List of trigonometric identities

In trigonometry, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle.

These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.

Pythagorean identities

Main article: Pythagorean trigonometric identity

 

Trigonometric functions and their reciprocals on the unit circle. All of the right-angled triangles are similar, i.e. the ratios between their corresponding sides are the same. For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them. The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. The triangle shaded blue illustrates the identity ${\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta }$ , and the red triangle shows that ${\displaystyle \tan ^{2}\theta +1=\sec ^{2}\theta }$ .

The basic relationship between the sine and cosine is given by the Pythagorean identity:

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

 

where ${\displaystyle \sin ^{2}\theta }$  means ${\displaystyle (\sin \theta )^{2}}$  and ${\displaystyle \cos ^{2}\theta }$  means ${\displaystyle (\cos \theta )^{2}.}$ 

This can be viewed as a version of the Pythagorean theorem, and follows from the equation ${\displaystyle x^{2}+y^{2}=1}$  for the unit circle. This equation can be solved for either the sine or the cosine:

$${\displaystyle {\begin{aligned}\sin \theta &=\pm {\sqrt {1-\cos ^{2}\theta }},\\\cos \theta &=\pm {\sqrt {1-\sin ^{2}\theta }}.\end{aligned}}}$$

 

where the sign depends on the quadrant of ${\displaystyle \theta .}$ 

Dividing this identity by ${\displaystyle \sin ^{2}\theta }$ , ${\displaystyle \cos ^{2}\theta }$ , or both yields the following identities:

$${\displaystyle {\begin{aligned}&1+\cot ^{2}\theta =\csc ^{2}\theta \\&1+\tan ^{2}\theta =\sec ^{2}\theta \\&\sec ^{2}\theta +\csc ^{2}\theta =\sec ^{2}\theta \csc ^{2}\theta \end{aligned}}}$$

 

Using these identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign):

Each trigonometric function in terms of each of the other five.^[1]

in terms of	${\displaystyle \sin \theta }$	${\displaystyle \csc \theta }$	${\displaystyle \cos \theta }$	${\displaystyle \sec \theta }$	${\displaystyle \tan \theta }$	${\displaystyle \cot \theta }$
${\displaystyle \sin \theta =}$	${\displaystyle \sin \theta }$	${\displaystyle {\frac {1}{\csc \theta }}}$	${\displaystyle \pm {\sqrt {1-\cos ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}$	${\displaystyle \pm {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \csc \theta =}$	${\displaystyle {\frac {1}{\sin \theta }}}$	${\displaystyle \csc \theta }$	${\displaystyle \pm {\frac {1}{\sqrt {1-\cos ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\sec \theta }{\sqrt {\sec ^{2}\theta -1}}}}$	${\displaystyle \pm {\frac {\sqrt {1+\tan ^{2}\theta }}{\tan \theta }}}$	${\displaystyle \pm {\sqrt {1+\cot ^{2}\theta }}}$
${\displaystyle \cos \theta =}$	${\displaystyle \pm {\sqrt {1-\sin ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}$	${\displaystyle \cos \theta }$	${\displaystyle {\frac {1}{\sec \theta }}}$	${\displaystyle \pm {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \sec \theta =}$	${\displaystyle \pm {\frac {1}{\sqrt {1-\sin ^{2}\theta }}}}$	${\displaystyle \pm {\frac {\csc \theta }{\sqrt {\csc ^{2}\theta -1}}}}$	${\displaystyle {\frac {1}{\cos \theta }}}$	${\displaystyle \sec \theta }$	${\displaystyle \pm {\sqrt {1+\tan ^{2}\theta }}}$	${\displaystyle \pm {\frac {\sqrt {1+\cot ^{2}\theta }}{\cot \theta }}}$
${\displaystyle \tan \theta =}$	${\displaystyle \pm {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}$	${\displaystyle \pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}$	${\displaystyle \pm {\sqrt {\sec ^{2}\theta -1}}}$	${\displaystyle \tan \theta }$	${\displaystyle {\frac {1}{\cot \theta }}}$
${\displaystyle \cot \theta =}$	${\displaystyle \pm {\frac {\sqrt {1-\sin ^{2}\theta }}{\sin \theta }}}$	${\displaystyle \pm {\sqrt {\csc ^{2}\theta -1}}}$	${\displaystyle \pm {\frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}}}$	${\displaystyle \pm {\frac {1}{\sqrt {\sec ^{2}\theta -1}}}}$	${\displaystyle {\frac {1}{\tan \theta }}}$	${\displaystyle \cot \theta }$

Reflections, shifts, and periodicity

By examining the unit circle, one can establish the following properties of the trigonometric functions.

