# What is the Taylor series for sin 2x?

Sep 13, 2015

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

#### Explanation:

Let us start from here:

$\frac{d}{\mathrm{dt}} \sin t = \cos t$

$\frac{d}{\mathrm{dt}} \cos t = - \sin t$

$\sin 0 = 0$

$\cos 0 = 1$

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 $\left(1 , 0\right)$. The position of the point at time $t$ is $\left(\cos t , \sin t\right)$ and its velocity (which is tangential) is $\left(- \sin t , \cos t\right)$.

The general formula for the Taylor series for $f \left(t\right)$ 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:

${f}^{\left(0\right)} \left(t\right) = \sin t , {f}^{\left(1\right)} \left(t\right) = \cos t ,$

${f}^{\left(2\right)} \left(t\right) = - \sin t , {f}^{\left(3\right)} \left(t\right) = - \cos t , \ldots$

So:

${f}^{\left(2 k\right)} \left(0\right) = {\left(- 1\right)}^{k} \sin \left(0\right) = 0$

${f}^{\left(2 k + 1\right)} \left(0\right) = {\left(- 1\right)}^{k} \cos \left(0\right) = {\left(- 1\right)}^{k}$

So we can write:

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

Now substitute $t = 2 x$ to get:

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