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There are several equivalent ways for defining trigonometric functions, and the proof of the trigonometric identities between them depend on the chosen definition. The oldest and somehow the most elementary definition is based on the geometry of
right triangle A right triangle (American English) or right-angled triangle (British), or more formally an orthogonal triangle, formerly called a rectangled triangle ( grc, ὀρθόσγωνία, lit=upright angle), is a triangle in which one angle is a right an ...
s. The proofs given in this article use this definition, and thus apply to non-negative angles not greater than a right angle. For greater and negative
angle In Euclidean geometry, an angle is the figure formed by two rays, called the '' sides'' of the angle, sharing a common endpoint, called the ''vertex'' of the angle. Angles formed by two rays lie in the plane that contains the rays. Angles a ...
s, see Trigonometric functions. Other definitions, and therefore other proofs are based on the Taylor series of sine and cosine, or on the
differential equation In mathematics, a differential equation is an equation that relates one or more unknown functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, an ...
f''+f=0 to which they are solutions.


Elementary trigonometric identities


Definitions

The six trigonometric functions are defined for every real number, except, for some of them, for angles that differ from 0 by a multiple of the right angle (90°). Referring to the diagram at the right, the six trigonometric functions of θ are, for angles smaller than the right angle: : \sin \theta = \frac = \frac : \cos \theta = \frac = \frac : \tan \theta = \frac = \frac : \cot \theta = \frac = \frac : \sec \theta = \frac = \frac : \csc \theta = \frac = \frac


Ratio identities

In the case of angles smaller than a right angle, the following identities are direct consequences of above definitions through the division identity : \frac = \frac . They remain valid for angles greater than 90° and for negative angles. : \tan \theta = \frac = \frac = \frac : \cot \theta =\frac = \frac = \frac = \frac : \sec \theta = \frac = \frac : \csc \theta = \frac = \frac : \tan \theta = \frac = \frac = \frac = \frac Or : \tan \theta = \frac = \frac = \frac = \frac : \cot \theta = \frac


Complementary angle identities

Two angles whose sum is π/2 radians (90 degrees) are ''complementary''. In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining: : \sin\left( \pi/2-\theta\right) = \cos \theta : \cos\left( \pi/2-\theta\right) = \sin \theta : \tan\left( \pi/2-\theta\right) = \cot \theta : \cot\left( \pi/2-\theta\right) = \tan \theta : \sec\left( \pi/2-\theta\right) = \csc \theta : \csc\left( \pi/2-\theta\right) = \sec \theta


Pythagorean identities

Identity 1: :\sin^2\theta + \cos^2\theta = 1 The following two results follow from this and the ratio identities. To obtain the first, divide both sides of \sin^2\theta + \cos^2\theta = 1 by \cos^2\theta; for the second, divide by \sin^2\theta. :\tan^2\theta + 1\ = \sec^2\theta :\sec^2\theta - \tan^2\theta = 1 Similarly :1\ + \cot^2\theta = \csc^2\theta :\csc^2\theta - \cot^2\theta = 1 Identity 2: The following accounts for all three reciprocal functions. : \csc^2\theta + \sec^2\theta - \cot^2\theta = 2\ + \tan^2\theta Proof 2: Refer to the triangle diagram above. Note that a^2+b^2=h^2 by
Pythagorean theorem In mathematics, the Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse (the side opposite ...
. :\csc^2\theta + \sec^2\theta = \frac + \frac = \frac + \frac = 2\ + \frac + \frac Substituting with appropriate functions - : 2\ + \frac + \frac = 2\ + \tan^2\theta+ \cot^2\theta Rearranging gives: : \csc^2\theta + \sec^2\theta - \cot^2\theta = 2\ + \tan^2\theta


Angle sum identities


Sine

Draw a horizontal line (the ''x''-axis); mark an origin O. Draw a line from O at an angle \alpha above the horizontal line and a second line at an angle \beta above that; the angle between the second line and the ''x''-axis is \alpha + \beta. Place P on the line defined by \alpha + \beta at a unit distance from the origin. Let PQ be a line perpendicular to line OQ defined by angle \alpha, drawn from point Q on this line to point P. \therefore OQP is a right angle. Let QA be a perpendicular from point A on the ''x''-axis to Q and PB be a perpendicular from point B on the ''x''-axis to P. \therefore OAQ and OBP are right angles. Draw R on PB so that QR is parallel to the ''x''-axis. Now angle RPQ = \alpha (because OQA = \frac - \alpha, making RQO = \alpha, RQP = \frac-\alpha, and finally RPQ = \alpha) :RPQ = \tfrac - RQP = \tfrac - (\tfrac - RQO) = RQO = \alpha :OP = 1 :PQ = \sin \beta :OQ = \cos \beta :\frac = \sin \alpha, so AQ = \sin \alpha \cos \beta :\frac = \cos \alpha, so PR = \cos \alpha \sin \beta :\sin (\alpha + \beta) = PB = RB+PR = AQ+PR = \sin \alpha \cos \beta + \cos \alpha \sin \beta By substituting -\beta for \beta and using Symmetry, we also get: :\sin (\alpha - \beta) = \sin \alpha \cos (-\beta) + \cos \alpha \sin (-\beta) :\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta


