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

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Answer 1 · 2016-09-25T09:04:45

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

Consider the derivatives of $f(x)=sinx$ :

$f'(x)= cosx, f''(x)-sinx, f'''(x) = -cosx, f''''(x)=sinx....$

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 $f(x) =sinx$

Note that since $sin0 = 0$ 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 $cos0=1$ 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:

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