# Using Euler's Formulas to Obtain Trigonometric Identities

### Written by tutor Jeffery D.

In this lesson we will explore the derivation of several trigonometric identities, namely

cos (*x* + *y*) = cos *x* cos *y* - sin *x* sin *y*

and

sin (*x* + *y*) = sin *x* cos *y* + sin *x* cos *y*

also

cos 2*x* = cos^{2} *x* - sin^{2} *x*

along with

sin 2*x* = 2 sin *x* cos *x*

and lastly, DeMoivre's Formula,

(cos *x* + *i* sin *x*)^{n} = cos *nx* + *i* sin *nx*

using Euler's Formula. To get a good understanding of what is going on, you will need a previous knowledge of series expansions and complex numbers! You may want to refresh your knowledge of those subjects first.

## Power series expansions

We start by examining the power series expansion of the functions *e ^{x}*, sin

*x*, and cos

*x*. The power series of a function is commonly derived from the Taylor series of a function for the case where

*a*= 0. This case, where

*a*= 0 is called the MacLaurin Series. The Taylor series:

The case where *a* = 0 is the MacLaurin Series:

These series are used to approximate the values of functions around a certain point. That is all I'll say about that. The power series of *e ^{x}*; cos

*x*and sin

*x*comes from their MacLaurin Series representation:

for all x.

for all x.

for all x.

## Complex numbers and *e*^{x}

^{x}

A complex number is a number of the form *a* + *bi* where *i* is a root of the equation *x ^{2}* + 1 = 0 and

*a*and

*b*are real numbers. Making note of this we can use

*i*in our power series of

*ex*since it is true for all

*x*.

for all x.

Keeping in mind that *x ^{2}* + 1 = 0 →

*x*=

*i*and so √-1 =

*i*→

*i*= -1,

^{2}*i*= -

^{3}*i*,

*etc.*So, applying the powers selectively we obtain

We can now rearrange the terms and factor out *i* so that that we have

Now, if we look back at our series representations of cos *x* and sin *x* we have

*e ^{ix}* = cos

*x*+

*i*sin

*x*

This conclusion is huge. It is known as **Euler's formula**. From here we can deduce some of the trigonometric identities as well as come up with formulas for general cases. Let us examine a simple derivation first:

*e ^{ix}e^{iy}* = (cos

*x*+

*i*sin

*x*)(cos

*y*+

*i*sin

*y*)

But, recall that *e ^{x}e^{y}* =

*e*. Therefore, we have

^{x+y}*e ^{ix+iy}* = cos (

*x*+

*y*) +

*i*sin (

*x*+

*y*) = (cos

*x*+

*i*sin

*x*)(cos

*y*+

*i*sin

*y*)

= cos

*x*cos

*y*+

*i*sin

*x*cos

*y*+

*i*sin

*y*cos

*x*+

*i*

^{2}sin

*x*sin

*y*

And now we can rearrange this so that the complex part and the real part is separate.

And so we have

*e ^{ix+iy}* = cos (

*x*+

*y*) +

*i*sin (

*x*+

*y*) = (cos

*x*+

*i*sin

*x*)(cos

*y*+

*i*sin

*y*)

= cos

*x*cos

*y*+

*i*sin

*x*cos

*y*+

*i*sin

*y*cos

*x*+

*i*sin

^{2}*x*sin

*y*

= (cos

*x*cos

*y*- sin

*x*sin

*y*) + (

*i*sin

*x*cos

*y*+

*i*sin

*y*cos

*x*)

Taking the real parts and equating them we obtain the familiar trigonometric sum formula:

cos (*x* + *y*) = cos *x* cos *y* - sin *x* sin *y*

and also

sin (*x* + *y*) = sin *x* cos *y* + sin *y* cos *x*

Now, suppose we have something like this:

*e ^{ix}e^{ix}* =

*e*=

^{ix+ix}*e*= cos (

^{i2x}*x*+

*x*) +

*i*sin(

*x*+

*x*)

= (cos

*x*+

*i*sin

*x*)(cos

*x*+

*i*sin

*x*)

= cos

*x*cos

*x*+

*i*sin

*x*cos

*x*+

*i*sin

*x*cos

*x*+

*i*sin

^{2}*x*sin

*x*

If we equate the real parts of the equation

cos 2*x* = cos^{2}*x* - sin^{2}*x*

And also we have

sin 2*x* = 2 sin *x* cos *x*

In general, we may obtain a formula for any multiple of an angle in this way. This leads us to another famous formula known as DeMoivre's Formula. DeMoivre's Formula can be derived by taking the *n*th case of Euler's Formula.

*e ^{inx}* = cos

*nx*+

*i*sin

*nx*

We are interested in showing that

(cos *x* + *i* sin *x*)^{n} = cos *nx* + *i* sin *nx*

which is *exactly* DeMoivre's Formula. It is obvious that this is true for any *n*. We can show that it is true for all *n* by using induction.

(cos *x* + *i* sin *x*)^{n+1} = (cos *x* + *i* sin *x*)^{n}(cos *x* + *i* sin *x*)

From this we apply what we know about the *n*th case from above.

= (cos *nx* + *i* sin *nx*)(cos *x* + *i* sin *x*)

We can now multiply.

= cos *nx* cos *x* + *i* sin *x* cos *nx* + *i* sin *nx* cos *x* + *i*^{2} sin *nx* sin *x*

And from our work above we have already shown that these can be simplified into our sum formulas as such:

cos (*nx* + *x*) + *i* sin (*nx* + *x*) = cos (*n* + 1)x + *i* sin (*n* + 1)*x*

And so, we have shown that

(cos *x* + *i* sin *x*)^{n} = cos *nx* + *i* sin *nx*

is true for all *n*. Thus we have shown that some very common trigonometric identities are related and can be derived from series expansions and complex numbers!