Proof based on right-angle triangles Any
similar triangles have the property that if we select the same
angle in all of them, the ratio of the two sides defining the angle is the same regardless of which similar triangle is selected, regardless of its actual size: the ratios depend upon the three angles, not the lengths of the sides. Thus for either of the similar
right triangles in the figure, the ratio of its horizontal side to its
hypotenuse is the same, namely . The elementary definitions of the sine and cosine functions in terms of the sides of a right triangle are: \begin{alignat}{3} \sin \theta &= \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac{b}{c} \\ \cos \theta &= \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac{a}{c} \end{alignat} The Pythagorean identity follows by
squaring both definitions above, 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. This definition is valid for all angles, due to the definition of defining and for the unit circle and thus and for a circle of radius and reflecting our triangle in the and setting and . Alternatively, the identities found at
Trigonometric symmetry, shifts, and periodicity may be employed. By the
periodicity identities we can say if the formula is true for then it is true for all
real . Next we
prove the identity in the range . To do this we let , will now be in the range . We can then make use of squared versions of some basic shift identities (squaring conveniently removes the minus signs): \sin^2\theta + \cos^2\theta = \sin^2\left(t + \tfrac{\pi}{2}\right) + \cos^2\left(t + \tfrac{\pi}{2}\right) = \cos^2 t + \sin^2 t = 1. Finally, it remains is to prove the formula for ; this can be done by squaring the symmetry identities to get \sin^2\theta = \sin^2(-\theta)\text{ and }\cos^2\theta = \cos^2(-\theta).
Related identities applied to the blue triangle shows the identity , and applied to the red triangle shows that . The two identities \begin{align} 1 + \tan^2 \theta &= \sec^2 \theta \\ 1 + \cot^2 \theta &= \csc^2 \theta \end{align} 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. \begin{align} \tan \theta &= \frac{b}{a}\,, \\ \sec \theta &= \frac{c}{a}\,. \end{align} In this way, this trigonometric identity involving the tangent and the secant follows from the Pythagorean theorem. The angle opposite the leg of length 1 (this angle can be labeled ) 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 also follows from the Pythagorean theorem. The following table gives the identities with the factor or
divisor that relates them to the main identity.
Proof using the unit circle (bottom) The unit circle centered at the origin in the
Euclidean plane is defined by the equation: :x^2 + y^2 = 1. Given an angle
θ, there is a unique point
P on the unit circle at an anticlockwise angle of
θ from the
x-axis, and the
x- and
y-coordinates of
P are: x = \cos\theta \ \text{ and }\ y = \sin\theta. Consequently, from the equation for the unit circle, \cos^2 \theta + \sin^2 \theta = 1, the Pythagorean identity. In the figure, the point has a -coordinate, and is appropriately given by , which is a
negative number: . Point has a positive -coordinate, and . As increases from zero to the full circle , the sine and cosine change signs in the various quadrants to keep and with the correct signs. The figure shows how the sign of the sine function varies as the angle changes quadrant. Because the - and -axes are perpendicular, this Pythagorean identity is 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.
Proof using power series The trigonometric functions may also be defined using
power series, namely for (an angle measured in
radians): y'' + y = 0 satisfying respectively , and , . It follows from the theory of
ordinary differential equations that the first solution, sine, has the second, cosine, as its
derivative, and it follows from this that the derivative of cosine is the negative of the sine. The identity is equivalent to the assertion that the function z = \sin^2 x + \cos^2 x is constant and equal to 1.
Differentiating using the
chain rule gives: \frac{d}{dx} z = 2 \sin x \cos x + 2 \cos x(-\sin x) = 0, so is constant. A calculation confirms that , and is a constant so for all , so the Pythagorean identity is established. A similar proof can be completed using power series as above to establish that the sine has as its derivative the cosine, and the cosine has as its derivative the negative sine. In fact, the definitions by ordinary differential equation and by power series lead to similar derivations of most identities. This proof of the identity has no direct connection with
Euclid's demonstration of the Pythagorean theorem.
Proof using Euler's formula Factoring \cos^2 \theta + \sin^2 \theta as the complex
difference of two squares and using
Euler's formula e^{i\theta} = \cos\theta + i\sin\theta, \begin{align} \cos^2 \theta + \sin^2 \theta &= \cos^2 \theta - i^2 \sin^2 \theta \\[3mu] &= (\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) \\[3mu] &= e^{i\theta}e^{-i\theta} \\ &= e^{i\theta-i\theta} = e^0 = 1 \end{align} ==See also==