Question #662ba

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Answer 1 · 2017-04-09T22:36:45

$sin^2x = x^2- 1/3x^4+...$

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

$sin^2x = (1-cos2x)/2$

As the MacLaurin series of $cos t$ is:

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

substituting $t = 2x$ 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:

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