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

##### 1 Answer

#### 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

The general formula for the Taylor series for

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

In the case of

#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

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