We need to find the first, second, third, etc derivatives and evaluate them at x = 0. Starting with:

f(x) = sin x

f(0) = 0

First dervative:

f '(x) = cos x

f '(0) = cos 0 = 1

Second dervative:

f ''(x) = −sin x

f ''(0) = −sin 0 = 0

Third dervative:

f '''(x) = −cos x

f '''(0) = −cos 0 = −1

Fourth dervative:

f iv(x) = sin x

f iv(0) = sin 0 = 0

We observe that this pattern will continue forever.

Now to substitute the values of these derivatives into the Maclaurin Series:

`f(x)` `~~f(0)+f’(0)x` `+(f’’(0))/(2!)x^2` `+(f’’’(0))/(3!)x^3` `+(f^("iv")(0))/(4!)x^4+...`

We have:

`sin\ x=0+(1)(x)` `+0` `+(-1)/(3!)x^3` `+0` `+1/(5!)x^5` `+0` `+(-1)/(7!)x^7+...`

This gives us:

`sin\ x=x-1/6x^3+1/120x^5` `-1/5040x^7+...`