Question

What is the Taylor series for `sin 2x`?

Answer 1

`sin 2x = sum_(k=0)^oo (-1)^k/((2k+1)!) (2x)^(2k+1)`

Explanation:

Let us start from here:

`d/(dt) sin t = cos t`

`d/(dt) cos t = -sin t`

`sin 0 = 0`

`cos 0 = 1`

Incidentally, a picture of this would be a point moving anticlockwise around the unit circle at a speed of `1` radian per second, starting from `(1, 0)`. The position of the point at time `t` is `(cos t, sin t)` and its velocity (which is tangential) is `(-sin t, cos t)`.

The general formula for the Taylor series for `f(t)` at `0` is:

`f(t) = sum_(n=0)^oo f^((n))(0)/(n!) t^n`

In the case of `sin t`, we find that only the terms for odd values of `n` are non-zero, and the signs on them are alternating:

`f^((0))(t) = sin t, f^((1))(t) = cos t,`

`f^((2))(t) = -sin t, f^((3))(t) = -cos t,...`

So:

`f^((2k))(0) = (-1)^k sin(0) = 0`

`f^((2k+1))(0) = (-1)^k cos(0) = (-1)^k`

So we can write:

`sin t = sum_(k=0)^oo (-1)^k/((2k+1)!) t^(2k+1)`

Now substitute `t = 2x` to get:

`sin 2x = sum_(k=0)^oo (-1)^k/((2k+1)!) (2x)^(2k+1)`