Question 662ba

Apr 9, 2017

${\sin}^{2} x = {x}^{2} - \frac{1}{3} {x}^{4} + \ldots$

Explanation:

To find the MacLaurin series of ${\sin}^{2} x$ we can use the trigonometric identity:

${\sin}^{2} x = \frac{1 - \cos 2 x}{2}$

As the MacLaurin series of $\cos t$ is:

cost = sum_(n=0)^oo (-1)^n t^(2n)/((2n)!)

substituting $t = 2 x$ we have:

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

and:

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

Extracting the term of index $n = 0$ from the sum we get:

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

and finally:

sin^2x =sum_(n=1)^oo (-1)^(n+1) 2^(2n-1) x^(2n)/((2n)!) = x^2- 1/3x^4+...#