Proofs of trigonometric identities

Proofs of trigonometric identities are used to show relations between trigonometric functions. This article will list trigonometric identities and prove them.

1 Elementary trigonometric identities
2 Identities involving calculus
3 References

Elementary trigonometric identities

Definitions

Referring to the diagram at the right, the six trigonometric functions of θ are:

$\sin \theta = \frac {\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac {a}{h}$

$\cos \theta = \frac {\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac {b}{h}$

$\tan \theta = \frac {\mathrm{opposite}}{\mathrm{adjacent}} = \frac {a}{b}$

$\cot \theta = \frac {\mathrm{adjacent}}{\mathrm{opposite}} = \frac {b}{a}$

$\sec \theta = \frac {\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac {h}{b}$

$\csc \theta = \frac {\mathrm{hypotenuse}}{\mathrm{opposite}} = \frac {h}{a}$

Ratio identities

The following identities are trivial algebraic consequences of these definitions and the division identity.
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$\frac {a}{b}= \frac {\left(\frac {a}{c}\right)} {\left(\frac {b}{c}\right) }.$

$\tan \theta = \frac \mathrm{opposite}\mathrm{adjacent} = \frac { \left( \frac \mathrm{opposite} \mathrm{hypotenuse} \right) } { \left( \frac \mathrm{adjacent} \mathrm{hypotenuse} \right) } = \frac {\sin \theta} {\cos \theta}.$

$\cot \theta = \frac {\cos \theta}{\sin \theta}.$

$\cot \theta =\frac \mathrm{adjacent} \mathrm{opposite} = \frac { \left( \frac \mathrm{adjacent}\mathrm{adjacent} \right) } { \left( \frac \mathrm{opposite}\mathrm{adjacent} \right) } = \frac {1}{\tan \theta}.$

$\sec \theta = \frac {1}{\cos \theta}$

$\csc \theta = \frac {1}{\sin \theta}$

$\tan \theta = \frac \mathrm{opposite}\mathrm{adjacent} = \frac {\left(\frac {\mathrm{opposite} \times \mathrm{hypotenuse}}{\mathrm{opposite} \times \mathrm{adjacent}} \right) } { \left( \frac {\mathrm{adjacent} \times \mathrm{hypotenuse}} {\mathrm{opposite} \times \mathrm{adjacent} } \right) } = \frac { \left( \frac \mathrm{hypotenuse} \mathrm{adjacent} \right)} { \left( \frac \mathrm{hypotenuse} \mathrm{opposite} \right) } = \frac {\sec \theta}{\csc \theta}.$

$\cot \theta = \frac {\csc \theta}{\sec \theta}.$

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:

$\sin^2(x) %2B \cos^2(x) = 1\,$

Proof 1:

Refer to the triangle diagram above. Note that $a^2%2Bb^2=h^2$ by Pythagorean theorem.

$\sin^2(x) %2B \cos^2(x) = \frac{a^2}{h^2} %2B \frac{b^2}{h^2} = \frac{a^2%2Bb^2}{h^2} = \frac{h^2}{h^2} = 1.\,$

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of $\sin^2(x) %2B \cos^2(x) = 1$ by $\cos^2(x)$ ; for the second, divide by $\sin^2(x)$ .

$\tan^2(x) %2B 1\ = \sec^2(x)$

$\sec^2(x) - \tan^2(x) = 1\$

Similarly

$1\ %2B \cot^2(x) = \csc^2(x)$

$\csc^2(x) - \cot^2(x) = 1\$

Proof 2:

Differentiating the left-hand side of the identity yields:

$2 \sin x \cdot \cos x - 2 \sin x \cdot \cos x = 0$

Integrating this shows that the original identity is equal to a constant, and this constant can be found by plugging in any arbitrary value of x.

Identity 2:

The following accounts for all three reciprocal functions.

$\csc^2(x) %2B \sec^2(x) - \cot^2(x) = 2\ %2B \tan^2(x)$

Proof 1:

Refer to the triangle diagram above. Note that $a^2%2Bb^2=h^2$ by Pythagorean theorem.

$\csc^2(x) %2B \sec^2(x) = \frac{h^2}{a^2} %2B \frac{h^2}{b^2} = \frac{a^2%2Bb^2}{a^2} %2B \frac{a^2%2Bb^2}{b^2} = 2\ %2B \frac{b^2}{a^2} %2B \frac{a^2}{b^2}$

Substituting with appropriate functions -

$2\ %2B \frac{b^2}{a^2} %2B \frac{a^2}{b^2} = 2\ %2B \tan^2(x)%2B \cot^2(x)$

Rearranging gives:

$\csc^2(x) %2B \sec^2(x) - \cot^2(x) = 2\ %2B \tan^2(x)$

Angle sum identities

Sine

Draw the angles α and β. Place P on the line defined by α + β at unit distance from the origin.

Let PQ be a perpendicular from P to the line defined by the angle α. OQP is a right angle.

Let QA be a perpendicular from Q to the x axis, and PB be a perpendicular from P to the x axis. OAQ is a right angle.

