Question #13a86

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Answer 1 · 2017-03-13T00:33:37

$-1/6$

The typical Maclaurin series (a Taylor series about $x=0$ ) for $sin(x)$ is:

$sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...$

So the same series for $sin(x^2)$ can be found by replacing $x$ with $x^2$ in the series:

$sin(x^2)=sum_(n=0)^oo(-1)^n/((2n+1)!)(x^2)^(2n+1)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(4n+2)$

Writing out the first terms of this series gives:

$sin(x^2)=x^2-x^6/(3!)+x^10/(5!)-x^14/(7!)+...$

So the coefficient of the $x^6$ term is $-1//3!$ or $-1//6$ .