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

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

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Answer 1 · 2015-05-19T19:07:50

You can start by using the trig identity of ${sin}^{2} x = \frac{1 - cos 2 x}{2}$ $sin^2x = (1 - cos2x)/2$

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

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

So we can then sub in $2 x$ $2x$ in place of $x$ $x$ to solve for $cos 2 x$ $cos2x$

$cos 2 x = \infty \sum n = 0 {(- 1)}^{n} \frac{{(2 x)}^{2 n}}{(2 n)!}$ $cos2x = sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)$

thus we get:

${sin}^{2} x = \frac{1 - cos 2 x}{2} = \frac{1}{2} - \frac{1}{2} \infty \sum n = 0 {(- 1)}^{n} \frac{{(2 x)}^{2 n}}{(2 n)!}$ $sin^2x = (1-cos2x)/2 =1/2- 1/2sum_(n=0)^oo (-1)^n ((2x)^(2n))/((2n)!)$

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

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