Cosine

Using the figure above, :OP = 1 :PQ = \sin \beta :OQ = \cos \beta :\frac = \cos \alpha, so OA = \cos \alpha \cos \beta :\frac = \sin \alpha, so RQ = \sin \alpha \sin \beta :\cos (\alpha + \beta) = OB = OA-BA = OA-RQ = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta By substituting -\beta for \beta and using Symmetry, we also get: :\cos (\alpha - \beta) = \cos \alpha \cos (-\beta) - \sin \alpha \sin (-\beta), :\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta Also, using the complementary angle formulae, : \begin \cos (\alpha + \beta) & = \sin\left( \pi/2-(\alpha + \beta)\right) \\ & = \sin\left( (\pi/2-\alpha) - \beta\right) \\ & = \sin\left( \pi/2-\alpha\right) \cos \beta - \cos\left( \pi/2-\alpha\right) \sin \beta \\ & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \end


Tangent and cotangent

From the sine and cosine formulae, we get :\tan (\alpha + \beta) = \frac = \frac Dividing both numerator and denominator by \cos \alpha \cos \beta , we get :\tan (\alpha + \beta) = \frac Subtracting \beta from \alpha , using \tan (- \beta) = -\tan \beta , :\tan (\alpha - \beta) = \frac = \frac Similarly from the sine and cosine formulae, we get :\cot (\alpha + \beta) = \frac = \frac Then by dividing both numerator and denominator by \sin \alpha \sin \beta , we get :\cot (\alpha + \beta) = \frac Or, using \cot \theta = \frac , :\cot (\alpha + \beta) = \frac = \frac = \frac Using \cot (- \beta) = -\cot \beta , :\cot (\alpha - \beta) = \frac = \frac


Double-angle identities

From the angle sum identities, we get :\sin (2 \theta) = 2 \sin \theta \cos \theta and :\cos (2 \theta) = \cos^2 \theta - \sin^2 \theta The Pythagorean identities give the two alternative forms for the latter of these: :\cos (2 \theta) = 2 \cos^2 \theta - 1 :\cos (2 \theta) = 1 - 2 \sin^2 \theta The angle sum identities also give :\tan (2 \theta) = \frac = \frac :\cot (2 \theta) = \frac = \frac It can also be proved using
Euler's formula Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's formula states that for a ...
: e^=\cos \varphi +i \sin \varphi Squaring both sides yields : e^=(\cos \varphi +i \sin \varphi)^ But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields : e^=\cos 2\varphi +i \sin 2\varphi It follows that :(\cos \varphi +i \sin \varphi)^=\cos 2\varphi +i \sin 2\varphi. Expanding the square and simplifying on the left hand side of the equation gives :i(2 \sin \varphi \cos \varphi) + \cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi +i \sin 2\varphi. Because the imaginary and real parts have to be the same, we are left with the original identities :\cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi, and also :2 \sin \varphi \cos \varphi = \sin 2\varphi.


Half-angle identities

The two identities giving the alternative forms for cos 2θ lead to the following equations: :\cos \frac = \pm\, \sqrt\frac, :\sin \frac = \pm\, \sqrt\frac. The sign of the square root needs to be chosen properly—note that if 2 is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore, the correct sign to use depends on the value of θ. For the tan function, the equation is: :\tan \frac = \pm\, \sqrt\frac. Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to: :\tan \frac = \frac. Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is: :\tan \frac = \frac. This also gives: :\tan \frac = \csc \theta - \cot \theta. Similar manipulations for the cot function give: :\cot \frac = \pm\, \sqrt\frac = \frac = \frac = \csc \theta + \cot \theta.


