Pythagorean trigonometric identity

From Wikipedia, the free encyclopedia

Jump to: navigation, search

The Pythagorean trigonometric identity is a trigonometric identity expressing the Pythagorean theorem in terms of trigonometric functions. Along with the sum-of-angles formulae (see Angle sum and difference identities) it is the basic relation among the sin and cos functions from which all others may be derived (see other definitions for the relevant theorem).

Contents

[edit] Statement of the identity

Mathematically, the Pythagorean identity states:

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

(Note that sin2 x means (sin x)2.)

If the length of the hypotenuse of a right triangle is 1, then the lengths of the legs are the sine and cosine of one of the angles. Therefore this trigonometric identity follows from the Pythagorean theorem.

The identities

1 + \tan^2 x = \sec^2 x\,

and

1 + \cot^2 x = \csc^2 x\,

are also called Pythagorean trigonometric identities. If one leg of a right triangle has length 1, then the tangent of the angle adjacent to that leg is the length of the other leg, and the secant of the angle is the length of the hypotenuse. In that way, this trigonometric identity involving the tangent and the secant follows from the Pythagorean theorem. The angle opposite the leg of length 1 has cotangent equal to the length of the other leg, and cosecant equal to the length of the hypotenuse. In that way, this trigonometric identity involving the cotangent and the cosecant follows from the Pythagorean theorem.

Any of these three trigonometric identities can be derived from any of the others using simple algebra.

[edit] Proofs and their relationships to the Pythagorean theorem

[edit] Using right-angled triangles

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

\sin x = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}
\cos x = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

the theorem follows by squaring both and adding; the left-hand side of the identity then becomes

\frac{\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 ½π radians (not inclusive) and therefore this argument does not prove the identity for all angles. Values of 0 and ½π are trivially proven by direct evaluation of sin and cos at those angles.

To complete the proof, the identities found at Trigonometric symmetry, shifts, and periodicity must be employed. By the periodicity identities 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 ½π < x ≤ π, to do this we let t = x - ½π, t will now be in the range 0 < x ≤ ½π. 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+\frac{1}{2}\pi\right) + \cos^2\left(t+\frac{1}{2}\pi\right) \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)\,.

[edit] Using the unit circle

If the trigonometric functions are defined in terms of the unit circle, the proof is immediate: given an angle θ, there is a unique point P on the unit circle centered at the origin in the Euclidean plane at an angle θ from the x-axis, and cos θ, sin θ 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, hence 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.

[edit] Using power series

The trigonometric functions may also be defined using power series, namely (for x an angle measured in radians):

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

Using the formal multiplication law for power series at Multiplication and division of power series (suitably modified to account for the form of the series here) we obtain

\begin{align} \sin^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)} \\ & = \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n} \\ & = \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \end{align}
\begin{align} \cos^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)} \\ & = \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n} \\ & = \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n} \end{align}

Note that in the expression for sin2, n must be at least 1, while in the expression for cos2, the constant term is equal to 1. The remaining terms of their sum are (with common factors removed)

\sum_{i = 0}^n {2n \choose 2i} - \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} = \sum_{j = 0}^{2n} (-1)^j {2n \choose j} = (1 - 1)^{2n} = 0

by the binomial theorem. The Pythagorean theorem is not closely related to the Pythagorean identity when the trigonometric functions are defined in this way; instead, in combination with the theorem, the identity now shows that these power series parameterize the unit circle, which we used in the previous section. Note that this definition actually constructs the sin and cos functions in a rigorous fashion and proves that they are differentiable, so that in fact it subsumes the previous two.

[edit] Using the differential equation

It is possible (see Definitions of trigonometric functions via differential equations) to define the sin and cos functions as the two unique solutions to the differential equation

y'' + y = 0

satisfying respectively y(0) = 0,y'(0) = 1 and y(0) = 1,y'(0) = 0. It follows from the theory of ordinary differential equations that the former solution, sin, has the latter, cos, as its derivative, and it follows from this that the derivative of cos is −sin. To prove the Pythagorean identity it suffices to show that the function

z = sin2x + cos2x

is constant and equal to 1. However, differentiating it and applying the two facts just mentioned we see that z' = 0 so z is constant, and z(0) = 1.

This form of the identity likewise has no direct connection with the Pythagorean theorem.

[edit] See also

[edit] External links

Retrieved from "http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity"