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

Definitions

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

Ratio identities

The following identities are trivial algebraic consequences of these definitions and the division identity.

c is whatever value (not necessarily trigonometric), only to understand the simple demonstrations above.
That is because not appear in the graph.

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:

Pythagorean identities

Identity 1:


Proof 1:

Refer to the triangle diagram above. Note that by Pythagorean theorem

Pythagorean theorem

In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...

.

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of by ; for the second, divide by .


Similarly


Proof 2:

Differentiating the left-hand side of the identity yields:


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.


Proof 1:

Refer to the triangle diagram above. Note that by Pythagorean theorem

Pythagorean theorem

In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...

.

Substituting with appropriate functions -


Rearranging gives:

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
)



, so
, so


By substituting for and using Symmetry, we also get:


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


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


Also using the following properties of exponential functions:


Evaluating the product:

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

Cosine

Using the figure above,



, so
, so


By substituting for and using Symmetry, we also get:


Also, using the complementary angle formulae,

Tangent and cotangent

From the sine and cosine formulae, we get


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



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

Double-angle identities

From the angle sum identities, we get


and


The Pythagorean identities give the two alternative forms for the latter of these:



The angle sum identities also give



It can also be proved using Euler's formula

Euler's formula

Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the deep relationship between the trigonometric functions and the complex exponential function...

Squaring both sides yields


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


It follows that
.

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

Because the imaginary and real parts have to be the same, we are left with the original identities
,

and also
.

Half-angle identities

The two identities giving alternative forms for cos 2θ give these:



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


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


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


This also gives


Similar manipulations for the cot function give

Prosthaphaeresis identities

• 

• 

• 

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.




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


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


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


Hence

Preliminaries

These can be seen from looking at the diagrams.

Sine and angle ratio identity

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

Cosine and angle ratio identity

Proof:




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

Proof:

As in the preceding proof,


The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.