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 {\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac {a}{h} : \cos \theta = \frac {\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac {b}{h} : \tan \theta = \frac {\mathrm{opposite}}{\mathrm{adjacent}} = \frac {a}{b} : \cot \theta = \frac {\mathrm{adjacent}}{\mathrm{opposite}} = \frac {b}{a} : \sec \theta = \frac {\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac {h}{b} : \csc \theta = \frac {\mathrm{hypotenuse}}{\mathrm{opposite}} = \frac {h}{a}
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 {a}{b}= \frac {\left(\frac {a}{h}\right)} {\left(\frac {b}{h}\right) }. They remain valid for angles greater than 90° and for negative angles. : \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac { \left( \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} \right) } { \left( \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}\right) } = \frac {\sin \theta} {\cos \theta} : \cot \theta =\frac{\mathrm{adjacent}}{\mathrm{opposite}} = \frac { \left( \frac{\mathrm{adjacent}}{\mathrm{adjacent}} \right) } { \left( \frac {\mathrm{opposite}}{\mathrm{adjacent}} \right) } = \frac {1}{\tan \theta} = \frac {\cos \theta}{\sin \theta} : \sec \theta = \frac {1}{\cos \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} : \csc \theta = \frac {1}{\sin \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{opposite}} : \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac{\left(\frac{\mathrm{opposite} \times \mathrm{hypotenuse}}{\mathrm{opposite} \times \mathrm{adjacent}} \right) } { \left( \frac {\mathrm{adjacent} \times \mathrm{hypotenuse}} {\mathrm{opposite} \times \mathrm{adjacent} } \right) } = \frac{\left( \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} \right)} { \left( \frac{\mathrm{hypotenuse}}{\mathrm{opposite}} \right)} = \frac {\sec \theta}{\csc \theta} Or : \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\left( \frac{1}{\csc \theta} \right) }{\left( \frac{1}{\sec \theta} \right) } = \frac{\left( \frac{\csc \theta \sec \theta}{\csc \theta} \right) }{\left( \frac{\csc \theta \sec \theta}{\sec \theta} \right) } = \frac{\sec \theta}{\csc \theta} : \cot \theta = \frac {\csc \theta}{\sec \theta}
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. :\csc^2\theta + \sec^2\theta = \frac{h^2}{a^2} + \frac{h^2}{b^2} = \frac{a^2+b^2}{a^2} + \frac{a^2+b^2}{b^2} = 2\ + \frac{b^2}{a^2} + \frac{a^2}{b^2} Substituting with appropriate functions - : 2\ + \frac{b^2}{a^2} + \frac{a^2}{b^2} = 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{\pi}{2} - \alpha, making RQO = \alpha, RQP = \frac{\pi}{2}-\alpha, and finally RPQ = \alpha) :RPQ = \tfrac{\pi}{2} - RQP = \tfrac{\pi}{2} - (\tfrac{\pi}{2} - RQO) = RQO = \alpha :OP = 1 :PQ = \sin \beta :OQ = \cos \beta :\frac{AQ}{OQ} = \sin \alpha, so AQ = \sin \alpha \cos \beta :\frac{PR}{PQ} = \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 the
reflection identities of
even and odd functions, 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{OA}{OQ} = \cos \alpha, so OA = \cos \alpha \cos \beta :\frac{RQ}{PQ} = \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 the
reflection identities of
even and odd functions, 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{align} \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{align}
Tangent and cotangent From the sine and cosine formulae, we get :\tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} Dividing both numerator and denominator by \cos \alpha \cos \beta , we get :\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} Subtracting \beta from \alpha , using \tan (- \beta) = -\tan \beta , :\tan (\alpha - \beta) = \frac{\tan \alpha + \tan (-\beta)}{1 - \tan \alpha \tan (-\beta)} = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} Similarly, from the sine and cosine formulae, we get :\cot (\alpha + \beta) = \frac{\cos (\alpha + \beta)}{\sin (\alpha + \beta)} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta + \cos \alpha \sin \beta} Then by dividing both numerator and denominator by \sin \alpha \sin \beta , we get :\cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} Or, using \cot \theta = \frac{1}{\tan \theta} , :\cot (\alpha + \beta) = \frac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta} = \frac{\frac{1}{\tan \alpha \tan \beta} - 1}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}} = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} Using \cot (- \beta) = -\cot \beta , :\cot (\alpha - \beta) = \frac{\cot \alpha \cot (-\beta) - 1}{ \cot \alpha + \cot (-\beta) } = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}
