# Question 13a86

Mar 13, 2017

$- \frac{1}{6}$

#### Explanation:

The typical Maclaurin series (a Taylor series about $x = 0$) for $\sin \left(x\right)$ 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 $\sin \left({x}^{2}\right)$ 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 $- 1 / 6$.