Reflections

 

Transformation of coordinates (a,b) when shifting the reflection angle ${\displaystyle \alpha }$  in increments of ${\displaystyle {\frac {\pi }{4}}}$ .

When the direction of a Euclidean vector is represented by an angle ${\displaystyle \theta ,}$  this is the angle determined by the free vector (starting at the origin) and the positive ${\displaystyle x}$ -unit vector. The same concept may also be applied to lines in a Euclidean space, where the angle is that determined by a parallel to the given line through the origin and the positive ${\displaystyle x}$ -axis. If a line (vector) with direction ${\displaystyle \theta }$  is reflected about a line with direction ${\displaystyle \alpha ,}$  then the direction angle ${\displaystyle \theta ^{\prime }}$  of this reflected line (vector) has the value

$${\displaystyle \theta ^{\prime }=2\alpha -\theta .}$$

 

The values of the trigonometric functions of these angles ${\displaystyle \theta ,\;\theta ^{\prime }}$  for specific angles ${\displaystyle \alpha }$  satisfy simple identities: either they are equal, or have opposite signs, or employ the complementary trigonometric function. These are also known as reduction formulae.^[2]

${\displaystyle \theta }$  reflected in ${\displaystyle \alpha =0}$ ^[3] odd/even identities	${\displaystyle \theta }$  reflected in ${\displaystyle \alpha ={\frac {\pi }{4}}}$	${\displaystyle \theta }$  reflected in ${\displaystyle \alpha ={\frac {\pi }{2}}}$	${\displaystyle \theta }$  reflected in ${\displaystyle \alpha ={\frac {3\pi }{4}}}$	${\displaystyle \theta }$  reflected in ${\displaystyle \alpha =\pi }$  compare to ${\displaystyle \alpha =0}$
${\displaystyle \sin(-\theta )=-\sin \theta }$	${\displaystyle \sin \left({\tfrac {\pi }{2}}-\theta \right)=\cos \theta }$	${\displaystyle \sin(\pi -\theta )=+\sin \theta }$	${\displaystyle \sin \left({\tfrac {3\pi }{2}}-\theta \right)=-\cos \theta }$	${\displaystyle \sin(2\pi -\theta )=-\sin(\theta )=\sin(-\theta )}$
${\displaystyle \cos(-\theta )=+\cos \theta }$	${\displaystyle \cos \left({\tfrac {\pi }{2}}-\theta \right)=\sin \theta }$	${\displaystyle \cos(\pi -\theta )=-\cos \theta }$	${\displaystyle \cos \left({\tfrac {3\pi }{2}}-\theta \right)=-\sin \theta }$	${\displaystyle \cos(2\pi -\theta )=+\cos(\theta )=\cos(-\theta )}$
${\displaystyle \tan(-\theta )=-\tan \theta }$	${\displaystyle \tan \left({\tfrac {\pi }{2}}-\theta \right)=\cot \theta }$	${\displaystyle \tan(\pi -\theta )=-\tan \theta }$	${\displaystyle \tan \left({\tfrac {3\pi }{2}}-\theta \right)=+\cot \theta }$	${\displaystyle \tan(2\pi -\theta )=-\tan(\theta )=\tan(-\theta )}$
${\displaystyle \csc(-\theta )=-\csc \theta }$	${\displaystyle \csc \left({\tfrac {\pi }{2}}-\theta \right)=\sec \theta }$	${\displaystyle \csc(\pi -\theta )=+\csc \theta }$	${\displaystyle \csc \left({\tfrac {3\pi }{2}}-\theta \right)=-\sec \theta }$	${\displaystyle \csc(2\pi -\theta )=-\csc(\theta )=\csc(-\theta )}$
${\displaystyle \sec(-\theta )=+\sec \theta }$	${\displaystyle \sec \left({\tfrac {\pi }{2}}-\theta \right)=\csc \theta }$	${\displaystyle \sec(\pi -\theta )=-\sec \theta }$	${\displaystyle \sec \left({\tfrac {3\pi }{2}}-\theta \right)=-\csc \theta }$	${\displaystyle \sec(2\pi -\theta )=+\sec(\theta )=\sec(-\theta )}$
${\displaystyle \cot(-\theta )=-\cot \theta }$	${\displaystyle \cot \left({\tfrac {\pi }{2}}-\theta \right)=\tan \theta }$	${\displaystyle \cot(\pi -\theta )=-\cot \theta }$	${\displaystyle \cot \left({\tfrac {3\pi }{2}}-\theta \right)=+\tan \theta }$	${\displaystyle \cot(2\pi -\theta )=-\cot(\theta )=\cot(-\theta )}$

