\cos ^{2}x}}}$	For	${\displaystyle x\in \left[0,\pi \right[\quad =\quad [0^{\circ },180^{\circ }[}$
${\displaystyle \sin x\;=\;-{\sqrt {1-\cos ^{2}x}}}$	For	${\displaystyle x\in \left[\pi ,2\pi \right[\quad =\quad [180^{\circ },360^{\circ }[}$
${\displaystyle \sin x\;=\;{\frac {\tan x}{\sqrt {1+\tan ^{2}x}}}}$	For	${\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}$
${\displaystyle \sin x\;=\;-{\frac {\tan x}{\sqrt {1+\tan ^{2}x}}}}$	For	${\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}$
${\displaystyle \cos x\;=\;{\sqrt {1-\sin ^{2}x}}}$	For	${\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left[{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;[270^{\circ },360^{\circ }[}$
${\displaystyle \cos x\;=\;-{\sqrt {1-\sin ^{2}x}}}$	For	${\displaystyle x\in \left[{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad [90^{\circ },270^{\circ }[}$
${\displaystyle \cos x={\frac {1}{\sqrt {1+\tan ^{2}x}}}}$	For	${\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}$
${\displaystyle \cos x=-{\frac {1}{\sqrt {1+\tan ^{2}x}}}}$	For	${\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}$
${\displaystyle \tan x={\frac {\sqrt {1-\cos ^{2}x}}{\cos x}}}$	For	${\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {\pi }{2}},\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]90^{\circ },180^{\circ }[}$
${\displaystyle \tan x=-{\frac {\sqrt {1-\cos ^{2}x}}{\cos x}}}$	For	${\displaystyle x\in \left[\pi ,{\frac {3\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [180^{\circ },270^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}$
${\displaystyle \tan x={\frac {\sin x}{\sqrt {1-\sin ^{2}x}}}}$	For	${\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}$
${\displaystyle \tan x=-{\frac {\sin x}{\sqrt {1-\sin ^{2}x}}}}$	For	${\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}$

Sign of the trigonometric functions

${\displaystyle \sin x>0\quad {\text{für}}\quad x\in \left]0^{\circ },180^{\circ }\right[}$

${\displaystyle \sin x<0\quad {\text{für}}\quad x\in \left]180^{\circ },360^{\circ }\right[}$

${\displaystyle \cos x>0\quad {\text{für}}\quad x\in \left[0^{\circ },90^{\circ }\right[\cup \left]270^{\circ },360^{\circ }\right]}$

${\displaystyle \cos x<0\quad {\text{für}}\quad x\in \left]90^{\circ },270^{\circ }\right[}$

${\displaystyle \tan x>0\quad {\text{für}}\quad x\in \left]0^{\circ },90^{\circ }\right[\cup \left]180^{\circ },270^{\circ }\right[}$

${\displaystyle \tan x<0\quad {\text{für}}\quad x\in \left]90^{\circ },180^{\circ }\right[\cup \left]270^{\circ },360^{\circ }\right[}$

The signs of , and agree with those of their reciprocal functions , and . ${\displaystyle \cot }$ ${\displaystyle \sec }$ ${\displaystyle \csc }$ ${\displaystyle \tan }$ ${\displaystyle \cos }$ ${\displaystyle \sin }$

Important functional values

Representation of important function values ​​of sine and cosine on the unit circle

