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Linear combinations

$asin x+bcos x=sqrt{a^2+b^2}cdotsin(x+varphi),$

where

$varphi= left{ egin{matrix} { m arctan}(b/a),&&mbox{if }age0; ; \ { m arctan}(b/a) pm pi,&&mbox{if }a<0. ; end{matrix} ight. ;$

Angle sum and difference identities

$sin(x pm y) = sin(x) cos(y) pm cos(x) sin(y),$

$cos(x pm y) = cos(x) cos(y) mp sin(x) sin(y),$

$an(x pm y) = rac{ an(x) pm an(y)}{1 mp an(x) an(y)}$

Tangents of sums of finitely many terms

Let x_i = tan(?_i ), for i = 1, ..., n. Let e_k be the kth-degree elementary symmetric polynomial in the variables x_i, i = 1, ..., n, k = 0, ..., n. Then

$an( heta_1+cdots+ heta_n) = rac{e_1 - e_3 + e_5 -cdots}{e_0 - e_2 + e_4 - cdots},$

the number of terms depending on n.

Double-angle formulæ

$sin(2x) = 2 sin (x) cos(x) ,$

$cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x) = rac{1 - an^2(x)} {1 + an^2(x)} ,$

$an(2x) = rac{2 an (x)} {1 - an^2(x)},$

$cot(2x) = rac{cot(x) - an(x)}{2},$

The double-angle formula can also be used to find Pythagorean triples. If (a, b, c) are the lengths of the sides of a right triangle, then (a² ? b², 2ab, c²) also form a right triangle, where angle B is the angle being doubled. If a² ? b² is negative, take its opposite and use the supplement of 2B in place of 2B.

Triple-angle formulæ

$sin(3x)= 3 sin(x)- 4 sin^3(x) ,$

$cos(3x)= 4 cos^3(x) - 3 cos(x) ,$

$an(3x)= rac{3 an x - an^3 x}{1 - 3 an^2(x)}$

de Moivre's formula:

$cos(nx)+isin(nx)=(cos(x)+isin(x))^n ,$

The Dirichlet kernel D_n(x) is the function occurring on both sides of the next identity:

$1+2cos(x)+2cos(2x)+2cos(3x)+cdots+2cos(nx) = rac{ sinleft(left(n+ rac{1}{2} ight)x ight) }{ sin(x/2) }.$

The convolution of any integrable function of period 2? with the Dirichlet kernel coincides with the function's nth-degree Fourier approximation. The same holds for any measure or generalized function.

Power-reduction formulæ

$sin^2(x) = {1 - cos(2x) over 2}$

$cos^2(x) = {1 + cos(2x) over 2}$

$sin^2(x) cos^2(x) = {1 - cos(4 x) over 8}$

$sin^3(x) = rac{3 sin(x) - sin(3 x)}{4}$

$cos^3(x) = rac{3 cos(x) + cos(3 x)}{4}$

Half-angle formulæ

$cosleft( rac{x}{2} ight) = pm, sqrt{ rac{1 + cos(x)}{2}}$

$sinleft( rac{x}{2} ight) = pm, sqrt{ rac{1 - cos(x)}{2}}$

These may also be called the half-angle formulæ. Then

$anleft( rac{x}{2} ight) = {sin (x/2) over cos (x/2)} = pm, sqrt{1 - cos x over 1 + cos x}. qquad qquad (1)$

Multiply both numerator and denominator inside the radical by 1 + cos x, then simplify (using a Pythagorean identity):

$anleft( rac{x}{2} ight) = pm, sqrt{(1 - cos x) (1 + cos x) over (1 + cos x) (1 + cos x)} = pm, sqrt{1 - cos^2 x over (1 + cos x)^2}$

$= {sin x over 1 + cos x}.$

Likewise, multiplying both numerator and denominator inside the radical ? in equation (1) ? by
1 ? cos x, then simplifying:

$anleft( rac{x}{2} ight) = pm, sqrt{(1 - cos x) (1 - cos x) over (1 + cos x) (1 - cos x)} = pm, sqrt{(1 - cos x)^2 over (1 - cos^2 x)}$