Shifts and periodicity

 

Transformation of coordinates (a,b) when shifting the angle ${\displaystyle \theta }$  in increments of ${\displaystyle {\frac {\pi }{2}}}$ .

Shift by one quarter period	Shift by one half period	Shift by full periods^[4]	Period
${\displaystyle \sin(\theta \pm {\tfrac {\pi }{2}})=\pm \cos \theta }$	${\displaystyle \sin(\theta +\pi )=-\sin \theta }$	${\displaystyle \sin(\theta +k\cdot 2\pi )=+\sin \theta }$	${\displaystyle 2\pi }$
${\displaystyle \cos(\theta \pm {\tfrac {\pi }{2}})=\mp \sin \theta }$	${\displaystyle \cos(\theta +\pi )=-\cos \theta }$	${\displaystyle \cos(\theta +k\cdot 2\pi )=+\cos \theta }$	${\displaystyle 2\pi }$
${\displaystyle \csc(\theta \pm {\tfrac {\pi }{2}})=\pm \sec \theta }$	${\displaystyle \csc(\theta +\pi )=-\csc \theta }$	${\displaystyle \csc(\theta +k\cdot 2\pi )=+\csc \theta }$	${\displaystyle 2\pi }$
${\displaystyle \sec(\theta \pm {\tfrac {\pi }{2}})=\mp \csc \theta }$	${\displaystyle \sec(\theta +\pi )=-\sec \theta }$	${\displaystyle \sec(\theta +k\cdot 2\pi )=+\sec \theta }$	${\displaystyle 2\pi }$
${\displaystyle \tan(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\tan \theta \pm 1}{1\mp \tan \theta }}}$	${\displaystyle \tan(\theta +{\tfrac {\pi }{2}})=-\cot \theta }$	${\displaystyle \tan(\theta +k\cdot \pi )=+\tan \theta }$	${\displaystyle \pi }$
${\displaystyle \cot(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\cot \theta \mp 1}{1\pm \cot \theta }}}$	${\displaystyle \cot(\theta +{\tfrac {\pi }{2}})=-\tan \theta }$	${\displaystyle \cot(\theta +k\cdot \pi )=+\cot \theta }$	${\displaystyle \pi }$

Signs

The sign of trigonometric functions depends on quadrant of the angle. If ${\displaystyle {-\pi }<\theta \leq \pi }$  and sgn is the sign function,

$${\displaystyle {\begin{aligned}\operatorname {sgn}(\sin \theta )=\operatorname {sgn}(\csc \theta )&={\begin{cases}+1&{\text{if}}\ \ 0<\theta <\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <0\\0&{\text{if}}\ \ \theta \in \{0,\pi \}\end{cases}}\\[5mu]\operatorname {sgn}(\cos \theta )=\operatorname {sgn}(\sec \theta )&={\begin{cases}+1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },{\tfrac {1}{2}}\pi {\bigr \}}\end{cases}}\\[5mu]\operatorname {sgn}(\tan \theta )=\operatorname {sgn}(\cot \theta )&={\begin{cases}+1&{\text{if}}\ \ {-\pi }<\theta <-{\tfrac {1}{2}}\pi \ \ {\text{or}}\ \ 0<\theta <{\tfrac {1}{2}}\pi \\-1&{\text{if}}\ \ {-{\tfrac {1}{2}}\pi }<\theta <0\ \ {\text{or}}\ \ {\tfrac {1}{2}}\pi <\theta <\pi \\0&{\text{if}}\ \ \theta \in {\bigl \{}{-{\tfrac {1}{2}}\pi },0,{\tfrac {1}{2}}\pi ,\pi {\bigr \}}\end{cases}}\end{aligned}}}$$

 

The trigonometric functions are periodic with common period ${\displaystyle 2\pi ,}$  so for values of θ outside the interval ${\displaystyle ({-\pi },\pi ],}$  they take repeating values (see § Shifts and periodicity above).

