What is the taylor series in sigma notation of: sin^2(x)?

May 19, 2015

You can start by using the trig identity of ${\sin}^{2} x = \frac{1 - \cos 2 x}{2}$

we know the Maclurin series of $\cos x$ is sum_(n=0)^oo (-1)^n (x^(2n))/((2n)!)

Keep in mind here that 0!=1, so the case of n=0 is still valid.

So we can then sub in $2 x$ in place of $x$ to solve for $\cos 2 x$

cos2x = sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)

thus we get:

sin^2x = (1-cos2x)/2 =1/2- 1/2sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)