${\displaystyle \alpha }$ (°)	${\displaystyle \alpha }$ (wheel)	${\displaystyle \sin \alpha }$	${\displaystyle \cos \alpha }$	${\displaystyle \tan \alpha }$	${\displaystyle \cot \alpha }$
${\displaystyle 0^{\circ }}$	${\displaystyle \,0}$	${\displaystyle \,0}$	${\displaystyle \,1}$	${\displaystyle \,0}$	${\displaystyle \pm \infty }$
${\displaystyle 15^{\circ }}$	${\displaystyle {\tfrac {\pi }{12}}}$	${\displaystyle {\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})}$	${\displaystyle {\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})}$	${\displaystyle 2-{\sqrt {3}}}$	${\displaystyle 2+{\sqrt {3}}}$
${\displaystyle 18^{\circ }}$	${\displaystyle {\tfrac {\pi }{10}}}$	${\displaystyle {\tfrac {1}{4}}\left({\sqrt {5}}-1\right)}$	${\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}$	${\displaystyle {\sqrt {5+2{\sqrt {5}}}}}$
${\displaystyle 30^{\circ }}$	${\displaystyle {\tfrac {\pi }{6}}}$	${\displaystyle {\tfrac {1}{2}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}$	${\displaystyle {\tfrac {1}{3}}{\sqrt {3}}}$	${\displaystyle {\sqrt {3}}}$
${\displaystyle 36^{\circ }}$	${\displaystyle {\tfrac {\pi }{5}}}$	${\displaystyle {\tfrac {1}{4}}{\sqrt {10-2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{4}}\left(1+{\sqrt {5}}\right)}$	${\displaystyle {\sqrt {5-2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{5}}{\sqrt {25+10{\sqrt {5}}}}}$
${\displaystyle 45^{\circ }}$	${\displaystyle {\tfrac {\pi }{4}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}$	${\displaystyle 1\,}$	${\displaystyle 1\,}$
${\displaystyle 54^{\circ }}$	${\displaystyle {\tfrac {3\pi }{10}}}$	${\displaystyle {\tfrac {1}{4}}\left(1+{\sqrt {5}}\right)}$	${\displaystyle {\tfrac {1}{4}}{\sqrt {10-2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{5}}{\sqrt {25+10{\sqrt {5}}}}}$	${\displaystyle {\sqrt {5-2{\sqrt {5}}}}}$
${\displaystyle 60^{\circ }}$	${\displaystyle {\tfrac {\pi }{3}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}$	${\displaystyle {\tfrac {1}{2}}}$	${\displaystyle {\sqrt {3}}}$	${\displaystyle {\tfrac {1}{3}}{\sqrt {3}}}$
${\displaystyle 72^{\circ }}$	${\displaystyle {\tfrac {2\pi }{5}}}$	${\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{4}}\left({\sqrt {5}}-1\right)}$	${\displaystyle {\sqrt {5+2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}$
${\displaystyle 75^{\circ }}$	${\displaystyle {\tfrac {5\pi }{12}}}$	${\displaystyle {\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})}$	${\displaystyle {\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})}$	${\displaystyle 2+{\sqrt {3}}}$	${\displaystyle 2-{\sqrt {3}}}$
${\displaystyle 90^{\circ }}$	${\displaystyle {\tfrac {\pi }{2}}}$	${\displaystyle \,1}$	${\displaystyle \,0}$	${\displaystyle \pm \infty }$	${\displaystyle \,0}$
${\displaystyle 108^{\circ }}$	${\displaystyle {\tfrac {3\pi }{5}}}$	${\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}$	${\displaystyle {\tfrac {1}{4}}\left(1-{\sqrt {5}}\right)}$	${\displaystyle -{\sqrt {5+2{\sqrt {5}}}}}$	${\displaystyle -{\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}$
${\displaystyle 120^{\circ }}$	${\displaystyle {\tfrac {2\pi }{3}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}$	${\displaystyle -{\tfrac {1}{2}}}$	${\displaystyle -{\sqrt {3}}}$	${\displaystyle -{\tfrac {1}{3}}{\sqrt {3}}}$
${\displaystyle 135^{\circ }}$	${\displaystyle {\tfrac {3\pi }{4}}}$	${\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}$	${\displaystyle -{\tfrac {1}{2}}{\sqrt {2}}}$	${\displaystyle -1\,}$	${\displaystyle -1\,}$
${\displaystyle 180^{\circ }}$	${\displaystyle \pi \,}$	${\displaystyle \,0}$	${\displaystyle \,-1}$	${\displaystyle \,0}$	${\displaystyle \pm \infty }$
${\displaystyle 270^{\circ }}$	${\displaystyle {\tfrac {3\pi }{2}}}$	${\displaystyle \,-1}$	${\displaystyle \,0}$	${\displaystyle \pm \infty }$	${\displaystyle \,0}$
${\displaystyle 360^{\circ }}$	${\displaystyle 2\pi }$	${\displaystyle \,0}$	${\displaystyle \,1}$	${\displaystyle \,0}$	${\displaystyle \pm \infty }$

Many other values ​​can be represented.

Symmetries

The trigonometric functions have simple symmetries:

${\displaystyle \sin(-x)=-\sin x\;}$

${\displaystyle \cos(-x)=+\cos x\;}$

${\displaystyle \tan(-x)=-\tan x\;}$

${\displaystyle \cot(-x)=-\cot x\;}$

${\displaystyle \sec(-x)=+\sec x\;}$

${\displaystyle \csc(-x)=-\csc x\;}$

Phase shifts

${\displaystyle \sin \left(x+{\frac {\pi }{2}}\right)=\cos x\;\quad {\text{bzw.}}\quad \sin \left(x+90^{\circ }\right)=\cos x\;}$

${\displaystyle \cos \left(x+{\frac {\pi }{2}}\right)=-\sin x\;\quad {\text{bzw.}}\quad \cos \left(x+90^{\circ }\right)=-\sin x\;}$

${\displaystyle \tan \left(x+{\frac {\pi }{2}}\right)=-\cot x\;\quad {\text{bzw.}}\quad \tan \left(x+90^{\circ }\right)=-\cot x\;}$

${\displaystyle \cot \left(x+{\frac {\pi }{2}}\right)=-\tan x\;\quad {\text{bzw.}}\quad \cot \left(x+90^{\circ }\right)=-\tan x\;}$

Reduction to acute angles

${\displaystyle \sin x\ \;=\;\;\;\sin \left(\pi -x\right)\,\quad {\text{bzw.}}\quad \sin x\ =\;\;\;\sin \left(180^{\circ }-x\right)}$