Angle sum and difference identities

See also: Proofs of trigonometric identities § Angle sum identities, and Small-angle approximation § Angle sum and difference

 

Illustration of angle addition formulae for the sine and cosine of acute angles. Emphasized segment is of unit length.

 

Diagram showing the angle difference identities for ${\displaystyle \sin(\alpha -\beta )}$  and ${\displaystyle \cos(\alpha -\beta )}$ .

These are also known as the angle addition and subtraction theorems (or formulae).

$${\displaystyle {\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\sin(\alpha -\beta )&=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\cos(\alpha -\beta )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta \end{aligned}}}$$

 

The angle difference identities for ${\displaystyle \sin(\alpha -\beta )}$  and ${\displaystyle \cos(\alpha -\beta )}$  can be derived from the angle sum versions by substituting ${\displaystyle -\beta }$  for ${\displaystyle \beta }$  and using the facts that ${\displaystyle \sin(-\beta )=-\sin(\beta )}$  and ${\displaystyle \cos(-\beta )=\cos(\beta )}$ . They can also be derived by using a slightly modified version of the figure for the angle sum identities, both of which are shown here.

These identities are summarized in the first two rows of the following table, which also includes sum and difference identities for the other trigonometric functions.

Sine	${\displaystyle \sin(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle \sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$ ^[5]^[6]
Cosine	${\displaystyle \cos(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle \cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$ ^[6]^[7]
Tangent	${\displaystyle \tan(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$ ^[6]^[8]
Cosecant	${\displaystyle \csc(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\sec \alpha \csc \beta \pm \csc \alpha \sec \beta }}}$ ^[9]
Secant	${\displaystyle \sec(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\csc \alpha \csc \beta \mp \sec \alpha \sec \beta }}}$ ^[9]
Cotangent	${\displaystyle \cot(\alpha \pm \beta )}$			${\displaystyle =}$	${\displaystyle {\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}$ ^[6]^[10]
Arcsine	${\displaystyle \arcsin x\pm \arcsin y}$			${\displaystyle =}$	${\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}\pm y{\sqrt {1-x^{2}}}\right)}$ ^[11]
Arccosine	${\displaystyle \arccos x\pm \arccos y}$			${\displaystyle =}$	${\displaystyle \arccos \left(xy\mp {\sqrt {\left(1-x^{2}\right)\left(1-y^{2}\right)}}\right)}$ ^[12]
Arctangent	${\displaystyle \arctan x\pm \arctan y}$			${\displaystyle =}$	${\displaystyle \arctan \left({\frac {x\pm y}{1\mp xy}}\right)}$ ^[13]
Arccotangent	${\displaystyle \operatorname {arccot} x\pm \operatorname {arccot} y}$			${\displaystyle =}$	${\displaystyle \operatorname {arccot} \left({\frac {xy\mp 1}{y\pm x}}\right)}$

Sines and cosines of sums of infinitely many angles

When the series ${\textstyle \sum _{i=1}^{\infty }\theta _{i}}$  converges absolutely then

$${\displaystyle {\begin{aligned}{\sin }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggl )}&=\sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\!\!\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}\\{\cos }{\biggl (}\sum _{i=1}^{\infty }\theta _{i}{\biggr )}&=\sum _{{\text{even}}\ k\geq 0}(-1)^{\frac {k}{2}}\,\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}{\biggl (}\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}{\biggr )}.\end{aligned}}}$$

 

Because the series ${\textstyle \sum _{i=1}^{\infty }\theta _{i}}$  converges absolutely, it is necessarily the case that ${\textstyle \lim _{i\to \infty }\theta _{i}=0,}$  ${\textstyle \lim _{i\to \infty }\sin \theta _{i}=0,}$  and ${\textstyle \lim _{i\to \infty }\cos \theta _{i}=1.}$  In particular, in these two identities an asymmetry appears that is not seen in the case of sums of finitely many angles: in each product, there are only finitely many sine factors but there are cofinitely many cosine factors. Terms with infinitely many sine factors would necessarily be equal to zero.

