List of trigonometric identities

From Wikipedia, the free encyclopedia

In mathematics, trigonometric identities are equalities that involve trigonometric functions that are true for all values of the occurring variables. 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 trick involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.

All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O.

The unit circle

1 Notation
2 Definitions
3 Periodicity, symmetry, and shifts
- 3.1 Periodicity
- 3.2 Symmetry
- 3.3 Shifts
- 3.4 Linear combinations
4 Pythagorean identities
5 Angle sum and difference identities
- 5.1 Tangents of sums of finitely many terms
6 Double-angle formulae
7 Triple-angle formulae
8 Multiple-angle formulae
9 Power-reduction formulae
10 Half-angle formulae
11 Product-to-sum identities
- 11.1 Ptolemy's theorem
12 Sum-to-product identities
13 Other sums of trigonometric functions
14 Inverse trigonometric functions
15 Trigonometric conversions
16 Relation to the complex exponential function
- 16.1 Pedagogy and "cis"
17 Infinite product formulæ
18 The Gudermannian function
19 Identities without variables
20 Calculus
- 20.1 Implications
21 Geometric proofs
- 21.1 sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
- 21.2 cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
22 Proofs of cos(x − y) and sin(x − y) formulæ
- 22.1 sin(x − y) = sin(x) cos(y) − cos(x) sin(y)
- 22.2 cos(x − y) = cos(x) cos(y) + sin(x) sin(y)
23 Exponential definitions
24 See also
25 External links

[edit] Notation

The following notations hold for all six trigonometric functions: sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). For brevity, only the sine case is given in the table.

Notation	Reading	Description	Definition
sin²(`x`)	"sine squared [of] `x`"	the square of sine; sine to the second power	sin²(x) = (sin(`x`))²
arcsin(`x`)	"arcsine [of] `x`"	the inverse function for sine	arcsin(`x`) = `y` if and only if sin(`y`) = `x` and $-{\pi \over 2} \le y \le {\pi \over 2}$
(sin(`x`))⁻¹	"sine [of] `x`, to the [power of] minus-one"	the reciprocal of sine; the multiplicative inverse of sine	(sin(`x`))⁻¹ = 1 / sin(x) = csc(x)

arcsin(x) can also be written sin⁻¹(x); this must not be confused with (sin(x))⁻¹.

[edit] Definitions

$\begin{align} \cos(x) &= \sin\left( x + \frac {\pi} {2}\right)\\ \tan(x) &= \frac{\sin(x)}{\cos(x)} &\quad \cot(x)&= \frac{\cos(x)}{\sin(x)} = \frac{1}{\tan(x)}\\ \sec(x) &= \frac{1}{\cos(x)} &\quad \csc(x)&= \frac{1}{\sin(x)} \end{align}$

For more information, including definitions based on the sides of a right triangle, see trigonometric function.

[edit] Periodicity, symmetry, and shifts

These are most easily shown from the unit circle:

[edit] Periodicity

The sine, cosine, secant, and cosecant functions have period 2π (a full circle): If $k$ is any integer then

$\begin{align} \sin(x) &= \sin(x + 2k\pi) \\ \cos(x) &= \cos(x + 2k\pi) \\ \sec(x) &= \sec(x + 2k\pi) \\ \csc(x) &= \csc(x + 2k\pi) \\ \end{align}$

The tangent and cotangent functions have period π (a half-circle): If $k$ is any integer then

$\begin{align} \tan(x) &= \tan(x + k\pi) \\ \cot(x) &= \cot(x + k\pi) \\ \end{align}$

[edit] Symmetry

The symmetries along x → −x, x → π/2 − x and x → π − x for the trigonometric functions are:

