Pythagorean trigonometric identity
Template:Short description Template:Use American English The Pythagorean trigonometric identity, also called simply the Pythagorean identity, is an identity expressing the Pythagorean theorem in terms of trigonometric functions. Along with the sum-of-angles formulae, it is one of the basic relations between the sine and cosine functions.
The identity is
- <math>\sin^2 \theta + \cos^2 \theta = 1.</math>
As usual, <math>\sin^2 \theta</math> means <math display="inline">(\sin\theta)^2</math>.
Proofs and their relationships to the Pythagorean theoremEdit
Proof based on right-angle trianglesEdit
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 Template:Math.
The elementary definitions of the sine and cosine functions in terms of the sides of a right triangle are:
<math display="block">\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}</math>
The Pythagorean identity follows by squaring both definitions above, and adding; the left-hand side of the identity then becomes
<math display="block">\frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2}</math>
which by the Pythagorean theorem is equal to 1. This definition is valid for all angles, due to the definition of defining Template:Math and Template:Math for the unit circle and thus Template:Math and Template:Math for a circle of radius Template:Mvar and reflecting our triangle in the Template:Nowrap and setting Template:Math and Template:Math.
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 Template:Math then it is true for all real Template:Mvar. Next we prove the identity in the range Template:Math. To do this we let Template:Math, Template:Mvar will now be in the range Template:Math. We can then make use of squared versions of some basic shift identities (squaring conveniently removes the minus signs):
<math display="block">\sin^2\theta + \cos^2\theta = \sin^2\left(t + \tfrac{1}{2}\pi\right) + \cos^2\left(t + \tfrac{1}{2}\pi\right) = \cos^2 t + \sin^2 t = 1.</math>
Finally, it remains is to prove the formula for Template:Math; this can be done by squaring the symmetry identities to get
<math display="block">\sin^2\theta = \sin^2(-\theta)\text{ and }\cos^2\theta = \cos^2(-\theta).</math>
Related identitiesEdit
The two identities <math display="block">\begin{align} 1 + \tan^2 \theta &= \sec^2 \theta \\ 1 + \cot^2 \theta &= \csc^2 \theta \end{align}</math> are also called Pythagorean trigonometric identities.<ref name= Leff> Template:Cite book</ref> 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.
<math display="block">\begin{align} \tan \theta &= \frac{b}{a}\,, \\ \sec \theta &= \frac{c}{a}\,. \end{align}</math>
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 Template:Math) 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.
| Divisor | Divisor Equation | Derived Identity | Derived Identity (Alternate) |
|---|---|---|---|
| Template:Math | <math> \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}</math> | <math> \tan^2 \theta + 1 = \sec^2 \theta</math> | <math> \begin{align}
\sec^2\theta - \tan^2\theta = 1\\ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1\\ \end{align}</math> |
| Template:Math | <math> \frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}</math> | <math> 1 + \cot^2 \theta = \csc^2 \theta</math> | <math> \begin{align}
\csc^2\theta - \cot^2\theta = 1\\ (\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1\\ \end{align}</math> |
Proof using the unit circleEdit
{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}}
The unit circle centered at the origin in the Euclidean plane is defined by the equation:<ref name=circle>
This result can be found using the distance formula <math>d = \sqrt{x^2 +y^2}</math> for the distance from the origin to the point Template:Nowrap See Template:Cite book This approach assumes Pythagoras' theorem. Alternatively, one could simply substitute values and determine that the graph is a circle.
</ref>
- <math>x^2 + y^2 = 1.</math>
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:<ref name=Shaw>
</ref>
<math display="block">x = \cos\theta \ \text{ and }\ y = \sin\theta.</math>
Consequently, from the equation for the unit circle, <math display="block">\cos^2 \theta + \sin^2 \theta = 1,</math> the Pythagorean identity.
In the figure, the point Template:Mvar has a Template:Em Template:Mvar-coordinate, and is appropriately given by Template:Math, which is a negative number: Template:Math. Point Template:Mvar has a positive Template:Mvar-coordinate, and Template:Math. As Template:Mvar increases from zero to the full circle Template:Math, the sine and cosine change signs in the various quadrants to keep Template:Mvar and Template:Mvar with the correct signs. The figure shows how the sign of the sine function varies as the angle changes quadrant.
Because the Template:Mvar- and Template:Mvar-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 seriesEdit
The trigonometric functions may also be defined using power series, namely for Template:Mvar (an angle measured in radians):<ref name=Hamilton>Template:Cite book</ref><ref name=Krantz>Template:Cite book</ref>
<math display="block">\begin{align}
\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}.
\end{align}</math>
Using the multiplication formula for power series at Multiplication and division of power series (suitably modified to account for the form of the series here) we obtain
<math display="block">\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},\\ \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}</math>
In the expression for Template:Math, Template:Mvar must be at least 1, while in the expression for Template:Math, the constant term is equal to 1. The remaining terms of their sum are (with common factors removed)
<math display="block">\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</math>
by the binomial theorem. Consequently, <math display="block">\sin^2 x + \cos^2 x = 1,</math> which is the Pythagorean trigonometric identity.
When the trigonometric functions are defined in this way, the identity in combination with the Pythagorean theorem shows that these power series parameterize the unit circle, which we used in the previous section. This definition constructs the sine and cosine functions in a rigorous fashion and proves that they are differentiable, so that in fact it subsumes the previous two.
Proof using the differential equationEdit
Sine and cosine can be defined as the two solutions to the differential equation:<ref name=Debnath>Template:Cite book</ref>
<math display="block">y + y = 0</math>
satisfying respectively Template:Math, Template:Math and Template:Math, Template:Math. 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
<math display="block">z = \sin^2 x + \cos^2 x</math>
is constant and equal to 1. Differentiating using the chain rule gives:
<math display="block">\frac{d}{dx} z = 2 \sin x \cos x + 2 \cos x(-\sin x) = 0,</math> so Template:Mvar is constant. A calculation confirms that Template:Math, and Template:Mvar is a constant so Template:Math for all Template:Mvar, 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 formulaEdit
Using Euler's formula <math>e^{i\theta} = \cos\theta + i\sin\theta</math> and factoring <math>\cos^2 \theta + \sin^2 \theta</math> as the complex difference of two squares,
<math display="block">\begin{align} 1 &= e^{i\theta}e^{-i\theta} \\[3mu] &= (\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) \\[3mu] &= \cos^2 \theta + \sin^2 \theta. \end{align}</math>
See alsoEdit
NotesEdit
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