When only finitely many of the angles ${\displaystyle \theta _{i}}$  are nonzero then only finitely many of the terms on the right side are nonzero because all but finitely many sine factors vanish. Furthermore, in each term all but finitely many of the cosine factors are unity.

Tangents and cotangents of sums

Let ${\displaystyle e_{k}}$  (for ${\displaystyle k=0,1,2,3,\ldots }$ ) be the kth-degree elementary symmetric polynomial in the variables

$${\displaystyle x_{i}=\tan \theta _{i}}$$

 

for ${\displaystyle i=0,1,2,3,\ldots ,}$  that is,

$${\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{i}x_{i}&&=\sum _{i}\tan \theta _{i}\\[6pt]e_{2}&=\sum _{i<j}x_{i}x_{j}&&=\sum _{i<j}\tan \theta _{i}\tan \theta _{j}\\[6pt]e_{3}&=\sum _{i<j<k}x_{i}x_{j}x_{k}&&=\sum _{i<j<k}\tan \theta _{i}\tan \theta _{j}\tan \theta _{k}\\&\ \ \vdots &&\ \ \vdots \end{aligned}}}$$

 

Then

$${\displaystyle {\begin{aligned}{\tan }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {{\sin }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}{{\cos }{\bigl (}\sum _{i}\theta _{i}{\bigr )}/\prod _{i}\cos \theta _{i}}}\\[10pt]&={\frac {\displaystyle \sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}{\displaystyle \sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{1,2,3,\dots \}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}\\[10pt]{\cot }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {e_{0}-e_{2}+e_{4}-\cdots }{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}$$

 

using the sine and cosine sum formulae above.

The number of terms on the right side depends on the number of terms on the left side.

For example:

$${\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\tan \theta _{1}+\tan \theta _{2}}{1\ -\ \tan \theta _{1}\tan \theta _{2}}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\[8pt]&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}$$

 

and so on. The case of only finitely many terms can be proved by mathematical induction.^[14] The case of infinitely many terms can be proved by using some elementary inequalities.^[15]

Secants and cosecants of sums

$${\displaystyle {\begin{aligned}{\sec }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]{\csc }{\Bigl (}\sum _{i}\theta _{i}{\Bigr )}&={\frac {\prod _{i}\sec \theta _{i}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}$$

 

where ${\displaystyle e_{k}}$  is the kth-degree elementary symmetric polynomial in the n variables ${\displaystyle x_{i}=\tan \theta _{i},}$  ${\displaystyle i=1,\ldots ,n,}$  and the number of terms in the denominator and the number of factors in the product in the numerator depend on the number of terms in the sum on the left.^[16] The case of only finitely many terms can be proved by mathematical induction on the number of such terms.

For example,

$${\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }}.\end{aligned}}}$$

 

Ptolemy's theorem

Main article: Ptolemy's theorem

See also: History of trigonometry § Classical antiquity

 

Diagram illustrating the relation between Ptolemy's theorem and the angle sum trig identity for sine. Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β.

Ptolemy's theorem is important in the history of trigonometric identities, as it is how results equivalent to the sum and difference formulas for sine and cosine were first proved. It states that in a cyclic quadrilateral ${\displaystyle ABCD}$ , as shown in the accompanying figure, the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. In the special cases of one of the diagonals or sides being a diameter of the circle, this theorem gives rise directly to the angle sum and difference trigonometric identities.^[17] The relationship follows most easily when the circle is constructed to have a diameter of length one, as shown here.

By Thales's theorem, ${\displaystyle \angle DAB}$  and ${\displaystyle \angle DCB}$  are both right angles. The right-angled triangles ${\displaystyle DAB}$  and ${\displaystyle DCB}$  both share the hypotenuse ${\displaystyle {\overline {BD}}}$  of length 1. Thus, the side ${\displaystyle {\overline {AB}}=\sin \alpha }$ , ${\displaystyle {\overline {AD}}=\cos \alpha }$ , ${\displaystyle {\overline {BC}}=\sin \beta }$  and ${\displaystyle {\overline {CD}}=\cos \beta }$ .