${\displaystyle \cos x\ \,=-\cos \left(\pi -x\right)\quad {\text{bzw.}}\quad \cos x\ =-\cos \left(180^{\circ }-x\right)}$

${\displaystyle \tan x\ =-\tan \left(\pi -x\right)\quad {\text{bzw.}}\quad \tan x\ =-\tan \left(180^{\circ }-x\right)}$

Represented by the tangent of the half angle

With the label , the following relationships apply to anything ${\displaystyle t=\tan {\tfrac {x}{2}}}$ ${\displaystyle x}$

${\displaystyle \sin x={\frac {2t}{1+t^{2}}},}$		${\displaystyle \cos x={\frac {1-t^{2}}{1+t^{2}}},}$
${\displaystyle \tan x={\frac {2t}{1-t^{2}}},}$		${\displaystyle \cot x={\frac {1-t^{2}}{2t}},}$
${\displaystyle \sec x={\frac {1+t^{2}}{1-t^{2}}},}$		${\displaystyle \csc x={\frac {1+t^{2}}{2t}}.}$

Addition theorems

For sine and cosine, the addition theorems can be derived from the concatenation of two rotations about the angle or . This is fundamentally possible; Reading the formulas from the product of two rotary matrices of the plane is much easier . Alternatively, the addition theorems follow from the application of Euler's formula to the relationship . The results for the double sign are obtained by applying the symmetries . ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle \mathbb {R} ^{2}}$ ${\displaystyle \textstyle e^{i(x+y)}=e^{ix}\cdot e^{iy}}$

${\displaystyle \sin(x\pm y)=\sin x\cdot \cos y\pm \cos x\cdot \sin y}$

${\displaystyle \cos(x\pm y)=\cos x\cdot \cos y\mp \sin x\cdot \sin y}$

By expanding with or and simplifying the double fraction: ${\displaystyle \textstyle {1 \over \cos x\cos y}}$ ${\displaystyle \textstyle {1 \over \sin x\sin y}}$

${\displaystyle \tan(x\pm y)={\frac {\sin(x\pm y)}{\cos(x\pm y)}}={\frac {\tan x\pm \tan y}{1\mp \tan x\;\tan y}}}$

${\displaystyle \cot(x\pm y)={\frac {\cos(x\pm y)}{\sin(x\pm y)}}={\frac {\cot x\cot y\mp 1}{\cot y\pm \cot x}}}$

For the double angle functions follow from this , for the phase shifts . ${\displaystyle x=y}$ ${\displaystyle y=\pi /2}$

${\displaystyle \sin(x+y)\cdot \sin(x-y)=\cos ^{2}y-\cos ^{2}x=\sin ^{2}x-\sin ^{2}y}$

${\displaystyle \cos(x+y)\cdot \cos(x-y)=\cos ^{2}y-\sin ^{2}x=\cos ^{2}x-\sin ^{2}y}$

Addition theorems for arc functions

The following addition theorems apply to the arc functions

Summands	Molecular formula	Scope
${\displaystyle \arcsin x+\arcsin y=}$	${\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle xy\leq 0}$ or ${\displaystyle x^{2}+y^{2}\leq 1}$
	${\displaystyle \pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle x>0}$and and ${\displaystyle y>0}$ ${\displaystyle x^{2}+y^{2}>1}$
	${\displaystyle -\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle x<0}$and and ${\displaystyle y<0}$ ${\displaystyle x^{2}+y^{2}>1}$
${\displaystyle \arcsin x-\arcsin y=}$	${\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle xy\geq 0}$ or ${\displaystyle x^{2}+y^{2}\leq 1}$
	${\displaystyle \pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle x>0}$and and ${\displaystyle y<0}$ ${\displaystyle x^{2}+y^{2}>1}$
	${\displaystyle -\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}$	${\displaystyle x<0}$and and ${\displaystyle y>0}$ ${\displaystyle x^{2}+y^{2}>1}$
${\displaystyle \arccos x+\arccos y=}$	${\displaystyle \arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}$	${\displaystyle x+y\geq 0}$
	${\displaystyle 2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}$	${\displaystyle x+y<0}$
${\displaystyle \arccos x-\arccos y=}$	${\displaystyle -\arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}$	${\displaystyle x+y\geq 0}$
	${\displaystyle \arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}$	${\displaystyle x+y<0}$
${\displaystyle \arctan x+\arctan y=}$	${\displaystyle \arctan \left({\frac {x+y}{1-xy}}\right)}$	${\displaystyle xy<1}$
	${\displaystyle \pi +\arctan \left({\frac {x+y}{1-xy}}\right)}$	${\displaystyle x>0}$ and ${\displaystyle xy>1}$
	${\displaystyle -\pi +\arctan \left({\frac {x+y}{1-xy}}\right)}$	${\displaystyle x<0}$ and ${\displaystyle xy>1}$
${\displaystyle \arctan x-\arctan y=}$	${\displaystyle \arctan \left({\frac {x-y}{1+xy}}\right)}$	${\displaystyle xy>-1}$
	${\displaystyle \pi +\arctan \left({\frac {x-y}{1+xy}}\right)}$	${\displaystyle x>0}$ and ${\displaystyle xy<-1}$
	${\displaystyle -\pi +\arctan \left({\frac {x-y}{1+xy}}\right)}$	${\displaystyle x<0}$ and ${\displaystyle xy<-1}$