Miscellaneous – the triple tangent identity

If \psi + \theta + \phi = \pi = half circle (for example, \psi, \theta and \phi are the angles of a triangle), :\tan(\psi) + \tan(\theta) + \tan(\phi) = \tan(\psi)\tan(\theta)\tan(\phi). Proof: : \begin \psi & = \pi - \theta - \phi \\ \tan(\psi) & = \tan(\pi - \theta - \phi) \\ & = - \tan(\theta + \phi) \\ & = \frac \\ & = \frac \\ (\tan\theta \tan\phi - 1) \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi - \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi & = \tan\psi + \tan\theta + \tan\phi \\ \end


Miscellaneous – the triple cotangent identity

If \psi + \theta + \phi = \tfrac = quarter circle, : \cot(\psi) + \cot(\theta) + \cot(\phi) = \cot(\psi)\cot(\theta)\cot(\phi). Proof: Replace each of \psi , \theta , and \phi with their complementary angles, so cotangents turn into tangents and vice versa. Given :\psi + \theta + \phi = \tfrac :\therefore (\tfrac-\psi) + (\tfrac-\theta) + (\tfrac-\phi) = \tfrac - (\psi+\theta+\phi) = \tfrac - \tfrac = \pi so the result follows from the triple tangent identity.


Sum to product identities

* \sin \theta \pm \sin \phi = 2 \sin \left ( \frac2 \right ) \cos \left ( \frac2 \right ) * \cos \theta + \cos \phi = 2 \cos \left ( \frac2 \right ) \cos \left ( \frac2 \right ) * \cos \theta - \cos \phi = -2 \sin \left ( \frac2 \right ) \sin \left ( \frac2 \right )


Proof of sine identities

First, start with the sum-angle identities: :\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta :\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta By adding these together, :\sin (\alpha + \beta) + \sin (\alpha - \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta Similarly, by subtracting the two sum-angle identities, :\sin (\alpha + \beta) - \sin (\alpha - \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta - \sin \alpha \cos \beta + \cos \alpha \sin \beta = 2 \cos \alpha \sin \beta Let \alpha + \beta = \theta and \alpha - \beta = \phi, :\therefore \alpha = \frac2 and \beta = \frac2 Substitute \theta and \phi :\sin \theta + \sin \phi = 2 \sin \left( \frac2 \right) \cos \left( \frac2 \right) :\sin \theta - \sin \phi = 2 \cos \left( \frac2 \right) \sin \left( \frac2 \right) = 2 \sin \left( \frac2 \right) \cos \left( \frac2 \right) Therefore, :\sin \theta \pm \sin \phi = 2 \sin \left( \frac2 \right) \cos \left( \frac2 \right)


Proof of cosine identities

Similarly for cosine, start with the sum-angle identities: :\cos (\alpha + \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta :\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta Again, by adding and subtracting :\cos (\alpha + \beta) + \cos (\alpha - \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta + \cos \alpha \cos \beta + \sin \alpha \sin \beta = 2\cos \alpha \cos \beta :\cos (\alpha + \beta) - \cos (\alpha - \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta = -2 \sin \alpha \sin \beta Substitute \theta and \phi as before, :\cos \theta + \cos \phi = 2 \cos \left( \frac2 \right) \cos \left( \frac2 \right) :\cos \theta - \cos \phi = -2 \sin \left( \frac2 \right) \sin \left( \frac2 \right)


Inequalities

The figure at the right shows a sector of a circle with radius 1. The sector is of the whole circle, so its area is . We assume here that . :OA = OD = 1 :AB = \sin \theta :CD = \tan \theta The area of triangle is , or . The area of triangle is , or . Since triangle lies completely inside the sector, which in turn lies completely inside triangle , we have :\sin \theta < \theta < \tan \theta. This geometric argument relies on definitions of
arc length ARC may refer to: Business * Aircraft Radio Corporation, a major avionics manufacturer from the 1920s to the '50s * Airlines Reporting Corporation, an airline-owned company that provides ticket distribution, reporting, and settlement services * ...
and
area Area is the quantity that expresses the extent of a region on the plane or on a curved surface. The area of a plane region or ''plane area'' refers to the area of a shape or planar lamina, while ''surface area'' refers to the area of an open s ...
, which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than a provable property. For the sine function, we can handle other values. If , then . But (because of the Pythagorean identity), so . So we have :\frac < 1\ \ \ \mathrm\ \ \ 0 < \theta. For negative values of we have, by the symmetry of the sine function :\frac = \frac < 1. Hence :\frac < 1\quad \text\quad \theta \ne 0, and :\frac > 1\quad \text\quad 0 < \theta < \frac.