$\begin{align} \sin(-x) &= -\sin(x) & \sin\left(\tfrac{\pi}{2} - x\right) &= \cos(x) & \sin\left(\pi - x\right) &= +\sin(x) \\ \cos(-x) &= +\cos(x) & \cos\left(\tfrac{\pi}{2} - x\right) &= \sin(x) & \cos\left(\pi - x\right) &= -\cos(x) \\ \tan(-x) &= -\tan(x) & \tan\left(\tfrac{\pi}{2} - x\right) &= \cot(x) & \tan\left(\pi - x\right) &= -\tan(x) \\ \cot(-x) &= -\cot(x) & \cot\left(\tfrac{\pi}{2} - x\right) &= \tan(x) & \cot\left(\pi - x\right) &= -\cot(x) \\ \sec(-x) &= +\sec(x) & \sec\left(\tfrac{\pi}{2} - x\right) &= \csc(x) & \sec\left(\pi - x\right) &= -\sec(x) \\ \csc(-x) &= -\csc(x) & \csc\left(\tfrac{\pi}{2} - x\right) &= \sec(x) & \csc\left(\pi - x\right) &= +\csc(x) \end{align}$

[edit] Shifts

Among the simplest shifts (other than shifts by the period of each of these periodic functions) are shifts by π/2 and π:

$\begin{align} \sin\left(x + \tfrac{\pi}{2}\right) &= +\cos(x) & \sin\left(x + \pi\right) &= -\sin(x) \\ \cos\left(x + \tfrac{\pi}{2}\right) &= -\sin(x) & \cos\left(x + \pi\right) &= -\cos(x) \\ \tan\left(x + \tfrac{\pi}{2}\right) &= -\cot(x) & \tan\left(x + \pi\right) &= +\tan(x) \\ \cot\left(x + \tfrac{\pi}{2}\right) &= -\tan(x) & \cot\left(x + \pi\right) &= +\cot(x) \\ \sec\left(x + \tfrac{\pi}{2}\right) &= -\csc(x) & \sec\left(x + \pi\right) &= -\sec(x) \\ \csc\left(x + \tfrac{\pi}{2}\right) &= +\sec(x) & \csc\left(x + \pi\right) &= -\csc(x) \end{align}$

[edit] Linear combinations

For some purposes it is important to know that any linear combination of sine waves of the same period but different phase shifts is also a sine wave with the same period, but a different phase shift. In the case of a linear combination of a sine and cosine wave, we have

$a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot\sin(x+\varphi)\,$

where

$\varphi= \left\{ \begin{matrix} {\rm arctan}(b/a),&&\mbox{if }a\ge0; \; \\ \arctan(b/a) \pm \pi,&&\mbox{if }a<0. \; \end{matrix} \right. \;$

More generally, for an arbitrary phase shift, we have

$a\sin x+b\sin(x+\alpha)= c \sin(x+\beta)\,$

where

$c = \sqrt{a^2 + b^2 +2ab\cos \alpha},$

and

$\beta = {\rm arctan} \left(\frac{b\sin \alpha}{a + b\cos \alpha}\right).$

[edit] Pythagorean identities

These identities are based on the Pythagorean theorem. The first is sometimes simply called the Pythagorean trigonometric identity.

$\begin{align} \sin^2(x) + \cos^2(x) &= 1 \\ \tan^2(x) + 1 &= \sec^2(x) \\ \cot^2(x) + 1 &= \csc^2(x) \end{align}$

The second equation is obtained from the first by dividing both sides by cos²(x). To get the third equation, divide the first by sin²(x).

[edit] Angle sum and difference identities

These are also known as the addition and subtraction theorems or formulæ. The quickest way to prove these is Euler's formula. The tangent formula follows from the other two. A geometric proof of the sin(x + y) identity is given at the end of this article.

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

(When "+" is on the left side, then "+" is on the right, and vice versa.)

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

(When "+" is on the left side, then "-" is on the right, and vice versa.)

$\tan(x \pm y) = \frac{\tan(x) \pm \tan(y)}{1 \mp \tan(x)\tan(y)}$

[edit] Tangents of sums of finitely many terms

Let x_i = tan(θ_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

$\tan(\theta_1+\cdots+\theta_n) = \frac{e_1 - e_3 + e_5 -\cdots}{e_0 - e_2 + e_4 - \cdots},$

the number of terms depending on n.

For example,

$\begin{align} \tan(\theta_1 + \theta_2 + \theta_3) &{}= \frac{e_1 - e_3}{e_0 - e_2} = \frac{(x_1 + x_2 + x_3) \ - \ (x_1 x_2 x_3)}{ 1 \ - \ (x_1 x_2 + x_1 x_3 + x_2 x_3)}, \\ \\ \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &{}= \frac{e_1 - e_3}{e_0 - e_2 + e_4} \\ \\ &{}= \frac{(x_1 + x_2 + x_3 + x_4) \ - \ (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)}{ 1 \ - \ (x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4) \ + \ (x_1 x_2 x_3 x_4)},\end{align}$

and so on. The general case can be proved by mathematical induction.