Double angle functions

${\displaystyle \sin(2x)=2\sin x\;\cos x={\frac {2\tan x}{1+\tan ^{2}x}}}$

${\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x=1-2\sin ^{2}x=2\cos ^{2}x-1={\frac {1-\tan ^{2}x}{1+\tan ^{2}x}}}$

${\displaystyle \tan(2x)={\frac {2\tan x}{1-\tan ^{2}x}}={\frac {2}{\cot x-\tan x}}}$

${\displaystyle \cot(2x)={\frac {\cot ^{2}x-1}{2\cot x}}={\frac {\cot x-\tan x}{2}}}$

Trigonometric functions for further multiples

The formulas for multiples are usually calculated using the complex numbers from the Euler formula and the DeMoivre formula . This results in . Decomposition into real and imaginary parts then provides the formulas for and or the general series representation. ${\displaystyle z=r\left(\cos \phi +i\sin \phi \right)\iff z^{n}=r^{n}\left(\cos \phi +i\sin \phi \right)^{n}}$ ${\displaystyle z^{n}=r^{n}\left(\cos \left(n\phi \right)+i\sin \left(n\phi \right)\right)}$ ${\displaystyle \cos \left(n\phi \right)+i\sin \left(n\phi \right)=\left(\cos \phi +i\sin \phi \right)^{n}}$ ${\displaystyle \cos }$ ${\displaystyle \sin }$

The formula for is related to the Chebyshev polynomials . ${\displaystyle \cos(nx)}$ ${\displaystyle T_{n}(\cos x)=\cos(nx)}$

${\displaystyle \sin(3x)=3\sin x-4\sin ^{3}x\,}$

${\displaystyle =\;\sin x\left(4\cos ^{2}x-1\right)}$

${\displaystyle \sin(4x)=8\sin x\;\cos ^{3}x-4\sin x\;\cos x}$

${\displaystyle =\;\sin x\left(8\cos ^{3}x-4\cos x\right)}$

${\displaystyle \sin(5x)=5\sin x-20\sin ^{3}x+16\sin ^{5}x\;}$

${\displaystyle =\;\sin x\left(16\cos ^{4}x-12\cos ^{2}x+1\right)}$

${\displaystyle \sin(nx)=n\;\sin x\;\cos ^{n-1}x-{n \choose 3}\sin ^{3}x\;\cos ^{n-3}x+{n \choose 5}\sin ^{5}x\;\cos ^{n-5}x\;-\ldots +\ldots }$

${\displaystyle =\;\sum _{j=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{j}{n \choose 2j+1}\sin ^{2j+1}x\;\cos ^{n-2j-1}x}$

${\displaystyle =\;\sin x\sum _{k=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{k}{n-k-1 \choose k}2^{n-2k-1}\cos ^{n-2k-1}x}$

${\displaystyle \cos(3x)=4\cos ^{3}x-3\cos x\,}$

${\displaystyle \cos(4x)=8\cos ^{4}x-8\cos ^{2}x+1\,}$

${\displaystyle \cos(5x)=16\cos ^{5}x-20\cos ^{3}x+5\cos x\,}$

${\displaystyle \cos(6x)=32\cos ^{6}x-48\cos ^{4}x+18\cos ^{2}x-1\,}$

${\displaystyle \cos(nx)=\cos ^{n}x-{n \choose 2}\sin ^{2}x\;\cos ^{n-2}x+{n \choose 4}\sin ^{4}x\;\cos ^{n-4}x\;-\ldots +\ldots }$

${\displaystyle =\;\sum _{j=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{j}{n \choose 2j}\sin ^{2j}x\;\cos ^{n-2j}x}$

${\displaystyle \tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}}$

${\displaystyle \tan(4x)={\frac {4\tan x-4\tan ^{3}x}{1-6\tan ^{2}x+\tan ^{4}x}}}$

${\displaystyle \cot(3x)={\frac {\cot ^{3}x-3\cot x}{3\cot ^{2}x-1}}}$

${\displaystyle \cot(4x)={\frac {\cot ^{4}x-6\cot ^{2}x+1}{4\cot ^{3}x-4\cot x}}}$

Half-angle formulas

The half- angle formulas , which can be derived from the double-angle formulas by means of substitution, are used to calculate the function value of the half argument :

${\displaystyle \sin {\frac {x}{2}}={\sqrt {\frac {1-\cos x}{2}}}\quad {\text{für}}\quad x\in \left[0,2\pi \right]}$

${\displaystyle \cos {\frac {x}{2}}={\sqrt {\frac {1+\cos x}{2}}}\quad {\text{für}}\quad x\in \left[-\pi ,\pi \right]}$