$= {1 - cos x over sin x}.$

Thus, the pair of half-angle formulæ for the tangent are:

$anleft( rac{x}{2} ight) = rac{sin(x)}{1 + cos(x)} = rac{1-cos(x)}{sin(x)}.$

We also have

$anleft({x over 2} ight) = csc(x) - cot(x),$

$cotleft({x over 2} ight) = csc(x) + cot(x).$

If we set

$t = anleft( rac{x}{2} ight),$

then

$sin(x) = rac{2t}{1 + t^2}$

and

$cos(x) = rac{1 - t^2}{1 + t^2}$

and

$e^{i x} = rac{1 + i t}{1 - i t}.$

Product-to-sum identities

These can be proven by expanding their right-hand sides using the angle addition theorems.

$cosleft (x ight ) cosleft (y ight ) = {cosleft (x - y ight ) + cosleft (x + y ight ) over 2} ;$

$sinleft (x ight ) sinleft (y ight ) = {cosleft (x - y ight ) - cosleft (x + y ight ) over 2} ;$

$sinleft (x ight ) cosleft (y ight ) = {sinleft (x - y ight ) + sinleft (x + y ight ) over 2} ;$

Sum-to-product identities

Replace x by (x + y) / 2 and y by (x ? y) / 2 in the product-to-sum formulæ.

$cos(x) + cos(y) = 2 cosleft( rac{x + y}{2} ight) cosleft( rac{x - y}{2} ight) ;$

$sin(x) + sin(y) = 2 sinleft( rac{x + y}{2} ight) cosleft( rac{x - y}{2} ight) ;$

$cos(x) - cos(y) = -2 sinleft( {x + y over 2} ight) sinleft({x - y over 2} ight) ;$

$sin(x) - sin(y) = 2 cosleft({x + yover 2} ight) sinleft({x - yover 2} ight) ;$

If x, y, and z are the three angles of any triangle, or in other words

$mbox{if }x + y + z = pi = mbox{half circle,},$

$mbox{then } an(x) + an(y) + an(z) = an(x) an(y) an(z).,$

(If any of x, y, z is a right angle, one should take both sides to be ?. This is neither +? nor ??; for present purposes it makes sense to add just one point at infinity to the real line, that is approached by tan(?) as tan(?) either increases through positive values or decreases through negative values. This is a one-point compactification of the real line.)

$mbox{If }x + y + z = pi = mbox{half circle,},$

$mbox{then }sin(2x) + sin(2y) + sin(2z) = 4sin(x)sin(y)sin(z).,$

Other sums of trigonometric functions

Sum of sines and cosines with arguments in arithmetic progression:

$sin{varphi} + sin{(varphi + alpha)} + sin{(varphi + 2alpha)} + cdots + sin{(varphi + nalpha)}= rac{sin{left( rac{(n+1) alpha}{2} ight)} cdot sin{(varphi + rac{n alpha}{2})}}{sin{ rac{alpha}{2}}}.$

$cos{varphi} + cos{(varphi + alpha)} + cos{(varphi + 2alpha)} + cdots + cos{(varphi + nalpha)}= rac{sin{left( rac{(n+1) alpha}{2} ight)} cdot cos{(varphi + rac{n alpha}{2})}}{sin{ rac{alpha}{2}}}.$

For any a and b:

$a cos(x) + b sin(x) = sqrt{ a^2 + b^2 } cos(x - arctan(b, a)) ;$

where arctan(y, x) is the generalization of arctan(y/x) which covers the entire circular range (see also the account of this same identity in "symmetry, periodicity, and shifts" above for this generalization of arctan).

$an(x) + sec(x) = anleft({x over 2} + {pi over 4} ight).$

The above identity is sometimes convenient to know when thinking about the Gudermanian function.

If

x

,

y

, and

z

are the three angles of any triangle, i.e. if

x + y + z = ?

