# What would be the expansion of sin x in powers of x?

Sep 25, 2016

sum_"n=0->oo" (-1)^nx^(2n+1)/((2n+1)!)

#### Explanation:

Consider the derivatives of $f \left(x\right) = \sin x$:

$f ' \left(x\right) = \cos x , f ' ' \left(x\right) - \sin x , f ' ' ' \left(x\right) = - \cos x , f ' ' ' ' \left(x\right) = \sin x \ldots .$

From the Taylor/Mclaurin series expansion we have:
f(x) = f(0)+f'(0)x/(1!)+f''(0)x^2/(2!)+f'''(0)x^3/(3!)+ ........

In this example we have $f \left(x\right) = \sin x$

Note that since $\sin 0 = 0$ all even powers of $x$ will equal 0 in the series expansion.

Thus: f(x) = sinx = cos(0)x/(1!)-cos(0)x^3/(3!)+ cos0x^5/(5!)-.......

Now, since $\cos 0 = 1$ the series reduces to:

sinx = x/(1!)-x^3/(3!)+x^5/(5!)- .......

The series is infinite in odd powers of x with alternating sign and thus can be written as:

sum_"n=0->oo" (-1)^nx^(2n+1)/((2n+1)!)