By the inscribed angle theorem, the central angle subtended by the chord ${\displaystyle {\overline {AC}}}$  at the circle's center is twice the angle ${\displaystyle \angle ADC}$ , i.e. ${\displaystyle 2(\alpha +\beta )}$ . Therefore, the symmetrical pair of red triangles each has the angle ${\displaystyle \alpha +\beta }$  at the center. Each of these triangles has a hypotenuse of length ${\textstyle {\frac {1}{2}}}$ , so the length of ${\displaystyle {\overline {AC}}}$  is ${\textstyle 2\times {\frac {1}{2}}\sin(\alpha +\beta )}$ , i.e. simply ${\displaystyle \sin(\alpha +\beta )}$ . The quadrilateral's other diagonal is the diameter of length 1, so the product of the diagonals' lengths is also ${\displaystyle \sin(\alpha +\beta )}$ .

When these values are substituted into the statement of Ptolemy's theorem that ${\displaystyle |{\overline {AC}}|\cdot |{\overline {BD}}|=|{\overline {AB}}|\cdot |{\overline {CD}}|+|{\overline {AD}}|\cdot |{\overline {BC}}|}$ , this yields the angle sum trigonometric identity for sine: ${\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }$ . The angle difference formula for ${\displaystyle \sin(\alpha -\beta )}$  can be similarly derived by letting the side ${\displaystyle {\overline {CD}}}$  serve as a diameter instead of ${\displaystyle {\overline {BD}}}$ .^[17]

Multiple-angle and half-angle formulae

Tn is the nth Chebyshev polynomial	${\displaystyle \cos(n\theta )=T_{n}(\cos \theta )}$ ^[18]
de Moivre's formula, i is the imaginary unit	${\displaystyle \cos(n\theta )+i\sin(n\theta )=(\cos \theta +i\sin \theta )^{n}}$ ^[19]

Multiple-angle formulae

Double-angle formulae

 

Visual demonstration of the double-angle formula for sine. For the above isosceles triangle with unit sides and angle ${\displaystyle 2\theta }$ , the area 1/2 × base × height is calculated in two orientations. When upright, the area is ${\displaystyle \sin \theta \cos \theta }$ . When on its side, the same area is ${\textstyle {\frac {1}{2}}\sin 2\theta }$ . Therefore, ${\displaystyle \sin 2\theta =2\sin \theta \cos \theta .}$ 

Formulae for twice an angle.^[20]

• ${\displaystyle \sin(2\theta )=2\sin \theta \cos \theta =(\sin \theta +\cos \theta )^{2}-1={\frac {2\tan \theta }{1+\tan ^{2}\theta }}}$ 
• ${\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta ={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}}$ 
• ${\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$ 
• ${\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}={\frac {1-\tan ^{2}\theta }{2\tan \theta }}}$ 
• ${\displaystyle \sec(2\theta )={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}={\frac {1+\tan ^{2}\theta }{1-\tan ^{2}\theta }}}$ 
• ${\displaystyle \csc(2\theta )={\frac {\sec \theta \csc \theta }{2}}={\frac {1+\tan ^{2}\theta }{2\tan \theta }}}$ 

Triple-angle formulae

Formulae for triple angles.^[20]

• ${\displaystyle \sin(3\theta )=3\sin \theta -4\sin ^{3}\theta =4\sin \theta \sin \left({\frac {\pi }{3}}-\theta \right)\sin \left({\frac {\pi }{3}}+\theta \right)}$ 
• ${\displaystyle \cos(3\theta )=4\cos ^{3}\theta -3\cos \theta =4\cos \theta \cos \left({\frac {\pi }{3}}-\theta \right)\cos \left({\frac {\pi }{3}}+\theta \right)}$ 
• ${\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}=\tan \theta \tan \left({\frac {\pi }{3}}-\theta \right)\tan \left({\frac {\pi }{3}}+\theta \right)}$ 
• ${\displaystyle \cot(3\theta )={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}$ 
• ${\displaystyle \sec(3\theta )={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}}$ 
• ${\displaystyle \csc(3\theta )={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}}$ 

Multiple-angle formulae

Formulae for multiple angles.^[21]