Identities involving calculus


Preliminaries

:\lim_ = 0 :\lim_ = 1


Sine and angle ratio identity

:\lim_ = 1 In other words, the function sine is differentiable at 0, and its derivative is 1. Proof: From the previous inequalities, we have, for small angles :\sin \theta < \theta < \tan \theta, Therefore, :\frac < 1 < \frac, Consider the right-hand inequality. Since :\tan \theta = \frac :\therefore 1 < \frac Multiply through by \cos \theta :\cos \theta < \frac Combining with the left-hand inequality: :\cos \theta < \frac < 1 Taking \cos \theta to the limit as \theta \to 0 :\lim_ = 1 Therefore, :\lim_ = 1


Cosine and angle ratio identity

:\lim_\frac = 0 Proof: : \begin \frac & = \frac\\ & = \frac\\ & = \left( \frac \right) \times \sin \theta \times \left( \frac \right)\\ \end The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.


Cosine and square of angle ratio identity

: \lim_\frac = \frac Proof: As in the preceding proof, :\frac = \frac \times \frac \times \frac. The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.


Proof of compositions of trig and inverse trig functions

All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function :\sin arctan(x)\frac Proof: We start from :\sin^2\theta+\cos^2\theta=1 (I) Then we divide this equation (I) by \cos^2\theta :\cos^2\theta=\frac (II) :1-\sin^2\theta=\frac Then use the substitution \theta=\arctan(x): :1-\sin^2 arctan(x)\frac :\sin^2 arctan(x)\frac Then we use the identity \tan arctan(x)equiv x :\sin arctan(x)\frac (III) And initial Pythagorean trigonometric identity proofed... Similarly if we divide this equation (I) by \sin^2\theta :\sin^2\theta=\frac (II) :\sin^2\theta=\frac Then use the substitution \theta=\arctan(x): :\sin^2 arctan(x)\frac Then we use the identity \tan arctan(x)equiv x :\sin arctan(x)\frac (III) And initial Pythagorean trigonometric identity proofed... : arctan(x) arcsin(\frac)/math> :y=\frac :y^2=\frac (IV) Let we guess that we have to prove: :x=\frac :x^2=\frac (V) Replacing (V) into (IV) : :y^2=\frac :y^2=\frac So it's true: y^2=y^2 and guessing statement was true: x=\frac : arctan(x) arcsin(\frac) arcsin(y) arctan(\frac)/math> Now y can be written as x ; and we have rcsinexpressed through rctan.. : arcsin(x) arctan(\frac)/math> Similarly if we seek : arccos(x)/math>... :\cos arccos(x)x :\cos(\frac-(\frac- arccos(x))=x :\sin(\frac- arccos(x)=x :\frac- arccos(x) arcsin(x)/math> : arccos(x)\frac- arcsin(x)/math> From : arcsin(x)/math>... : arccos(x)\frac- arctan(\frac)/math> : arccos(x)\frac- arccot(\frac)/math> And finally we have rccosexpressed through rctan.. : arccos(x) arctan(\frac)/math>


See also

* List of trigonometric identities *
Bhaskara I's sine approximation formula In mathematics, Bhaskara I's sine approximation formula is a rational expression in one variable for the computation of the approximate values of the trigonometric sines discovered by Bhaskara I (c. 600 – c. 680), a seventh-century Indian m ...
* Generating trigonometric tables * Aryabhata's sine table *
Madhava's sine table Madhava's sine table is the table of trigonometric sines of various angles constructed by the 14th century Kerala mathematician-astronomer Madhava of Sangamagrama. The table lists the trigonometric sines of the twenty-four angles 3.75°, 7.50� ...
*
Table of Newtonian series In mathematics, a Newtonian series, named after Isaac Newton, is a sum over a sequence a_n written in the form :f(s) = \sum_^\infty (-1)^n a_n = \sum_^\infty \frac a_n where : is the binomial coefficient and (s)_n is the falling factorial. ...
*
Madhava series In mathematics, a Madhava series or Leibniz series is any one of the series in a collection of infinite series expressions all of which are believed to have been discovered by an Indian Mathematician and Astronomer Madhava of Sangamagrama (c.&nbs ...
*
Unit vector In mathematics, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1. A unit vector is often denoted by a lowercase letter with a circumflex, or "hat", as in \hat (pronounced "v-hat"). The term ''direction vec ...
(explains direction cosines) *
Euler's formula Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's formula states that for a ...


Notes


References

* E. T. Whittaker and
G. N. Watson George Neville Watson (31 January 1886 – 2 February 1965) was an English mathematician, who applied complex analysis to the theory of special functions. His collaboration on the 1915 second edition of E. T. Whittaker's ''A Course of Mode ...
. '' A Course of Modern Analysis'', Cambridge University Press, 1952 {{DEFAULTSORT:Trigonometric identities, Proofs of Trigonometry Article proofs