Use the Taylor series given in Key Idea 32 to verify the given identity:
\(sin(-x) = -sin(x)\)
\[ sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \\ sin(-x) = \sum_{n=0}^{\infty} (-1)^n \frac{(-x)^{2n+1}}{(2n+1)!} \\ sin(-x) = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)!} \\ sin(-x) = -\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \\ sin(-x) = -sin(x) \]
That was a little too easy. Prepare to have your minds blown:
Euler’s Identity states:
\[ e^{2i\pi} =1 \\ \text{where}\space i = \sqrt{-1} \]
Give it a second to sink in that 1 is equal to an irrational number to the power of the product of an imaginary number time another irrational number.
Note that \(i^n =-1\) when n is divisible by 2, but not 4, and \(i^n =1\) when n is divisible by 4. Furthermore \(i^n =-i\) when n is one more than divisible by 2 (e.g., 3,7,11,15,…), not 4, and \(i^n = i\) when n is one more than divisible by 4 (e.g., 5,9,13,17,…).
Recall that:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \\ \text{such that}\space e^{ix} = \sum_{n=0}^{\infty} i^n\frac{x^n}{n!} \\ e^{ix} = 1 + ix -\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\frac{x^6}{6!}-i\frac{x^7}{7!} +... \]
Look at the even powered terms, n = 0,2,4,6,…
\[ 1 -\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +... \\ cos(x) = 1 -\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +... \]
And now the odd powered terms, n = 1,3,5,7,…
\[ = ix -i\frac{x^3}{3!}+i\frac{x^5}{5!}-i\frac{x^7}{7!} +... \\ sin(x) = x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} +...\\ isin(x) = ix -i\frac{x^3}{3!}+i\frac{x^5}{5!}-i\frac{x^7}{7!} +... \\ \]
Putting this together:
\[ e^{ix} = cos(x) + isin(x) \\ let\space x = 2\pi \\ e^{2i\pi} = cos(2\pi) + isin(2\pi) \\ cos(2\pi) = 1 \space \&\space sin(2\pi)=0 \\ \boxed{e^{2i\pi} = 1} \]