${\displaystyle \tan {\frac {x}{2}}={\frac {1-\cos x}{\sin x}}={\frac {\sin x}{1+\cos x}}={\sqrt {\frac {1-\cos x}{1+\cos x}}}\quad {\text{für}}\quad x\in \left[0,\pi \right[}$

${\displaystyle \cot {\frac {x}{2}}={\frac {1+\cos x}{\sin x}}={\frac {\sin x}{1-\cos x}}={\sqrt {\frac {1+\cos x}{1-\cos x}}}\quad {\text{für}}\quad x\in \left]0,\pi \right]}$

In addition:

${\displaystyle \tan {\frac {x}{2}}={\frac {\tan x}{1+{\sqrt {1+\tan ^{2}x}}}}\quad {\text{für}}\quad x\in \left]-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right[}$

${\displaystyle \cot {\frac {x}{2}}=\cot x+{\sqrt {1+\cot ^{2}x}}\quad {\text{für}}\quad x\in \left]0,\pi \right[}$

See also: half-angle block

Sums of two trigonometric functions (identities)

Identities can be derived from the addition theorems , with the help of which the sum of two trigonometric functions can be represented as a product:

${\displaystyle \sin x+\sin y=2\sin {\frac {x+y}{2}}\cos {\frac {x-y}{2}}}$

${\displaystyle \sin x-\sin y=2\cos {\frac {x+y}{2}}\sin {\frac {x-y}{2}}}$

${\displaystyle \cos x+\cos y=2\cos {\frac {x+y}{2}}\cos {\frac {x-y}{2}}}$

${\displaystyle \cos x-\cos y=-2\sin {\frac {x+y}{2}}\sin {\frac {x-y}{2}}}$

${\displaystyle \left.{\begin{matrix}\tan x+\tan y={\dfrac {\sin(x+y)}{\cos x\cos y}}\\[1em]\tan x-\tan y={\dfrac {\sin(x-y)}{\cos x\cos y}}\end{matrix}}\right\}\Rightarrow \tan x\pm \tan y={\frac {\sin(x\pm y)}{\cos x\cos y}}}$

${\displaystyle \left.{\begin{matrix}\cot x+\cot y={\dfrac {\sin(y+x)}{\sin x\sin y}}\\[1em]\cot x-\cot y={\dfrac {\sin(y-x)}{\sin x\sin y}}\end{matrix}}\right\}\Rightarrow \cot x\pm \cot y={\frac {\sin(y\pm x)}{\sin x\sin y}}}$

This results in special cases:

${\displaystyle \cos x+\sin x={\sqrt {2}}\cdot \sin \left(x+{\frac {\pi }{4}}\right)={\sqrt {2}}\cdot \cos \left(x-{\frac {\pi }{4}}\right)}$

${\displaystyle \cos x-\sin x={\sqrt {2}}\cdot \cos \left(x+{\frac {\pi }{4}}\right)=-{\sqrt {2}}\cdot \sin \left(x-{\frac {\pi }{4}}\right)}$

Products of the trigonometric functions

Products of the trigonometric functions can be calculated using the following formulas:

${\displaystyle \sin x\;\sin y={\frac {1}{2}}{\Big (}\cos(x-y)-\cos(x+y){\Big )}}$

${\displaystyle \cos x\;\cos y={\frac {1}{2}}{\Big (}\cos(x-y)+\cos(x+y){\Big )}}$

${\displaystyle \sin x\;\cos y={\frac {1}{2}}{\Big (}\sin(x-y)+\sin(x+y){\Big )}}$

${\displaystyle \tan x\;\tan y={\frac {\tan x+\tan y}{\cot x+\cot y}}=-{\frac {\tan x-\tan y}{\cot x-\cot y}}}$

${\displaystyle \cot x\;\cot y={\frac {\cot x+\cot y}{\tan x+\tan y}}=-{\frac {\cot x-\cot y}{\tan x-\tan y}}}$

${\displaystyle \tan x\;\cot y={\frac {\tan x+\cot y}{\cot x+\tan y}}=-{\frac {\tan x-\cot y}{\cot x-\tan y}}}$

${\displaystyle \sin x\;\sin y\;\sin z={\frac {1}{4}}{\Big (}\sin(x+y-z)+\sin(y+z-x)+\sin(z+x-y)-\sin(x+y+z){\Big )}}$

${\displaystyle \cos x\;\cos y\;\cos z={\frac {1}{4}}{\Big (}\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)+\cos(x+y+z){\Big )}}$

${\displaystyle \sin x\;\sin y\;\cos z={\frac {1}{4}}{\Big (}-\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)-\cos(x+y+z){\Big )}}$

${\displaystyle \sin x\;\cos y\;\cos z={\frac {1}{4}}{\Big (}\sin(x+y-z)-\sin(y+z-x)+\sin(z+x-y)+\sin(x+y+z){\Big )}}$

