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Note: (written November 6 2016) I'm glad you found this essay I wrote long ago. It's archived here on my website, together with my other essays from a decade ago. If you want to see my more recent content, my blog is the place to find it.

# Different ways of viewing cos2(t) + sin2(t) = 1

## by Walter Vannini

Introduction
Triangles
Unit Circle
Root Mean Square
Rotation Matrix
Determinant
Orthonormal Basis
Combinatorial Identities
Exponential Function
Algebra
Factorization
Parametrized Curve Derivatives
Integral
Conservation Of Energy
Hyperbolic Analogues
Feedback

## Introduction

We will explore various interpretations of the above identity. Later parts of this essay don't often depend on earlier parts, so please feel free to skim past incomprehensible text until you find something you're comfortable with.

## Triangles

The usual way the identity is understood is via Pythagoras' theorem. In a right angled triangle with sides a,b,c, and angle t at the vertex where a and c meet, cos(t) is by definition a/c, sin(t) is by definition b/c , and so cos2(t) + sin2(t) is (a/c)2+(b/c)2, which by simple algebraic manipulation is (a2+b2)/c2. Pythagoras' theorem states that a2+b2 is c2, and so this simplifies to 1.

## Unit Circle

Traditionally (ie. the way I learned it), the next step is to realize that the unit circle centered at (0,0) in the (x,y) plane is defined by x2+y2=1. The above identity can then be interpreted as saying that the point (cos(t), sin(t)) is on the unit circle. Furthermore, this approach leads to a definition of cos(t) and sin(t) for all real t.

## Root Mean Square

Another way to understand this identity is via the trigonometric identities

 cos2(x) = 1/2 + (1/2) cos(2x) sin2(x) = 1/2 - (1/2) cos(2x)

Incidentally, these identities pretty much leap out at you from looking at the graphs of y = cos2(x) and y = sin2(x).

Another way to look at it is to remember that the root mean square of cos is 1/√2, as is the root mean square of sin. This means that cos2 is 1/2 + variation, and sin2 is 1/2 + variation, and amazingly the variation of cos2 is precisely the negative of the variation of sin2. Actually, this does fit in with the fact that sin and cos are simply phase-shifted versions of each other, so that sin2and cos2 are phase-shifted versions of each other, so their variations should be somehow related.
Anyway, we now have that cos2(t)+sin2(t) is of the form (1/2 + deviation) + (1/2 - deviation), and so this is 1.

## Rotation Matrix

Yet another way to understand the identity is via 2 by 2 matrices. The linear matrix that represents an anticlockwise rotation by angle t is
 cos(t) -sin(t) sin(t) cos(t)

## Determinant

The matrix
 a b c d
has determinant ad-bc, so the rotation matrix has determinant cos(t)cos(t)-(-sin(t))sin(t), which is cos(t)cos(t)+sin(t)sin(t) , ie cos2(t)+sin2(t). The determinant of a square matrix has a simple geometric interpretation. It is the area scaling factor. For a rotation, area is unchanged, so the determinant has to be 1. So the identity cos2(t)+sin2 (t)=1 can be interpreted as expressing the rather obvious fact that rotating an object in the (x,y) plane does not change its area.

## Orthonormal Basis

The rotation matrix sends the point (1,0) to (cos(x),sin(x)) and the point (0,1) to (-sin(x), cos(x)). The standard basis e1=(1,0), e2=(0,1) is an orthonormal basis, meaning that
e1.e1=1 (check: 1.1+0.0=1)
e1.e2=0 (check: 0.1+1.0=0)
e2.e2=1 (check: 0.0+1.1=1)

Since a rotation is a "rigid motion", it will transform an orthonormal basis to another orthonormal basis. This means that the above three equations will be true when e1 is (cos(t), sin(t)), and e2 is (-sin(t), cos(t)). The two equations e1.e1 = 1 and e2.e2 = 1 are each a restatement of cos2(t) + sin2(t) = 1, while e1.e2 = 0 is a statement of the identity cos(t)(-sin(t))+sin(t)cos(t) = 0. Strangely enough, even the e1.e2 statement is a manifestation of the identity, as is shown below.

## Combinatorial Identities

Another way to look at the identity is via the power series expansions of cos(x) and sin(x).
 cos(x) = 1 - x2/2! + x4/4! - x6/6! + x8/8! - … sin(x) = x - x3/3! + x5/5! - x7/7! + …

This means that
 cos2(x) = 1 - (1/2! + 1/2!) x2 + (1/4! + 1/2!2! + 1/4!) x4/4! - … sin2(x) = x2 - (1/3! + 1/3!) x4/4! + …

In this context, the identity cos2(t)+sin2(t)=1 is really encoding an infinite number of identities involving factorials, namely
 1/2! + 1/2! = 1 1/4! + 1/2!2! + 1/4! = 1/3! + 1/3! 1/6! + 1/4!2! + 1/2!4! + 1/6! = 1/5! + 1/3!3! + 1/5! …

These identities can be reexpressed in terms of combinatorial coefficients ( nCr = n!/(n-r)!r!)
 2C0 + 2C2 = 2C1 4C0 + 4C2 + 4C4 = 4C1 + 4C3 6C0 + 6C2 + 6C4 + 6C6 = 6C1 + 6C3 + 6C5 …

These are saying that the even combinatorial coefficients sum up to the same result as the odd combinatorial coefficients. This result can be proved directly. It is most easily seen by looking at Pascal's triangle, and adding the terms in a row in two different ways. Furthermore, using the power series expansions to extend the definition of cos and sin to the complex numbers, we now know that the identity is true for complex t. Thanks to Bob Uva for pointing this out.