• ${\displaystyle {\begin{aligned}\sin(n\theta )&=\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta =\sin \theta \sum _{i=0}^{(n+1)/2}\sum _{j=0}^{i}(-1)^{i-j}{n \choose 2i+1}{i \choose j}\cos ^{n-2(i-j)-1}\theta \\{}&=2^{(n-1)}\prod _{k=0}^{n-1}\sin(k\pi /n+\theta )\end{aligned}}}$ 
• ${\displaystyle \cos(n\theta )=\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta =\sum _{i=0}^{n/2}\sum _{j=0}^{i}(-1)^{i-j}{n \choose 2i}{i \choose j}\cos ^{n-2(i-j)}\theta }$ 
• ${\displaystyle \cos((2n+1)\theta )=(-1)^{n}2^{2n}\prod _{k=0}^{2n}\cos(k\pi /(2n+1)-\theta )}$ 
• ${\displaystyle \cos(2n\theta )=(-1)^{n}2^{2n-1}\prod _{k=0}^{2n-1}\cos((1+2k)\pi /(4n)-\theta )}$ 
• ${\displaystyle \tan(n\theta )={\frac {\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\tan ^{k}\theta }{\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\tan ^{k}\theta }}}$ 

Chebyshev method

The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the ${\displaystyle (n-1)}$ th and ${\displaystyle (n-2)}$ th values.^[22]

${\displaystyle \cos(nx)}$  can be computed from ${\displaystyle \cos((n-1)x)}$ , ${\displaystyle \cos((n-2)x)}$ , and ${\displaystyle \cos(x)}$  with

$${\displaystyle \cos(nx)=2\cos x\cos((n-1)x)-\cos((n-2)x).}$$

 

This can be proved by adding together the formulae

$${\displaystyle {\begin{aligned}\cos((n-1)x+x)&=\cos((n-1)x)\cos x-\sin((n-1)x)\sin x\\\cos((n-1)x-x)&=\cos((n-1)x)\cos x+\sin((n-1)x)\sin x\end{aligned}}}$$

 

It follows by induction that ${\displaystyle \cos(nx)}$  is a polynomial of ${\displaystyle \cos x,}$  the so-called Chebyshev polynomial of the first kind, see Chebyshev polynomials#Trigonometric definition.

Similarly, ${\displaystyle \sin(nx)}$  can be computed from ${\displaystyle \sin((n-1)x),}$  ${\displaystyle \sin((n-2)x),}$  and ${\displaystyle \cos x}$  with

$${\displaystyle \sin(nx)=2\cos x\sin((n-1)x)-\sin((n-2)x)}$$

 

This can be proved by adding formulae for ${\displaystyle \sin((n-1)x+x)}$  and ${\displaystyle \sin((n-1)x-x).}$ 

Serving a purpose similar to that of the Chebyshev method, for the tangent we can write:

$${\displaystyle \tan(nx)={\frac {\tan((n-1)x)+\tan x}{1-\tan((n-1)x)\tan x}}\,.}$$

 

Half-angle formulae

$${\displaystyle {\begin{aligned}\sin {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {1-\cos \theta }{2}}}\\[3pt]\cos {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {1+\cos \theta }{2}}}\\[3pt]\tan {\frac {\theta }{2}}&={\frac {1-\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1+\cos \theta }}=\csc \theta -\cot \theta ={\frac {\tan \theta }{1+\sec {\theta }}}\\[6mu]&=\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {-1+\operatorname {sgn}(\cos \theta ){\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}\\[3pt]\cot {\frac {\theta }{2}}&={\frac {1+\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1-\cos \theta }}=\csc \theta +\cot \theta =\operatorname {sgn}(\sin \theta ){\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}\\\sec {\frac {\theta }{2}}&=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1+\cos \theta }}}\\\csc {\frac {\theta }{2}}&=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {2}{1-\cos \theta }}}\\\end{aligned}}}$$

 

^[23][24]

Also

$${\displaystyle {\begin{aligned}\tan {\frac {\eta \pm \theta }{2}}&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}\\[3pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[3pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\end{aligned}}}$$

 

Table

See also: Tangent half-angle formula

These can be shown by using either the sum and difference identities or the multiple-angle formulae.