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{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2}{\cot \theta - \tan \theta} :\cot (2 \theta) = \frac{\cot^2 \theta - 1}{2 \cot \theta} = \frac{\cot \theta - \tan \theta}{2} It can also be proved using
Euler's formula : e^{i \varphi}=\cos \varphi +i \sin \varphi Squaring both sides yields : e^{i 2\varphi}=(\cos \varphi +i \sin \varphi)^{2} But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields : e^{i 2\varphi}=\cos 2\varphi +i \sin 2\varphi It follows that :(\cos \varphi +i \sin \varphi)^{2}=\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{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{2}, :\sin \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{2}. 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{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{1 + \cos \theta}. Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to: :\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}. Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is: :\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}. This also gives: :\tan \frac{\theta}{2} = \csc \theta - \cot \theta. Similar manipulations for the cot function give: :\cot \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 - \cos \theta} = \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{align} \psi & = \pi - \theta - \phi \\ \tan(\psi) & = \tan(\pi - \theta - \phi) \\ & = - \tan(\theta + \phi) \\ & = \frac{- \tan\theta - \tan\phi}{1 - \tan\theta \tan\phi} \\ & = \frac{\tan\theta + \tan\phi}{\tan\theta \tan\phi - 1} \\ (\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{align}
Miscellaneous – the triple cotangent identity If \psi + \theta + \phi = \tfrac{\pi}{2} = 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{\pi}{2} :\therefore (\tfrac{\pi}{2}-\psi) + (\tfrac{\pi}{2}-\theta) + (\tfrac{\pi}{2}-\phi) = \tfrac{3\pi}{2} - (\psi+\theta+\phi) = \tfrac{3\pi}{2} - \tfrac{\pi}{2} = \pi so the result follows from the triple tangent identity.
Sum to product identities • \sin \theta \pm \sin \phi = 2 \sin \left ( \frac{\theta\pm \phi}2 \right ) \cos \left ( \frac{\theta\mp \phi}2 \right ) • \cos \theta + \cos \phi = 2 \cos \left ( \frac{\theta+\phi}2 \right ) \cos \left ( \frac{\theta-\phi}2 \right ) • \cos \theta - \cos \phi = -2 \sin \left ( \frac{\theta+\phi}2 \right ) \sin \left ( \frac{\theta-\phi}2 \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 = \frac{\theta + \phi}2 and \beta = \frac{\theta - \phi}2 Substitute \theta and \phi :\sin \theta + \sin \phi = 2 \sin \left( \frac{\theta + \phi}2 \right) \cos \left( \frac{\theta - \phi}2 \right) :\sin \theta - \sin \phi = 2 \cos \left( \frac{\theta + \phi}2 \right) \sin \left( \frac{\theta - \phi}2 \right) = 2 \sin \left( \frac{\theta - \phi}2 \right) \cos \left( \frac{\theta + \phi}2 \right) Therefore, :\sin \theta \pm \sin \phi = 2 \sin \left( \frac{\theta\pm \phi}2 \right) \cos \left( \frac{\theta\mp \phi}2 \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( \frac{\theta+\phi}2 \right) \cos \left( \frac{\theta-\phi}2 \right) :\cos \theta - \cos \phi = -2 \sin \left( \frac{\theta+\phi}2 \right) \sin \left( \frac{\theta-\phi}2 \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 This geometric argument relies on definitions of
arc length and
area, 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{\sin \theta}{\theta} For negative values of we have, by the symmetry of the sine function :\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} Hence :\frac{\sin \theta}{\theta} and :\frac{\tan \theta}{\theta} > 1\quad \text{if }\quad 0 ==Identities involving calculus==