[edit] Double-angle formulae

These can be shown by substituting x = y in the addition theorems, and using the Pythagorean formula. Or use de Moivre's formula with n = 2.

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

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

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

$\cot(2x) = \frac{\cot(x) - \tan(x)}{2}\,$

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.

[edit] Triple-angle formulae

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

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

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

[edit] Multiple-angle formulae

If T_n is the nth Chebyshev polynomial then

$\cos(nx)=T_n(\cos(x)). \,$

If S_n is the nth spread polynomial, then

$\sin^2(n\theta) = S_n(\sin^2\theta).\,$

de Moivre's formula:

$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^n \,$

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

$1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots+2\cos(nx) = \frac{ \sin\left(\left(n+\frac{1}{2}\right)x\right) }{ \sin(x/2) }.$

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

[edit] Power-reduction formulae

Solve the second and third versions of the cosine double-angle formula for cos²(x) and sin²(x), respectively.

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

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

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

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

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

[edit] Half-angle formulae

Sometimes the formulae in the previous section are called half-angle formulæ. To see why, substitute x/2 for x in the power reduction formulae, then solve for cos(x/2) and sin(x/2) to get:

$\cos\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 + \cos(x)}{2}}$

$\sin\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 - \cos(x)}{2}}$

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

$\tan\left(\frac{x}{2}\right) = {\sin (x/2) \over \cos (x/2)} = \pm\, \sqrt{1 - \cos x \over 1 + \cos x}. \qquad \qquad (1)$

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

$\tan\left(\frac{x}{2}\right) = \pm\, \sqrt{(1 - \cos x) (1 + \cos x) \over (1 + \cos x) (1 + \cos x)} = \pm\, \sqrt{1 - \cos^2 x \over (1 + \cos x)^2}$

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

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

$\tan\left(\frac{x}{2}\right) = \pm\, \sqrt{(1 - \cos x) (1 - \cos x) \over (1 + \cos x) (1 - \cos x)} = \pm\, \sqrt{(1 - \cos x)^2 \over (1 - \cos^2 x)}$

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

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

$\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)} = \frac{1-\cos(x)}{\sin(x)}.$

We also have

$\tan\left({x \over 2}\right) = \csc(x) - \cot(x),$

$\cot\left({x \over 2}\right) = \csc(x) + \cot(x).$

If we set

$t = \tan\left(\frac{x}{2}\right),$

then

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

and

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

and

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

This substitution of t for tan(x/2), with the consequent replacement of sin(x) by 2t/(1 + t²) and cos(x) by (1 − t²)/(1 + t²) is useful in calculus for converting rational functions in sin(x) and cos(x) to functions of t in order to find their antiderivatives. For more information see tangent half-angle formula.

[edit] Product-to-sum identities

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

$\cos\left (x\right ) \cos\left (y\right ) = {\cos\left (x - y\right ) + \cos\left (x + y\right ) \over 2} \;$

$\sin\left (x\right ) \sin\left (y\right ) = {\cos\left (x - y\right ) - \cos\left (x + y\right ) \over 2} \;$

$\sin\left (x\right ) \cos\left (y\right ) = {\sin\left (x - y\right ) + \sin\left (x + y\right ) \over 2} \;$

[edit] Ptolemy's theorem

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

$\begin{align} \mbox{then } & \sin(w + x)\sin(x + y) \\ &{} = \sin(x + y)\sin(y + z) \\ &{} = \sin(y + z)\sin(z + w) \\ &{} = \sin(z + w)\sin(w + x) = \sin(w)\sin(y) + \sin(x)\sin(z). \end{align}$

(The first three equalities are trivial; the fourth is the substance of this identity.) Essentially this is Ptolemy's theorem adapted to the language of trigonometry.