, then

$cot(x)cot(y) + cot(y)cot(z) + cot(z)cot(x) = 1.,$

Inverse trigonometric functions

$arcsin(x)+arccos(x)=pi/2;$

$arctan(x)+arccot(x)=pi/2.;$

$arctan(x)+arctan(1/x)=left{egin{matrix} pi/2, & mbox{if }x > 0 \ -pi/2, & mbox{if }x < 0 end{matrix} ight.$

$arctan(x)+arctan(y)=arctanleft( rac{x+y}{1-xy} ight)+left{egin{matrix} pi, & mbox{if }x,y>0 \ -pi, & mbox{if }x,y<0 \ 0, & mbox{otherwise } end{matrix} ight.$

$sin[arccos(x)]=sqrt{1-x^2} ,$

$cos[arcsin(x)]=sqrt{1-x^2} ,$

$sin[arctan(x)]=$

$cos[arctan(x)]=$

$an[arcsin (x)]= rac{x}{sqrt{1 - x^2}}$

$an[arccos (x)]= rac{sqrt{1 - x^2}}{x}$

Trigonometric conversions


? ?	sin	cos	tan	csc	sec	cot
sin	$x$	$sqrt{1 - x^2}$	${x over sqrt{1 + x^2}}$	${1 over x}$	${sqrt{1 - x^2} over x}$	${1 over sqrt{1+x^2}}$
cos	$sqrt{1 - x^2}$	$x$	${1 over sqrt{1 + x^2}}$	${sqrt{x^2 - 1} over x}$	${1 over x}$	${x over sqrt{1 + x^2}}$
tan	${x over sqrt{1 - x^2}}$	${sqrt{1 - x^2} over x}$	$x$	${1 over sqrt{x^2 - 1}}$	$sqrt{x^2 - 1}$	${1 over x}$
csc	${1 over x}$	${1 over sqrt{1 - x^2}}$	${sqrt{1 + x^2} over x}$	$x$	${x over sqrt{x^2 - 1}}$	$sqrt{1 + x^2}$
sec	${1 over sqrt{1 - x^2}}$	${1 over x}$	$sqrt{1 + x^2}$	${x over sqrt{x^2 - 1}}$	$x$	${sqrt{1 + x^2} over x}$
cot	${sqrt{1 - x^2} over x}$	${x over sqrt{1 - x^2}}$	${1 over x}$	$sqrt{x^2 - 1}$	${1 over sqrt{x^2 - 1}}$	$x$

Relation to the complex exponential function

$e^{ix} = cos(x) + isin(x),$

$e^{-ix} = cos(x) - isin(x),$

$cos(x) = rac{e^{ix} + e^{-ix}}{2} ;$

$sin(x) = rac{e^{ix} - e^{-ix}}{2i} ;$

where i² = ?1.

Pedagogy and "cis"

Occasionally one sees the notation

$operatorname{cis}(x) = cos(x) + isin(x),,$

i.e. "cis" abbreviates "cos + i sin".

"Why", a mathematician may ask, "should one introduce such a notation, rather than writing simply e^ix?". In some contexts, this notation may serve the pedagogical purpose of emphasizing that one has not yet proved that this is an exponential function. In doing trigonometry without complex numbers, one may prove the two identities

$cos(x+y) = cos(x)cos(y) - sin(x)sin(y) = c_1 c_2 - s_1 s_2,,$

$sin(x+y) = sin(x)cos(y) + cos(x)sin(y) = s_1 c_2 + c_1 s_2.,$

Similarly in treating multiplication of complex numbers (with no involvement of trigonometry), one may observe that the real and imaginary parts of the product of c₁ + is₁ and c₂ + is₂ are respectively

$c_1 c_2 - s_1 s_2,,$

$s_1 c_2 + c_1 s_2.,$

Thus one sees this same pattern arising in two disparate contexts:

trigonometry without complex numbers, and
complex numbers without trigonometry.