From the double angle function for it also follows: ${\displaystyle \sin(2x)}$

${\displaystyle \sin x\;\cos x={\frac {1}{2}}\sin(2x)}$

Powers of the trigonometric functions

Sine

${\displaystyle \sin ^{2}x={\frac {1}{2}}\ {\Big (}1-\cos(2x){\Big )}}$

${\displaystyle \sin ^{3}x={\frac {1}{4}}\ {\Big (}3\,\sin x-\sin(3x){\Big )}}$

${\displaystyle \sin ^{4}x={\frac {1}{8}}\ {\Big (}3-4\,\cos(2x)+\cos(4x){\Big )}}$

${\displaystyle \sin ^{5}x={\frac {1}{16}}\ {\Big (}10\,\sin x-5\,\sin(3x)+\sin(5x){\Big )}}$

${\displaystyle \sin ^{6}x={\frac {1}{32}}\ {\Big (}10-15\,\cos(2x)+6\,\cos(4x)-\cos(6x){\Big )}}$

${\displaystyle \sin ^{n}x={\frac {1}{2^{n}}}\,\sum _{k=0}^{n}{n \choose k}\,\cos {\Big (}(n-2k)(x-{\frac {\pi }{2}}\ ){\Big )}\ ;\quad n\in \mathbb {N} }$

${\displaystyle \sin ^{n}x={\frac {1}{2^{n}}}\,{n \choose {\frac {n}{2}}}+{\frac {1}{2^{n-1}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{{\frac {n}{2}}-k}\,{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ gerade }}}$

${\displaystyle \sin ^{n}x={\frac {1}{2^{n-1}}}\,\sum _{k=0}^{\frac {n-1}{2}}(-1)^{{\frac {n-1}{2}}-k}\,{n \choose k}\,\sin {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ ungerade}}}$

cosine

${\displaystyle \cos ^{2}x={\frac {1}{2}}\ {\Big (}1+\cos(2x){\Big )}}$

${\displaystyle \cos ^{3}x={\frac {1}{4}}\ {\Big (}3\,\cos x+\cos(3x){\Big )}}$

${\displaystyle \cos ^{4}x={\frac {1}{8}}\ {\Big (}3+4\,\cos(2x)+\cos(4x){\Big )}}$

${\displaystyle \cos ^{5}x={\frac {1}{16}}\ {\Big (}10\,\cos x+5\,\cos(3x)+\cos(5x){\Big )}}$

${\displaystyle \cos ^{6}x={\frac {1}{32}}\ {\Big (}10+15\,\cos(2x)+6\,\cos(4x)+\cos(6x){\Big )}}$

${\displaystyle \cos ^{n}x={\frac {1}{2^{n}}}\,\sum _{k=0}^{n}{n \choose k}\,\cos((n-2k)x);\quad n\in \mathbb {N} }$

${\displaystyle \cos ^{n}x={\frac {1}{2^{n}}}\,{n \choose {\frac {n}{2}}}+{\frac {1}{2^{n-1}}}\sum _{k=0}^{{\frac {n}{2}}-1}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ gerade }}}$

${\displaystyle \cos ^{n}x={\frac {1}{2^{n-1}}}\,\sum _{k=0}^{\frac {n-1}{2}}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ ungerade}}}$

tangent

${\displaystyle \tan ^{2}x={\frac {1-\cos(2x)}{1+\cos(2x)}}=\sec ^{2}(x)-1}$

Conversion into other trigonometric functions

${\displaystyle \sin(\arccos x)=\cos(\arcsin x)={\sqrt {1-x^{2}}}}$

${\displaystyle \sin(\arctan x)=\cos(\operatorname {arccot} x)={\frac {x}{\sqrt {1+x^{2}}}}}$

${\displaystyle \sin(\operatorname {arccot} x)=\cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}}$

${\displaystyle \tan(\arcsin x)=\cot(\arccos x)={\frac {x}{\sqrt {1-x^{2}}}}}$

${\displaystyle \tan(\arccos x)=\cot(\arcsin x)={\frac {\sqrt {1-x^{2}}}{x}}}$

${\displaystyle \tan(\operatorname {arccot} x)=\cot(\arctan x)={\frac {1}{x}}}$

Further formulas for the case α + β + γ = 180 °

The following formulas apply to any plane triangles and follow after long term transformations , as long as the functions in the formulas are well-defined (the latter only applies to formulas in which the tangent and cotangent occur). ${\displaystyle \alpha +\beta +\gamma =180^{\circ }}$

${\displaystyle \tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \cdot \tan \beta \cdot \tan \gamma \,}$

${\displaystyle \cot \beta \cdot \cot \gamma +\cot \gamma \cdot \cot \alpha +\cot \alpha \cdot \cot \beta =1}$

${\displaystyle \cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}=\cot {\frac {\alpha }{2}}\cdot \cot {\frac {\beta }{2}}\cdot \cot {\frac {\gamma }{2}}}$

${\displaystyle \tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}+\tan {\frac {\gamma }{2}}\tan {\frac {\alpha }{2}}+\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}=1}$