[edit] Sum-to-product identities

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

$\cos(x) + \cos(y) = 2 \cos\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) \;$

$\sin(x) + \sin(y) = 2 \sin\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) \;$

$\cos(x) - \cos(y) = -2 \sin\left( {x + y \over 2}\right) \sin\left({x - y \over 2}\right) \;$

$\sin(x) - \sin(y) = 2 \cos\left({x + y\over 2}\right) \sin\left({x - y\over 2}\right) \;$

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

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

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

(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(θ) as tan(θ) either increases through positive values or decreases through negative values. This is a one-point compactification of the real line.)

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

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

[edit] Other sums of trigonometric functions

Sum of sines and cosines with arguments in arithmetic progression:

$\sin{\varphi} + \sin{(\varphi + \alpha)} + \sin{(\varphi + 2\alpha)} + \cdots + \sin{(\varphi + n\alpha)}=\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \sin{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}.$

$\cos{\varphi} + \cos{(\varphi + \alpha)} + \cos{(\varphi + 2\alpha)} + \cdots + \cos{(\varphi + n\alpha)}=\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \cos{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}.$

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).

$\tan(x) + \sec(x) = \tan\left({x \over 2} + {\pi \over 4}\right).$

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 $x + y + z = π$ , then

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

[edit] Inverse trigonometric functions

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

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

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

$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)+\left\{\begin{matrix} \pi, & \mbox{if }x,y>0 \\ -\pi, & \mbox{if }x,y<0 \\ 0, & \mbox{otherwise } \end{matrix}\right.$

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

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

$\sin[\arctan(x)]=\frac{x}{\sqrt{1+x^2}}$

$\cos[\arctan(x)]=\frac{1}{\sqrt{1+x^2}}$

$\tan[\arcsin (x)]=\frac{x}{\sqrt{1 - x^2}}$

$\tan[\arccos (x)]=\frac{\sqrt{1 - x^2}}{x}$

[edit] Trigonometric conversions

Every trigonometric function can be related directly to every other trigonometric function. Such relations can be expressed by means of inverse trigonometric functions as follows: let φ and ψ represent a pair of trigonometric functions, and let arcψ be the inverse of ψ, such that ψ(arcψ(x)) = x. Then φ(arcψ(x)) can be expressed as an algebraic formula in terms of x. Such formulæ are shown in the table below: φ can be made equal to the head of one of the rows, and ψ can be equated to the head of a column.

Warning: In this table, the entries in any column (not in any row) are the six trigonometric functions of one angle.

Table of conversion formulæ $\varphi(\operatorname{arc}\,\psi(x))$
φ \ ψ	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\$

One procedure that can be used to obtain the elements of this table is as follows:
Given trigonometric functions φ and ψ, what is φ(arcψ(x)) equal to?

Find an equation that relates φ(u) and ψ(u) to each other:

$f(\varphi(u), \psi(u)) = 0 \$
Let $u = \operatorname{arc}\psi(x)$ , so that:

$f(\varphi({\rm arc}\psi(x)),\psi({\rm arc}\psi(x)) = 0 \$

$f(\varphi({\rm arc}\psi(x)),x) = 0 \$
Solve the last equation for φ(arcψ(x)).

Example. What is cot(arccsc(x)) equal to? First, find an equation which relations the functions cot and csc to each other, such as

$\cot^2 u + 1 = \csc^2 u \$ .

Second, let u = arccsc(x):

$\cot^2(\arccsc(x)) + 1 = \csc^2(\arccsc(x)) \$ ,

$\cot^2(\arccsc(x)) + 1 = x^2 \$ .

Third, solve this equation for cot(arccsc(x)):

$\cot^2(\arccsc(x)) = x^2 - 1, \$

$\cot(\arccsc(x)) = \pm\sqrt{x^2 - 1},$

and this is the formula which shows up in the sixth row and fourth column of the table.

[edit] Relation to the complex exponential function

$e^{ix} = \cos(x) + i\sin(x)\,$ (Euler's formula),

$e^{-ix} = \cos(x) - i\sin(x)\,$

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

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

where i² = −1.

[edit] Pedagogy and "cis"

Occasionally one sees the notation

$\operatorname{cis}(x) = \cos(x) + i\sin(x),\,$

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.