Infinite product formulæ

For applications to special functions, the following infinite product formulæ for trigonometric functions are useful:

$sin x = x prod_{n = 1}^inftyleft(1 - rac{x^2}{pi^2 n^2} ight)$

$sinh x = x prod_{n = 1}^inftyleft(1 + rac{x^2}{pi^2 n^2} ight)$

$cos x = prod_{n = 1}^inftyleft(1 - rac{x^2}{pi^2(n - rac{1}{2})^2} ight)$

$cosh x = prod_{n = 1}^inftyleft(1 + rac{x^2}{pi^2(n - rac{1}{2})^2} ight)$

$rac{sin x}{x} = prod_{n = 1}^inftycosleft( rac{x}{2^n} ight)$

Identities without variables

The curious identity

$cos 20^circcdotcos 40^circcdotcos 80^circ= rac{1}{8}$

is a special case of an identity that contains one variable:

$prod_{j=0}^{k-1}cos(2^j x)= rac{sin(2^k x)}{2^ksin(x)}.$

The following is perhaps not as readily generalized to an identity containing variables:

$cos 24^circ+cos 48^circ+cos 96^circ+cos 168^circ= rac{1}{2}$ .

Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:

$cosleft( rac{2pi}{21} ight) ,+, cosleft(2cdot rac{2pi}{21} ight) ,+, cosleft(4cdot rac{2pi}{21} ight)$

$,+, cosleft( 5cdot rac{2pi}{21} ight) ,+, cosleft( 8cdot rac{2pi}{21} ight) ,+, cosleft(10cdot rac{2pi}{21} ight)= rac{1}{2}.$

An efficient way to compute is based on the following identity without variables, due to Machin:

$rac{pi}{4} = 4 arctan rac{1}{5} - arctan rac{1}{239}$

or, alternatively, by using Euler's formula:

$rac{pi}{4} = 5 arctan rac{1}{7} + 2 arctan rac{3}{79}.$

$egin{matrix} sin 0 & = & sin 0^circ & = & 0 & = & cos 90^circ & = & cos left( rac {pi} {2} ight) \ \ sin left( rac {pi} {6} ight) & = & sin 30^circ & = & 1/2 & = & cos 60^circ & = & cos left( rac {pi} {3} ight) \ \ sin left( rac {pi} {4} ight) & = & sin 45^circ & = & sqrt{2}/2 & = & cos 45^circ & = & cos left( rac {pi} {4} ight) \ \ sin left( rac {pi} {3} ight) & = & sin 60^circ & = & sqrt{3}/2 & = & cos 30^circ & = & cos left( rac {pi} {6} ight) \ \ sin left( rac {pi} {2} ight) & = & sin 90^circ & = & 1 & = & cos 0^circ & = & cos 0 end{matrix}$

$sin{ rac{pi}{7}}= rac{sqrt{7}}{6}- rac{sqrt{7}}{189} sum_{j=0}^{infty} rac{(3j+1)!}{189^j j!,(2j+2)!} !$

$sin{ rac{pi}{18}}= rac{1}{6} sum_{j=0}^{infty} rac{(3j)!}{27^j j!,(2j+1)!} !$

With the golden ratio :

$cos left( rac {pi} {5} ight) = cos 36^circ={sqrt{5}+1 over 4} = varphi /2$

$sin left( rac {pi} {10} ight) = sin 18^circ = {sqrt{5}-1 over 4} = {varphi - 1 over 2} = {1 over 2varphi}$

Calculus

In calculus the relations stated below require angles to be measured in radians; the relations would become more complicated if angles were measured in another unit such as degrees. If the trigonometric functions are defined in terms of geometry, their derivatives can be found by verifying two limits. The first is:

$lim_{x ightarrow 0} rac{sin(x)}{x}=1,$

verified using the unit circle and squeeze theorem. It may be tempting to propose to use L'Hôpital's rule to establish this limit. However, if one uses this limit in order to prove that the derivative of the sine is the cosine, and then uses the fact that the derivative of the sine is the cosine in applying L'Hôpital's rule, one is reasoning circularly?a logical fallacy. The second limit is:

$lim_{x ightarrow 0} rac{1-cos(x)}{x}=0,$

verified using the identity tan(x/2) = (1 ? cos(x))/sin(x). Having established these two limits, one can use the limit definition of the derivative and the addition theorems to show that sin?(x) = cos(x) and cos?(x) = ?sin(x). If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term.