## Exponential Function

 Continuing along these lines, the fact that ex = 1 + x + x2/2! + x3/3! + x4/4! + … immediately tells us that eix = 1 + ix - x2/2! - ix3/3! + x4/4! + … which is, splitting into the even and odd powers eix = cos(x) + isin(x) At this point there are two ways to get to the identity.

## Algebra

The first is to realize that the single equation eix = cos(x)+isin(x) gives us the equation e-ix = cos(x)-isin(x) and from these two equations we can solve for cos(x) and sin(x) to get
cos(x) = (eix+e-ix)/2
sin(x) = (eix-e-ix)/2i.
Letting w represent eix, so that 1/w is e-ix, we have that cos2(x)+sin2(x) is (w+1/w)2/4 - (w-1/w)2/4. This simplifies to 1. So the trigonometric identity can be viewed as the algebraic identity (w+1/w)2 - (w-1/w)2 = 4.

## Factorization

The second is to be clever (in a different way), and realize we can factorize a2 + b2 as (a-ib)(a+ib). If you haven't seen this before, it's a restatement of a2-b2 = (a-b)(a+b). Using complex numbers, we have that -b2 is (ib)2, which leads to a2+b2 = a2-(-b2) = a2 - (ib)2 = (a-ib)(a+ib). Incidentally, it's of course also true that a2+b2 = (b-ia)(b+ia). Anyway, applying the factorization to cos2(x)+sin2(x), we get (cos(x)+isin(x))(cos(x)-isin(x)), which is eixe-ix. Using w to represent eix, the fact that cos2(x)+sin2(x)=1 is simply a convoluted way of saying w(1/w)=1.

## Parametrized Curve Derivatives

Consider the parametrized curve c(t) = (cos(t), sin(t)). The identity tells us that this parametrized curve is always on the unit circle about the origin. Differentiating, we have c'(t) = (-sin(t), cos(t)) . c'(t).c'(t) is (-sin(t))(-sin(t)) + cos(t)cos(t), which is just cos2(t)+sin2(t). The fact that this is 1 tells us that the parametrized curve is actually parametrized by arc length. So, the trigonometric identity can be viewed as simply stating the fact that radians traverse the unit circle at unit speed. Finally, differentiating again, we get c''(t) = (-cos(t), -sin(t)). Obviously, c''(t).c''(t) = cos2(t) + sin2(t), so that the identity tells us that moving uniformly along the circle results in an acceleration of constant magnitude.

## Integral

Finally, another approach to establishing that cos2(x)+sin2(x) is 1 is to realize that it is really stating two separate properties, namely that cos2(x)+sin2(x) is a constant, and that the constant happens to be 1, ie. for some value of x, cos2(x)+sin2(x) is 1. The latter property is quickly established: taking x=0, cos(x)=1 and sin(x)= 0. Clearly 12+02 =1.

Given a function f, to establish that f is a constant function, it suffices to establish that the derivative of f is zero. Applying this technique to f=cos2+sin2, the first step is using the chain rule to get that the derivative of cos2(x)+sin2(x) is 2cos(x)cos'x + 2sin(x)sin'(x). Via the derivative identities cos'(x) = -sin(x), and sin'(x) = cos(x), we have that 2cos(x)cos'x + 2sin(x)sin'(x)simplifies to 2cos(x)(-sin(x)) + 2sin(x)cos(x) , which is zero. So the identity can be viewed as the integrated version of the trivial identity
cos(x)(-sin(x)) + sin(x)(cos(x)) = 0, which we saw earlier in a completely different context.

## Conservation Of Energy

Yet another way to see that the cos2+sin2 is constant is to realize that it represents the sum of the potential and kinetic energies of the solution x=cos(t) to the equation for simple harmonic motion x''(t)+x(t)=0.

For a particle of mass 1, the kinetic energy is (1/2)x'(t)2, and the potential energy is (1/2)x(t)2 (upto an additive constant). Conservation of energy tells us that (1/2)x'(t)2+(1/2)x(t)2 is a constant, and so x'(t)2+x(t)2 is also a constant. Taking the solution x(t)=cos(t) (or x(t)=sin(t)) we get the identity.

I should mention that none of the above mathematics is dependent on the underlying physics. If x(t) is a solution to the second order equation x''(t)+F(x(t)) = 0, then (1/2)x'(t)2 + V(x(t)), where V is an antiderivative of F, is constant. Simple differentiation extablishes this.

Incidentally, since x(t)=acos(t)+bsin(t) is the general solution to the second order equation, we have that
(acos(t)+bsin(t))2+(-asin(t)+bcos(t))2 is the constant a2+b2

## Hyperbolic Analogues

The next step to understanding the identity is to compare and contrast it to the identity for hyperbolic cosine and hyperbolic sine, namely cosh2(t) - sinh2(t) = 1. I'll defer this to a later essay.

## Feedback

If you have corrections, additions, modifications, etc please let me know mailto:walterv@gbbservices.com

 April 2 2003 Posted April 7 2003 Last Updated