Pythagorean trigonometric identity

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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 trigonometric identity#Angle sum and difference identities) it is the basic relation among the sin and cos functions from which all others may be derived (see trigonometric function#Other definitions for the relevant theorem).

1 Statement of the identity
2 Proofs and their relationships to the Pythagorean theorem
3 See also

[edit] Statement of the identity

Mathematically, the Pythagorean identity states

$\left ( \sin(x) \right )^2 + \left ( \cos(x) \right )^2=1$

Or written in another way

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

From this form the two others listed here may be derived 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 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 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).

$\sin^2x+\cos^2x\equiv\sin^2(t+\frac{1}{2}\pi)+\cos^2x(t+\frac{1}{2}\pi)\equiv\cos^2t+\sin^2t\equiv1$

All that remains is to prove it for $- π < x < 0$ , this can be done by squaring the symmetry identities to get $\sin^2x\equiv\sin^2(-x)$ 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, 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.

[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 power series#Multiplication and division (suitably modified to account for the form of the series here) we obtain

$\sin^2(x)\,$	$= \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)}$
	$= \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n}$
	$= \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}$

$\cos^2(x)\,$	$= \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)}$
	$= \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n}$
	$= \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n}$

Note that in the expression for $sin 2$ , n must be at least 1, while in the expression for $cos 2$ , 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 is the first we have encountered which 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 trigonometric function#Definitions 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 = sin 2 (x) + cos 2 (x)

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.

Categories: Mathematical identities | Proofs | Trigonometry

Pythagorean trigonometric identity

From Wikipedia, the free encyclopedia

Contents

[edit] Statement of the identity

[edit] Proofs and their relationships to the Pythagorean theorem

[edit] Using right angled triangles

[edit] Using the unit circle

[edit] Using power series

[edit] Using the differential equation

[edit] See also

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