${d over dx}sin(x) = cos(x)$

The rest of the trigonometric functions can be differentiated using the above identities and the rules of differentiation. We have:

$egin{matrix} {d over dx} sin x =& cos x ,& {d over dx} arcsin x =& {1 over sqrt{1 - x^2} } \ \ {d over dx} cos x =& -sin x ,& {d over dx} arccos x =& {-1 over sqrt{1 - x^2}} \ \ {d over dx} an x =& sec^2 x ,& {d over dx} arctan x =& { 1 over 1 + x^2} \ \ {d over dx} cot x =& -csc^2 x ,& {d over dx} arccot x =& {-1 over 1 + x^2} \ \ {d over dx} sec x =& an x sec x ,& {d over dx} arcsec x =& { 1 over |x|sqrt{x^2 - 1}} \ \ {d over dx} csc x =& -csc x cot x ,& {d over dx} arccsc x =& {-1 over |x|sqrt{x^2 - 1}} end{matrix}$

Geometric proofs

These proofs apply directly only to acute angles, but the truth of these identities in the case of acute angles can be used to infer their truth in more general cases.

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

In the figure the angle x is part of right angled triangle ABC, and the angle y part of right angled triangle ACD. Then construct DG perpendicular to AB and construct CE parallel to AB.

Angle x = Angle BAC = Angle ACE = Angle CDE.

EG = BC.

$sin(x + y) ,$

$= rac {DG} {AD} ,$

$= rac {EG + DE} {AD} ,$

$= rac {BC + DE} {AD} ,$

$= rac {BC} {AD} + rac {DE} {AD} ,$

$= rac{BC}{AD} cdot rac{AC}{AC} + rac{DE}{AD} cdot rac{CD}{CD} ,$

$= rac{BC}{AC} cdot rac{AC}{AD} + rac{DE}{CD} cdot rac{CD}{AD} ,$

$= sin( x ) cos( y ) + cos( x ) sin( y ). ,$

cos(x + y) = cos(x) cos(y) ? sin(x) sin(y)

Using the above figure:

$cos(x + y) ,$

$= rac {AG} {AD} ,$

$= rac {AB - GB} {AD} ,$

$= rac {AB - EC} {AD} ,$

$= rac {AB} {AD} - rac {EC} {AD} ,$

$= rac{AB}{AD} cdot rac{AC}{AC} - rac{EC}{AD} cdot rac{CD}{CD} ,$

$= rac{AB}{AC} cdot rac{AC}{AD} - rac{EC}{CD} cdot rac{CD}{AD} ,$

$= cos( x ) cos( y ) - sin( x ) sin( y ). ,$

Proofs of cos(x ? y) and sin(x ? y) formulæ

The formulæ for cos(x ? y) and sin(x ? y) are easily proven using the formulæ for cos(x + y) and sin(x + y), respectively

sin(x ? y) = sin(x) cos(y) ? cos(x) sin(y)

To begin, we substitute y with ?y into the sin(x + y) formula:

$! sin(x+(-y)) = sin(x)cos(-y) + cos(x)sin(-y).$

Using the fact that sine is an odd function and cosine is an even function, we get

$! sin(x-y) = sin(x)cos(y) - cos(x)sin(y).$

cos(x ? y) = cos(x) cos(y) + sin(x) sin(y)

To begin, we substitute y with ?y into the cos(x + y) formula:

$! cos(x+(-y)) = cos(x)cos(-y) - sin(x)sin(-y).$

Using the fact that sine is an odd function and cosine is an even function, we get

$! cos(x-y) = cos(x)cos(y) + sin(x)sin(y).$

Exponential definitions

$operatorname{sin} ( heta) = rac{e^{i heta} - e^{-i heta}}{2i} ,$

$operatorname{cos} ( heta) = rac{e^{i heta} + e^{-i heta}}{2} ,$

$operatorname{tan} ( heta) = rac{operatorname{sin} ( heta)}{operatorname{cos} ( heta)} = rac{( rac{e^{i heta} - e^{-i heta}}{2i})}{( rac{e^{i heta} + e^{-i heta}}{2})} ,$

$operatorname{cot} ( heta) = rac{operatorname{cos} ( heta)}{operatorname{sin} ( heta)} = rac{( rac{e^{i heta} + e^{-i heta}}{2})}{( rac{e^{i heta} - e^{-i heta}}{2i})} ,$

$operatorname{sec} ( heta) = rac{1}{operatorname{cos} ( heta)} = rac{1}{( rac{e^{i heta} + e^{-i heta}}{2})} ,$

$operatorname{csc} ( heta) = rac{1}{operatorname{sin} ( heta)} = rac{1}{( rac{e^{i heta} - e^{-i heta}}{2i})} ,$

$operatorname{versin} ( heta) = 1 - operatorname{cos} ( heta) = 1 - rac{e^{i heta} + e^{-i heta}}{2} ,$

$operatorname{vercos} ( heta) = 1 - operatorname{sin} ( heta) = 1 - rac{e^{i heta} - e^{-i heta}}{2i} ,$

$operatorname{exsec} ( heta) = operatorname{sec} ( heta) - 1 = rac{1}{operatorname{cos} ( heta)} - 1 = rac{1}{( rac{e^{i heta} + e^{-i heta}}{2})} - 1 ,$

$operatorname{excsc} ( heta) = operatorname{csc} ( heta) - 1 = rac{1}{operatorname{sin} ( heta)} - 1 = rac{1}{( rac{e^{i heta} - e^{-i heta}}{2i})} - 1 ,$

$operatorname{sinh} ( heta) = -ioperatorname{sin} (i heta) = rac{e^{ heta} - e^{- heta}}{2} ,$

$operatorname{cosh} ( heta) = operatorname{cos} (i heta) = rac{e^{ heta} + e^{- heta}}{2} ,$

$operatorname{tanh} ( heta) = -ioperatorname{tan} (i heta) = rac{operatorname{sinh} ( heta)}{operatorname{cosh} ( heta)} = rac{e^ heta - e^{- heta}}{e^ heta + e^{- heta}} = rac{e^{2 heta} - 1}{e^{2 heta} + 1} ,$

$operatorname{coth} ( heta) = ioperatorname{cot} (i heta) = rac{operatorname{cosh} ( heta)}{operatorname{sinh} ( heta)} = rac{e^ heta + e^{- heta}}{e^ heta - e^{- heta}} = rac{e^{2 heta} + 1}{e^{2 heta} - 1} ,$

$operatorname{sech} ( heta) = rac{1}{operatorname{cosh} ( heta)} = operatorname{sec} (i heta) = rac{2}{e^{ heta} + e^{- heta}} ,$

$operatorname{csch} ( heta) = rac{1}{operatorname{sinh} ( heta)} = i operatorname{cos} (i heta) = rac{2}{e^{ heta} - e^{- heta}} ,$

$operatorname{versinh} ( heta) = 1 - operatorname{cos} (i heta) = 1 - rac{e^{ heta} + e^{- heta}}{2} ,$

$operatorname{vercosh} ( heta) = 1 + ioperatorname{sin} (i heta) = 1 - rac{e^{ heta} - e^{- heta}}{2} ,$

$operatorname{exsech} ( heta) = operatorname{sech} ( heta) - 1 = rac{1}{operatorname{cosh} ( heta)} - 1 = operatorname{sec} (i heta) = rac{2}{e^{ heta} + e^{- heta}} - 1 ,$

$operatorname{excsch} ( heta) = operatorname{csch} ( heta) - 1 = rac{1}{operatorname{sinh} ( heta)} - 1 = i operatorname{cos} (i heta) = rac{2}{e^{ heta} - e^{- heta}} - 1 ,$

$operatorname{arcsin} ( heta) = -i ln (i heta + sqrt{1 - heta^2}) ,$

$operatorname{arccos} ( heta) = -i ln ( heta + isqrt{1 - heta^2}) ,$

$operatorname{arctan} ( heta) = rac{ln ( rac{i + heta}{i - heta}) i}{2} ,$

$operatorname{arccot} ( heta) = operatorname{arctan} (- heta) = rac{i ln ( rac{i - heta}{i + heta})}{2} ,$