${\displaystyle \sin \alpha +\sin \beta +\sin \gamma =4\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}}$

${\displaystyle -\sin \alpha +\sin \beta +\sin \gamma =4\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}}$

${\displaystyle \cos \alpha +\cos \beta +\cos \gamma =4\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1}$

${\displaystyle -\cos \alpha +\cos \beta +\cos \gamma =4\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}-1}$

${\displaystyle \sin(2\alpha )+\sin(2\beta )+\sin(2\gamma )=4\sin \alpha \sin \beta \sin \gamma }$

${\displaystyle -\sin(2\alpha )+\sin(2\beta )+\sin(2\gamma )=4\sin \alpha \cos \beta \cos \gamma }$

${\displaystyle \cos(2\alpha )+\cos(2\beta )+\cos(2\gamma )=-4\cos \alpha \cos \beta \cos \gamma -1}$

${\displaystyle -\cos(2\alpha )+\cos(2\beta )+\cos(2\gamma )=-4\cos \alpha \sin \beta \sin \gamma +1}$

${\displaystyle \sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma =2\cos \alpha \cos \beta \cos \gamma +2}$

${\displaystyle -\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma =2\cos \alpha \sin \beta \sin \gamma }$

${\displaystyle \cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma =-2\cos \alpha \cos \beta \cos \gamma +1}$

${\displaystyle -\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma =-2\cos \alpha \sin \beta \sin \gamma +1}$

${\displaystyle -\sin ^{2}(2\alpha )+\sin ^{2}(2\beta )+\sin ^{2}(2\gamma )=-2\cos(2\alpha )\,\sin(2\beta )\,\sin(2\gamma )}$

${\displaystyle -\cos ^{2}(2\alpha )+\cos ^{2}(2\beta )+\cos ^{2}(2\gamma )=2\cos(2\alpha )\,\sin(2\beta )\,\sin(2\gamma )+1}$

${\displaystyle \sin ^{2}\left({\frac {\alpha }{2}}\right)+\sin ^{2}\left({\frac {\beta }{2}}\right)+\sin ^{2}\left({\frac {\gamma }{2}}\right)+2\sin \left({\frac {\alpha }{2}}\right)\,\sin \left({\frac {\beta }{2}}\right)\,\sin \left({\frac {\gamma }{2}}\right)=1}$

Sinusoid and linear combination with the same phase

${\displaystyle {\begin{aligned}a\sin \alpha +b\cos \alpha =&{\begin{cases}{\sqrt {a^{2}+b^{2}}}\sin \left(\alpha +\arctan \left({\tfrac {b}{a}}\right)\right)&{\text{, für alle }}a>0\\{\sqrt {a^{2}+b^{2}}}\cos \left(\alpha -\arctan \left({\tfrac {a}{b}}\right)\right)&{\text{, für alle }}b>0\end{cases}}\end{aligned}}}$

${\displaystyle {\begin{aligned}a\cos \alpha +b\sin \alpha =\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}}\cos \left(\alpha +\arctan \left(-{\tfrac {b}{a}}\right)\right)\end{aligned}}}$

${\displaystyle a\sin(x+\alpha )+b\sin(x+\beta )={\sqrt {a^{2}+b^{2}+2ab\cos(\alpha -\beta )}}\cdot \sin(x+\delta ),}$

in which ${\displaystyle \delta =\operatorname {atan2} (a\sin \alpha +b\sin \beta ,a\cos \alpha +b\cos \beta ).}$

Is more general

${\displaystyle \sum _{i}a_{i}\sin(x+\delta _{i})=a\sin(x+\delta ),}$

in which

${\displaystyle a^{2}=\sum _{i,j}a_{i}a_{j}\cos(\delta _{i}-\delta _{j})}$

and

${\displaystyle \delta =\operatorname {atan2} \left(\sum _{i}a_{i}\sin \delta _{i},\sum _{i}a_{i}\cos \delta _{i}\right).}$

Derivatives and antiderivatives

See the formula collection derivatives and antiderivatives

Series development

The sine (red) compared to its 7th Taylor polynomial (green)

As in other calculus , all angles are given in radians .

It can be shown that the cosine is the derivative of the sine and the derivative of the cosine is the negative sine. Having these derivatives, one can expand the Taylor series (easiest with the expansion point ) and show that the following identities hold for all of the real numbers . These series are used to define the trigonometric functions for complex arguments ( or denote the Bernoulli numbers ): ${\displaystyle x=0}$ ${\displaystyle x}$ ${\displaystyle B_{n}}$ ${\displaystyle \beta _{n}}$

${\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\\&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}}$

${\displaystyle {\begin{aligned}\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}\\&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}}$

${\displaystyle {\begin{aligned}\tan x&=\sum _{n=1}^{\infty }(-1)^{n}{\frac {2^{2n}(1-2^{2n})\beta _{2n}}{(2n)!}}x^{2n-1}=\sum _{n=1}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{n}}{(2n)!}}x^{2n-1}\\&=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+{\frac {62}{2835}}x^{9}+\,\cdots \qquad |x|<{\tfrac {\pi }{2}}\end{aligned}}}$

