The given function is $$f(x)= \sin (x) \cos (x)$$ Now first we rewrite the function, knowing that $ \sin x \cos x=\frac{1}{2}(2\sin(x)\cos(x))$ we can now rewrite our given into: $$\frac{1}{2}\sin(2x)$$ Then if we see the maclaurin series of $\sin(x)$ is: $$\sin(x)=x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$$ Then manipulating this series into our given results in the series being: $$\frac{1}{2}\left[2x+\frac{(2x)^{3}}{3!}-\frac{(2x)^{5}}{5!}-\frac{(2x)^{7}}{7!} \right]$$ $$f(x)=x-\frac{2}{3}x^3+\frac{2}{15}x^5-\frac{4}{315}x^7+...$$
Am I correct in my calculations?
$$\sin(x)=x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$$ You forgot the factorial at the denominator and the power of $2$: $$\sin(2x)=2x -\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\frac{2^7x^7}{7!}+....$$ $$\frac 12\sin(2x)=x -\frac{2^2x^3}{3!}+\frac{2^4x^5}{5!}-\frac{2^6x^7}{7!}+....$$ $$\frac 12\sin(2x)=x -\frac{2x^3}{3}+\frac{2x^5}{15}-\frac{4x^7}{315}+....$$