This coincidence can serve as a motivation for conjoining the two contexts and thereby discovering the trigonometric identity

$\operatorname{cis}(x+y) = \operatorname{cis}(x)\operatorname{cis}(y),\,$

and observing that this identity for cis of a sum is simpler than the identities for sin and cos of a sum. Having proved this identity, one can challenge the students to recall which familiar sort of function satisfies this same functional equation

$f(x+y) = f(x)f(y).\,$

The answer is exponential functions. That suggests that cis may be an exponential function

$\operatorname{cis}(x) = b^x.\,$

Then the question is: what is the base b? The definition of cis and the local behavior of sin and cos near zero suggest that

$\operatorname{cis}(0+dx) = \operatorname{cis}(0) + i\,dx,$

(where dx is an infinitesimal increment of x). Thus the rate of change at 0 is i, so the base should be eⁱ. Thus if this is an exponential function, then it must be

$\operatorname{cis}(x) = e^{ix}.\,$

[edit] Infinite product formulæ

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

$\sin x = x \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2 n^2}\right)$

$\sinh x = x \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2 n^2}\right)$

$\cos x = \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$

$\cosh x = \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$

$\frac{\sin x}{x} = \prod_{n = 1}^\infty\cos\left(\frac{x}{2^n}\right)$

[edit] The Gudermannian function

The Gudermannian function relates the circular and hyperbolic trigonometric functions without resorting to complex numbers; see that article for details.

[edit] Identities without variables

The curious identity

$\cos 20^\circ\cdot\cos 40^\circ\cdot\cos 80^\circ=\frac{1}{8}$

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

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

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

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

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

$\cos\left( \frac{2\pi}{21}\right) \,+\, \cos\left(2\cdot\frac{2\pi}{21}\right) \,+\, \cos\left(4\cdot\frac{2\pi}{21}\right)$

$\,+\, \cos\left( 5\cdot\frac{2\pi}{21}\right) \,+\, \cos\left( 8\cdot\frac{2\pi}{21}\right) \,+\, \cos\left(10\cdot\frac{2\pi}{21}\right)=\frac{1}{2}.$

The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.

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

$\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239}$

or, alternatively, by using Euler's formula:

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

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

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

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

With the golden ratio φ:

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

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

Also see exact trigonometric constants.

[edit] 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:

$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1,$

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:

$\lim_{x\rightarrow 0}\frac{1-\cos(x)}{x}=0,$

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:

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

The integral identities can be found in "list of integrals of trigonometric functions".

[edit] Implications

The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and fourier transformations.

[edit] Geometric proofs

These proofs apply directly only to acute angles, but the identities are still correct even when generalized to all angles. In this way, most of the trigonometric identities are deducible from elementary geometry, though many definitions and concepts have to be expanded.

[edit] 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.

$\sin(x + y) \,$

$= \frac {DG} {AD} \,$

$= \frac {EG + DE} {AD} \,$

$= \frac {BC + DE} {AD} \,$

$= \frac {BC} {AD} + \frac {DE} {AD} \,$

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

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

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

[edit] cos(x + y) = cos(x) cos(y) − sin(x) sin(y)

Using the above figure:

$\cos(x + y) \,$

$= \frac {AG} {AD} \,$

$= \frac {AB - GB} {AD} \,$

$= \frac {AB - EC} {AD} \,$

$= \frac {AB} {AD} - \frac {EC} {AD} \,$

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

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

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

[edit] 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

[edit] 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

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

[edit] 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

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

[edit] Exponential definitions

$\sin (\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} \,$

$\cos (\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2} \,$

$\tan (\theta) = \frac{\sin (\theta)}{\operatorname{cos} (\theta)} = \frac{(\frac{e^{i\theta} - e^{-i\theta}}{2i})}{(\frac{e^{i\theta} + e^{-i\theta}}{2})} \,$

$\cot (\theta) = \frac{\cos (\theta)}{\sin (\theta)} = \frac{(\frac{e^{i\theta} + e^{-i\theta}}{2})}{(\frac{e^{i\theta} - e^{-i\theta}}{2i})} \,$

$\sec (\theta) = \frac{1}{\cos (\theta)} = \frac{1}{(\frac{e^{i\theta} + e^{-i\theta}}{2})} \,$

$\csc (\theta) = \frac{1}{\sin (\theta)} = \frac{1}{(\frac{e^{i\theta} - e^{-i\theta}}{2i})} \,$