${\displaystyle {\begin{aligned}\cot x&={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n-1}2^{2n}\beta _{2n}}{(2n)!}}x^{2n-1}={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {2^{2n}B_{n}}{(2n)!}}x^{2n-1}\\&={\frac {1}{x}}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-{\frac {1}{4725}}x^{7}-\,\cdots ,\qquad 0<|x|<\pi \end{aligned}}}$

Product development

${\displaystyle \sin(x)=x\prod _{k=1}^{\infty }\left(1-{\frac {x^{2}}{k^{2}\pi ^{2}}}\right)}$

${\displaystyle \cos(x)=\prod _{k=1}^{\infty }\left(1-{\frac {4x^{2}}{(2k-1)^{2}\pi ^{2}}}\right)}$

${\displaystyle \sin(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{{\frac {\pi }{2}}+n\pi }}\right)}$

${\displaystyle \cos(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{{\frac {\pi }{2}}+n\pi }}\right)}$

${\displaystyle \tan(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)}$

${\displaystyle \csc(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi }}\right)}$

${\displaystyle \sec(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)}$

${\displaystyle \cot(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{x+n\pi }}\right)}$

Connection with the complex exponential function

Further, there is between the functions , and the complex exponential function of the following relationship: ${\displaystyle \sin x}$ ${\displaystyle \cos x}$ ${\displaystyle \exp(\mathrm {i} x)}$

${\displaystyle \exp(\pm \mathrm {i} x)=\cos x\pm \mathrm {i} \sin x=e^{\pm \mathrm {i} x}}$( Euler's formula )

There is still writing. ${\displaystyle \cos {x}+\mathrm {i} \sin {x}=:\operatorname {cis} (x)}$

Due to the symmetries mentioned above, the following also applies:

${\displaystyle \cos x={\frac {\exp(\mathrm {i} x)+\exp(-\mathrm {i} x)}{2}}}$

${\displaystyle \sin x={\frac {\exp(\mathrm {i} x)-\exp(-\mathrm {i} x)}{2\mathrm {i} }}}$

With these relationships, some addition theorems can be derived particularly easily and elegantly.

Spherical trigonometry

A collection of formulas for the right-angled and the general triangle on the spherical surface can be found in a separate chapter.

Literature, web links

• Abramowitz-Stegun : online (formulas, theory part - without the pure table part) ; an HTML or PDF version can (legally) be downloaded.

Individual evidence

1. ↑ Die Wurzel 2006/04 + 05, 104ff., Without proof
2. ↑ Joachim Mohr: Cosine, sine and tangent values , accessed on June 1, 2016
3. ^ ^{A b} Otto Forster: Analysis 1. Differential and integral calculus of a variable. vieweg 1983, p. 87.
4. ^ IN Bronstein, KA Semendjajew: Taschenbuch der Mathematik . 19th edition, 1979. BG Teubner Verlagsgesellschaft, Leipzig. P. 237.
5. ↑ Milton Abramowitz and Irene A. Stegun, March 22 , 2015 , (see above "Web Links ")
6. ↑ Milton Abramowitz and Irene A. Stegun, 4.3.27 , (see also above "Weblinks")
7. ↑ Milton Abramowitz and Irene A. Stegun, 4.3.29 , (see above "Weblinks")
8. ^ IS Gradshteyn and IM Ryzhik , Table of Integrals, Series, and Products , Academic Press, 5th edition (1994). ISBN 0-12-294755-X 1.333.4
9. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.331.3 (In this formula, however, Gradshteyn / Ryzhik contains a sign error )
10. ↑ ^{a b c d e f g h i j k l m n o} I. N. Bronstein, KA Semendjajew, Taschenbuch der Mathematik , BG Teubner Verlagsgesellschaft Leipzig. 19th edition 1979. 2.5.2.1.3
11. ↑ Milton Abramowitz and Irene A. Stegun, 4.3.28 , (see above "Weblinks")
12. ↑ Milton Abramowitz and Irene A. Stegun, March 4th , 30th , (see above "Weblinks")
13. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.335.4
14. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.335.5
15. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.331.3
16. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.321.1
17. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.321.2
18. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.321.3
19. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.321.4
20. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.321.5
21. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.323.1
22. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.323.2
23. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.323.3
24. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.323.4
25. ↑ IS Gradshteyn and IM Ryzhik, ibid 1.323.5
26. ^ Weisstein, Eric W .: Harmonic Addition Theorem. Retrieved January 20, 2018 . 
27. ↑ Milton Abramowitz and Irene A. Stegun, March 4th , 67 , (see above "Weblinks")
28. ↑ Milton Abramowitz and Irene A. Stegun, 4.3.70 , (see above "Weblinks")
29. ^ Herbert Amann, Joachim Escher: Analysis I, Birkhäuser Verlag, Basel 2006, 3rd edition, pp. 292 and 298