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.

Elementary trigonometric identities

Definitions

Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse.

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.
They rely on multiplying or dividing the numerator and denominator of fractions by a variable. Ie,

\frac {a}{b}= \frac {\left(\frac {a}{h}\right)} {\left(\frac {b}{h}\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{\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(x) + \cos^2(x) = 1\,

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

\tan^2(x) + 1\ = \sec^2(x)
1\ + \cot^2(x) = \csc^2(x)

Similarly

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

Identity 2:

The following accounts for all three reciprocal functions.

\csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x)

Proof 2:

Refer to the triangle diagram above. Note that a^2+b^2=h^2 by Pythagorean theorem.

\csc^2(x) + \sec^2(x) = \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(x)+ \cot^2(x)

Rearranging gives:

\csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x)

Angle sum identities

Sine

Illustration of the sum formula.

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 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 QR parallel to the x-axis.

Now angle RPQ = \alpha (because OQA = 90 - \alpha, making RQO = \alpha, RQP = 90-\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 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

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

e^{i\varphi}=\cos \varphi +i \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 + \beta)} = \cos (\alpha +\beta) + i \sin(\alpha +\beta)

Also using the following properties of exponential functions:

e^{i(\alpha + \beta)} = e^{i \alpha} e^{i\beta}= (\cos \alpha +i \sin \alpha) (\cos \beta + i \sin \beta)

Evaluating the product:

e^{i(\alpha + \beta)} = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)+i(\sin \alpha \cos \beta + \sin \beta \cos \alpha)

Equating real and imaginary parts:

\cos (\alpha +\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta
\sin (\alpha +\beta)=\sin \alpha \cos \beta + \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 + \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{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 π 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:[1]

\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.

Prosthaphaeresis identities

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

Illustration of the sine and tangent 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. It 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.[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\,

For negative values of θ we have, by symmetry of the sine function

\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} < 1\,

Hence

\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ \theta \ne 0\,
\frac{\tan \theta}{\theta} > 1\ \ \ \mathrm{if}\ \ \ 0 < \theta < \frac{\pi}{2}\,

Identities involving calculus

Preliminaries

\lim_{\theta \to 0}{\sin \theta} = 0\,
\lim_{\theta \to 0}{\cos \theta} = 1\,

Sine and angle ratio identity

\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1

Proof: From the previous inequalities, we have, for small angles

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

Therefore,

\frac{\sin \theta}{\theta} < 1 < \frac{\tan \theta}{\theta}\,,

Consider the right-hand inequality. Since

\tan \theta = \frac{\sin \theta}{\cos \theta}
\therefore 1 < \frac{\sin \theta}{\theta \cos \theta}

Multply through by \cos \theta

\cos \theta < \frac{\sin \theta}{\theta}

Combining with the left-hand inequality:

\cos \theta < \frac{\sin \theta}{\theta} < 1

Taking \cos \theta to the limit as \theta \to 0

\lim_{\theta \to 0}{\cos \theta} = 1\,

Therefore,

\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1

Cosine and angle ratio identity

\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta} = 0

Proof:

\begin{align} \frac{1 - \cos \theta}{\theta} & = \frac{1 - \cos^2 \theta}{\theta (1 + \cos \theta)}\\ & = \frac{\sin^2 \theta}{\theta (1 + \cos \theta)}\\ & = \left( \frac{\sin \theta}{\theta} \right) \times \sin \theta \times \left( \frac{1}{1 + \cos \theta} \right)\\ \end{align}

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_{\theta \to 0}\frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}

Proof:

As in the preceding proof,

\frac{1 - \cos \theta}{\theta^2} = \frac{\sin \theta}{\theta} \times \frac{\sin \theta}{\theta} \times \frac{1}{1 + \cos \theta}.\,

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{x}{\sqrt{1+x^2}}

Proof:

We start from

\sin^2\theta+\cos^2\theta=1

Then we divide this equation by \cos^2\theta

\cos^2\theta=\frac{1}{\tan^2\theta+1}

Then use the substitution \theta=\arctan(x), also use the Pythagorean trigonometric identity:

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

Then we use the identity \tan[\arctan(x)]\equiv x

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

See also

Notes

  1. ↑ http://mathlaoshi.com/tags/tangent-identity/ dead link
  2. ↑ Richman, Fred (March 1993). . "A Circular Argument". The College Mathematics Journal 24 (2): 160–162. doi:10.2307/2686787. Retrieved 3 November 2012.

References

  • E. T. Whittaker and G. N. Watson. A course of modern analysis, Cambridge University Press, 1952