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STATEMENT OF THE IDENTITY

Mathematically, the Pythagorean identity states

:\left ( \sin(x) ight )^2 + \left ( \cos(x) ight )^2=1

Or written in another way

:\sin^2(x) + \cos^2(x) = 1\qquad\qquad (1) \!

The identities

:1 + an^2(x) = \sec^2(x)\,

and

:1 + \cot^2(x) = \csc^2(x)\,

are also called Pythagorean trigonometric identities. They can be derived from (1) using simple algebra. Like (1), they also have simple geometric interpretations as instances of the Pythagorean Theorem .


PROOFS AND THEIR RELATIONSHIPS TO THE PYTHAGOREAN THEOREM


Using right-angled triangles

Using the elementary "definition" of the trigonometric functions in terms of the sides of a right triangle,

:\sin(x) = rac{\mathrm{opposite}}{\mathrm{hypotenuse}}
:\cos(x) = rac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

the theorem follows by squaring both and adding; the Left-hand Side of the identity then becomes

: rac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2}

which by the Pythagorean theorem is equal to 1. Note, however, that this definition is only valid for angles between 0 and ½''π'' radians (not inclusive) and therefore this argument does not prove the identity for all angles. Values of 0 and ½''π'' are trivially proven by direct evaluation of sin and cos at those angles.

To complete the proof, the identities found at Trigonometric Identity#Periodicity, Symmetry, And Shifts must be employed. By the periodicity identies we can say if the formula is true for -''π'' < x ≤ ''π'' then it is true for all real x. Next we prove the range ½''π'' < x ≤ ''π'', to do this we let ''t'' = ''x'' - ½''π'', t will now be in the range 0 < x ≤ ½''π''. We can then make use of squared versions of some basic shift identites (squaring conveniently removes the minus signs).

: \sin^2x+\cos^2x \equiv \sin^2\left(t+ rac{1}{2}\pi ight) + \cos^2x\left(t+ rac{1}{2}\pi ight) \equiv \cos^2t+\sin^2t \equiv 1.

All that remains is to prove it for −π < ''x'' < 0, this can be done by squaring the symmetry identities to get

: \sin^2x\equiv\sin^2(-x)\mbox{ and }\cos^2x\equiv\cos^2(-x).


Using the unit circle

If the trigonometric functions are defined in terms of the unit circle, the proof is immediate: given an angle heta, there is a unique point ''P'' on the unit circle centered at the origin in the Euclidean plane at an angle heta from the ''x''-axis, and \cos( heta), \sin( heta) are respectively the ''x''- and ''y''-coordinates of ''P''. By definition of the unit circle, the sum of the squares of these coordinates is ''1'', whence the identity.

The relationship to the Pythagorean theorem is through the fact that the unit circle is actually defined by the equation

:x^2 + y^2 = 1

Since the ''x''- and ''y''-axes are perpendicular, this fact is actually equivalent to the Pythagorean theorem for triangles with hypotenuse of length ''1'' (which is in turn equivalent to the full Pythagorean theorem by applying a similar-triangles argument). See Unit Circle for a short explanation.


Using power series

The trigonometric functions may also be defined using Power Series , namely (for ''x'' an angle measured in Radian s):

:\sin(x) = \sum_{n = 0}^\infty rac{(-1)^n}{(2n + 1)!} x^{2n + 1}
:\cos(x) = \sum_{n = 0}^\infty rac{(-1)^n}{(2n)!} x^{2n}

Using the formal multiplication law for power series at Power Series#Multiplication And Division (suitably modified to account for the form of the series here) we obtain