Draw QR parallel to the x-axis. Now angle RPQ = α (because $RPQ = \tfrac{\pi}{2} - RQP = \tfrac{\pi}{2} - (\tfrac{\pi}{2} - RQO) = RQO = \alpha$ )

$OP = 1\,$

$PQ = \sin \beta\,$

$OQ = \cos \beta\,$

$\frac{AQ}{OQ} = \sin \alpha\,$ , so $AQ = \sin \alpha \cos \beta\,$

$\frac{PR}{PQ} = \cos \alpha\,$ , so $PR = \cos \alpha \sin \beta\,$

$\sin (\alpha %2B \beta) = PB = RB%2BPR = AQ%2BPR = \sin \alpha \cos \beta %2B \cos \alpha \sin \beta\,$

By substituting $-\beta$ for $\beta$ and using Symmetry, we also get:

$\sin (\alpha - \beta) = \sin \alpha \cos -\beta %2B \cos \alpha \sin -\beta\,$

$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\,$

Another simple "proof" can be given using Euler's formula known from complex analysis: Euler's formula is:

$e^{i\varphi}=\cos \varphi %2Bi \sin \varphi$

Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles $\alpha$ and $\beta$ we have:

$e^{i (\alpha %2B \beta)} = \cos (\alpha %2B\beta) %2B i \sin(\alpha %2B\beta)$

Also using the following properties of exponential functions:

$e^{i(\alpha %2B \beta)} = e^{i \alpha} e^{i\beta}= (\cos \alpha %2Bi \sin \alpha) (\cos \beta %2B i \sin \beta)$

Evaluating the product:

$e^{i(\alpha %2B \beta)} = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)%2Bi(\sin \alpha \cos \beta %2B \sin \beta \cos \alpha)$

This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get:

$\cos (\alpha %2B\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\sin (\alpha %2B\beta)=\sin \alpha \cos \beta %2B \sin \beta \cos \alpha$

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 $RQ = \sin \alpha \sin \beta\,$

$\cos (\alpha %2B \beta) = OB = OA-BA = OA-RQ = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta\,$

By substituting $-\beta$ for $\beta$ and using Symmetry, we also get:

$\cos (\alpha - \beta) = \cos \alpha \cos - \beta\ - \sin \alpha \sin - \beta\,$

$\cos (\alpha - \beta) = \cos \alpha \cos \beta %2B \sin \alpha \sin \beta\,$

Also, using the complementary angle formulae,

$\cos (\alpha %2B \beta) = \sin\left( \pi/2-(\alpha %2B \beta)\right) = \sin\left( (\pi/2-\alpha) - \beta\right)\,$

$= \sin\left( \pi/2-\alpha\right) \cos \beta - \cos\left( \pi/2-\alpha\right) \sin \beta\,$

$= \cos \alpha \cos \beta - \sin \alpha \sin \beta\,$

Tangent and cotangent

From the sine and cosine formulae, we get

$\tan (\alpha %2B \beta) = \frac{\sin (\alpha %2B \beta)}{\cos (\alpha %2B \beta)}\,$

$= \frac{\sin \alpha \cos \beta %2B \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}\,$

Dividing both numerator and denominator by cos α cos β, we get

$\tan (\alpha %2B \beta) = \frac{\tan \alpha %2B \tan \beta}{1 - \tan \alpha \tan \beta}\,$

$\tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 %2B \tan \alpha \tan \beta}\,$

Similarly (using a division by sin α sin β), we get

$\cot (\alpha %2B \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha %2B \cot \beta}\,$

$\cot (\alpha - \beta) = \frac{\cot \alpha \cot \beta %2B 1}{\cot \beta - \cot \alpha}\,$

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

$e^{i \varphi}=\cos \varphi %2Bi \sin \varphi$

Squaring both sides yields

$e^{i 2\varphi}=(\cos \varphi %2Bi \sin \varphi)^{2}$

But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields

$e^{i 2\varphi}=\cos 2\varphi %2Bi \sin 2\varphi$

It follows that

$(\cos \varphi %2Bi \sin \varphi)^{2}=\cos 2\varphi %2Bi \sin 2\varphi$ .

Expanding the square and simplifying on the left hand side of the equation gives

$i(2 \sin \varphi \cos \varphi) %2B \cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi %2Bi \sin 2\varphi$ .

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 alternative forms for cos 2θ give these:

$\cos \frac{\theta}{2} = \pm\, \sqrt\frac{1 %2B \cos \theta}{2},\,$

$\sin \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{2}.\,$

One must choose the sign of the square root properly—note that if 2π 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, we have

$\tan \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{1 %2B \cos \theta}.\,$

If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get

$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 %2B \cos \theta}.\,$

If instead we multiply the numerator and denominator by (1 - cos θ), we get

$\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}.\,$

This also gives

$\tan \frac{\theta}{2} = \csc \theta - \cot \theta.\,$

Similar manipulations for the cot function give

$\cot \frac{\theta}{2} = \pm\, \sqrt\frac{1 %2B \cos \theta}{1 - \cos \theta} = \frac{1 %2B \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 - \cos \theta} = \csc \theta %2B \cot \theta.\,$

Prosthaphaeresis identities

$\sin \theta \pm \sin y = 2 \sin \frac{\theta\pm y}2 \cos \frac{\theta\mp y}2$

$\cos \theta %2B \cos y = 2 \cos \frac{\theta%2By}2 \cos \frac{\theta-y}2$

$\cos \theta - \cos y = -2 \sin \frac{\theta%2By}2 \sin \frac{\theta-y}2$

Inequalities

The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2π) of the whole circle, so its area is θ/2.

$OA = OD = 1\,$

$AB = \sin \theta\,$

$CD = \tan \theta\,$

The area of triangle OAD is AB/2, or sinθ/2. The area of triangle OCD is CD/2, or tanθ/2.

Since triangle OAD lies completely inside the sector, which in turn lies completely inside triangle OCD, we have

$\sin \theta < \theta < \tan \theta\,$

This geometric argument applies if 0<θ<π/2. For the sine function, we can handle other values. If θ>π/2, then θ>1. But sinθ≤1 (because of the Pythagorean identity), so sinθ<θ. So we have

$\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ 0 < \theta\,$