	Sine	Cosine	Tangent	Cotangent
Double-angle formula^[25]^[26]	${\displaystyle {\begin{aligned}\sin(2\theta )&=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}}$	${\displaystyle {\begin{aligned}\cos(2\theta )&=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}}$	${\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$	${\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}}$
Triple-angle formula^[18]^[27]	${\displaystyle {\begin{aligned}\sin(3\theta )&=-\sin ^{3}\theta +3\cos ^{2}\theta \sin \theta \\&=-4\sin ^{3}\theta +3\sin \theta \end{aligned}}}$	${\displaystyle {\begin{aligned}\cos(3\theta )&=\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta \\&=4\cos ^{3}\theta -3\cos \theta \end{aligned}}}$	${\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}}$	${\displaystyle \cot(3\theta )={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}$
Half-angle formula^[23]^[24]	${\displaystyle {\begin{aligned}&\sin {\frac {\theta }{2}}=\operatorname {sgn} \left(\sin {\frac {\theta }{2}}\right){\sqrt {\frac {1-\cos \theta }{2}}}\\\\&\left({\text{or }}\sin ^{2}{\frac {\theta }{2}}={\frac {1-\cos \theta }{2}}\right)\end{aligned}}}$	${\displaystyle {\begin{aligned}&\cos {\frac {\theta }{2}}=\operatorname {sgn} \left(\cos {\frac {\theta }{2}}\right){\sqrt {\frac {1+\cos \theta }{2}}}\\\\&\left({\text{or }}\cos ^{2}{\frac {\theta }{2}}={\frac {1+\cos \theta }{2}}\right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}\\[3pt]&={\frac {\sin \theta }{1+\cos \theta }}\\[3pt]&={\frac {1-\cos \theta }{\sin \theta }}\\[5pt]\tan {\frac {\eta +\theta }{2}}&={\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}\\[5pt]\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)&=\sec \theta +\tan \theta \\[5pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {\left|1-\tan {\frac {\theta }{2}}\right|}{\left|1+\tan {\frac {\theta }{2}}\right|}}\\[5pt]\tan {\frac {\theta }{2}}&={\frac {\tan \theta }{1+{\sqrt {1+\tan ^{2}\theta }}}}\\&{\text{for }}\theta \in \left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)\end{aligned}}}$	${\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}\\[3pt]&={\frac {\sin \theta }{1-\cos \theta }}\\[4pt]&={\frac {1+\cos \theta }{\sin \theta }}\end{aligned}}}$

The fact that the triple-angle formula for sine and cosine only involves powers of a single function allows one to relate the geometric problem of a compass and straightedge construction of angle trisection to the algebraic problem of solving a cubic equation, which allows one to prove that trisection is in general impossible using the given tools, by field theory. ^{[citation needed]}

A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x³ − 3x + d = 0, where ${\displaystyle x}$  is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle. However, the discriminant of this equation is positive, so this equation has three real roots (of which only one is the solution for the cosine of the one-third angle). None of these solutions are reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots.

Power-reduction formulae

Obtained by solving the second and third versions of the cosine double-angle formula.

Sine	Cosine	Other
${\displaystyle \sin ^{2}\theta ={\frac {1-\cos(2\theta )}{2}}}$	${\displaystyle \cos ^{2}\theta ={\frac {1+\cos(2\theta )}{2}}}$	${\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos(4\theta )}{8}}}$
${\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin(3\theta )}{4}}}$	${\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos(3\theta )}{4}}}$	${\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin(2\theta )-\sin(6\theta )}{32}}}$
${\displaystyle \sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}}$	${\displaystyle \cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}}$	${\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos(4\theta )+\cos(8\theta )}{128}}}$
${\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin(3\theta )+\sin(5\theta )}{16}}}$	${\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos(3\theta )+\cos(5\theta )}{16}}}$	${\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin(2\theta )-5\sin(6\theta )+\sin(10\theta )}{512}}}$

 

Cosine power-reduction formula: an illustrative diagram. The red, orange and blue triangles are all similar, and the red and orange triangles are congruent. The hypotenuse ${\displaystyle {\overline {AD}}}$  of the blue triangle has length ${\displaystyle 2\cos \theta }$ . The angle ${\displaystyle \angle DAE}$  is ${\displaystyle \theta }$ , so the base ${\displaystyle {\overline {AE}}}$  of that triangle has length ${\displaystyle 2\cos ^{2}\theta }$ . That length is also equal to the summed lengths of ${\displaystyle {\overline {BD}}}$  and ${\displaystyle {\overline {AF}}}$ , i.e. ${\displaystyle 1+\cos(2\theta )}$

April 10, 2024