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Proof of Euler’s Formula Using Derivatives

April 9th, 2009 |

Author: Merrin

Euler’s formula is so useful that we present a proof of it here so we can use it
right away. Euler’s formula was used to justify all of our trigonometric identity calculations. Usually a proof of Euler’s formula is presented later with infinite series, but I have seen a proof from wikipedia which only relies on the derivatives of sine and cosine. Getting Euler’s formula in now allows us to use it to easily prove many new results later.

Example 1. Prove Euler’s formula

$e^{ix}=\cos x +\ i \sin x$

Solution 1. This proof depends on knowing the derivatives of a constant function and the derivatives of exponentials and trigonometric functions. Take the quantity,

$f(x)=(\cos x + i \sin x)e^{-ix}$

We take

$\cfrac{d}{dx} f(x)=e^{-ix} \cfrac{d}{dx} (\cos x+i\sin x)+(\cos x+i\sin x)\cfrac{d}{dx} e^{-ix} \\\\ f'(x)=e^{-ix} (-\sin x+i\cos x)+(\cos x+i\sin x)(-ie^{-ix} ) \\\\ f'(x)=e^{-ix} ((-\sin x+\sin x)+i(\cos x-\cos x)) \\\\f'(x)=0$

Therefore f(x) must be a constant because its derivative is zero. We plug in f(0) to find the constant, and then we find Euler’s formula by multiplying through.

$f(0)=(\cos 0+i\sin 0)e^{0} =1 \\1=(\cos x+i\sin x)e^{-ix} \\e^{ix} =\cos x+i\sin x$

Example 2. Find the derivatives of sine and cosine using Euler’s formula.

Solution 2. We can take the derivative of Euler’s formula which is a complex exponential and relates the trigonometric functions. The imaginary number i does not bother us because we just use the chain rule on it as if it were any other constant.

$\cfrac{d}{dx}e^{ix}= \cfrac{d}{dx}e^{ix}$