$\operatorname{versin} (\theta) = 1 - \cos (\theta) = 1 - \frac{e^{i\theta} + e^{-i\theta}}{2} \,$

$\operatorname{vercos} (\theta) = 1 - \sin (\theta) = 1 - \frac{e^{i\theta} - e^{-i\theta}}{2i} \,$

$\operatorname{exsec} (\theta) = \operatorname{sec} (\theta) - 1 \ = \frac{1}{\cos (\theta)} - 1 = \frac{1}{(\frac{e^{i\theta} + e^{-i\theta}}{2})} - 1 \,$

$\operatorname{excsc} (\theta) = \operatorname{csc} (\theta) - 1 \ = \frac{1}{\sin (\theta)} - 1 = \frac{1}{(\frac{e^{i\theta} - e^{-i\theta}}{2i})} - 1 \,$

$\operatorname{sinh} (\theta) = -i\sin (i\theta) = \frac{e^{\theta} - e^{-\theta}}{2} \,$

$\operatorname{cosh} (\theta) = \cos (i\theta) = \frac{e^{\theta} + e^{-\theta}}{2} \,$

$\operatorname{tanh} (\theta) = -i\tan (i\theta) = \frac{\operatorname{sinh} (\theta)}{\operatorname{cosh} (\theta)} = \frac{e^\theta - e^{-\theta}}{e^\theta + e^{-\theta}} = \frac{e^{2\theta} - 1}{e^{2\theta} + 1} \,$

$\operatorname{coth} (\theta) = i\operatorname{cot} (i\theta) = \frac{\operatorname{cosh} (\theta)}{\operatorname{sinh} (\theta)} = \frac{e^\theta + e^{-\theta}}{e^\theta - e^{-\theta}} = \frac{e^{2\theta} + 1}{e^{2\theta} - 1} \,$

$\operatorname{sech} (\theta) = \frac{1}{\operatorname{cosh} (\theta)} = \operatorname{sec} (i\theta) = \frac{2}{e^{\theta} + e^{-\theta}} \,$

$\operatorname{csch} (\theta) = \frac{1}{\operatorname{sinh} (\theta)} = i \cos (i\theta) = \frac{2}{e^{\theta} - e^{-\theta}} \,$

$\operatorname{versinh} (\theta) = 1 - \cos (i\theta) = 1 - \frac{e^{\theta} + e^{-\theta}}{2} \,$

$\operatorname{vercosh} (\theta) = 1 + i\sin (i\theta) = 1 - \frac{e^{\theta} - e^{-\theta}}{2} \,$

$\operatorname{exsech} (\theta) = \operatorname{sech} (\theta) - 1 = \frac{1}{\operatorname{cosh} (\theta)} - 1 = \operatorname{sec} (i\theta) = \frac{2}{e^{\theta} + e^{-\theta}} - 1 \,$

$\operatorname{excsch} (\theta) = \operatorname{csch} (\theta) - 1 = \frac{1}{\operatorname{sinh} (\theta)} - 1 = i \cos (i\theta) = \frac{2}{e^{\theta} - e^{-\theta}} - 1 \,$

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

$\arccos (\theta) = -i \ln (\theta + i\sqrt{1 - \theta^2}) \,$

$\arctan (\theta) = \frac{\ln (\frac{i + \theta}{i - \theta}) i}{2} \,$

$\arccot (\theta) = \arctan (-\theta) = \frac{i \ln (\frac{i - \theta}{i + \theta})}{2} \,$

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

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

$\operatorname{arcversin} (\theta) = \arccos (1 - \theta) = -i \ln (1 - \theta + i\sqrt{1 - (1 - \theta)^2}) \,$

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

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

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

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

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

$\operatorname{arctanh} (\theta) = \frac{\ln (\frac{i + \theta}{i - \theta})}{2} \,$

$\operatorname{arccoth} (\theta) = \operatorname{arctanh} (-\theta) = \frac{\ln (\frac{i - \theta}{i + \theta})}{2} \,$

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

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

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

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

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

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

[edit] See also

[edit] External links

A one-page proof of many trigonometric identities using Euler's formula, by Connelly Barnes.

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Categories: Mathematical identities | Trigonometry