$operatorname{arcsec} ( heta) = operatorname{arccos} ( heta^{-1}) = -i ln ( heta^{-1} + sqrt{1 - heta^{-2}}i) ,$

$operatorname{arccsc} ( heta) = operatorname{arcsin} ( heta^{-1}) = -i ln (i heta^{-1} + sqrt{1 - heta^{-2}}) ,$

$operatorname{arcversin} ( heta) = operatorname{arccos} (1 - heta) = -i ln (1 - heta + isqrt{1 - (1 - heta)^2}) ,$

$operatorname{arcvercos} ( heta) = operatorname{arcsin} (1 - heta) = -i ln (i - i heta + sqrt{1 - (1 - heta)^2}) ,$

$operatorname{arcexsec} ( heta) = operatorname{arcsec} (1 + heta) = -i ln (( heta + 1)^{-1} + i sqrt{1 - (1 + heta)^2}) ,$

$operatorname{arcexcsc} ( heta) = operatorname{arccsc} (1 + heta) = -i ln (i ( heta + 1)^{-1} + sqrt{1 - (1 + heta)^2}) ,$

$operatorname{arcsinh} ( heta) = ln ( heta + sqrt{ heta^2 + 1}) ,$

$operatorname{arccosh} ( heta) = ln ( heta + sqrt{ heta^2 - 1}) ,$

$operatorname{arctanh} ( heta) = rac{ln ( rac{i + heta}{i - heta})}{2} ,$

$operatorname{arccoth} ( heta) = operatorname{arctanh} (- heta) = rac{ln ( rac{i - heta}{i + heta})}{2} ,$

$operatorname{arcsech} ( heta) = operatorname{arccosh} ( heta^{-1}) = ln ( heta^{-1} + sqrt{ heta^{-2} - 1}) ,$

$operatorname{arccsch} ( heta) = operatorname{arcsinh} ( heta^{-1}) = ln ( heta^{-1} + sqrt{ heta^{-2} + 1}) ,$

$operatorname{arcversinh} ( heta) = operatorname{arccosh} ( heta) - 1 = ln ( heta + sqrt{ heta^2 - 1}) - 1 ,$

$operatorname{arcvercosh} ( heta) = operatorname{arcsinh} ( heta) - 1 = ln ( heta + sqrt{ heta^2 + 1}) - 1,$

$operatorname{arcexsech} ( heta) = operatorname{arcsech} ( heta + 1) = operatorname{arccosh} (( heta + 1)^{-1}) = ln (( heta + 1)^{-1} + sqrt{( heta + 1)^{-2} - 1}) ,$

$operatorname{arcexcsch} ( heta) = operatorname{arccsch} ( heta + 1) = operatorname{arcsinh} (( heta + 1)^{-1}) = ln (( heta + 1)^{-1} + sqrt{( heta + 1)^{-2} + 1}) ,$

law of cosines

Fig. 1 - A triangle.

In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) is a statement about a general triangle which relates the lengths of its sides to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines states that

$c^2 = a^2 + b^2 - 2abcos(gamma) , ,$

$b^2 = a^2 + c^2 - 2accos(eta) , ,$

$a^2 = b^2 + c^2 - 2bccos(alpha) . ,$

Note that c is the side opposite the angle ? and a, b are the two sides enclosing ?.

law of sines

If the sides of the triangle are a, b and c and the angles opposite those sides are A, B and C, then the law of sines states:

${a over sin A}={b over sin B}={c over sin C}=2R,$

where R is the radius of the triangle's circumcircle. This law is useful when computing the remaining sides of a triangle if two angles and a side are known, a common problem in the technique of triangulation. It can also be used when two sides and one of the non-enclosed angles are known; in this case, the formula may give two possible values for the enclosed angle. When this happens, often only one result will cause all angles to be less than 180°; in other cases, there are two valid solutions to the triangle.

It can be shown that:

$2R = rac{abc} {2sqrt{s(s-a)(s-b)(s-c)}} = rac {2abc} {sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) }}$

where s is the semi-perimeter,

$s = rac{(a+b+c)} {2}.$

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