use the taylor series shown in the table to find the first four nonzero terms of the taylor series for the…

Explanation:

Step1: Recall Taylor - series of sin(x)

The Taylor - series of $\sin t$ centered at $0$ is given by $\sin t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}t^{2n + 1}=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+\cdots$.

Step2: Substitute $t = 8x^{9}$

Let $t = 8x^{9}$. Then $\sin(8x^{9})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}(8x^{9})^{2n + 1}$.

Step3: Find the first - four non - zero terms

When $n = 0$: $\frac{(-1)^{0}}{(2\times0 + 1)!}(8x^{9})^{2\times0+1}=\frac{1}{1!}\times8x^{9}=8x^{9}$. When $n = 1$: $\frac{(-1)^{1}}{(2\times1+1)!}(8x^{9})^{2\times1 + 1}=\frac{-1}{3!}\times8^{3}x^{27}=-\frac{512}{6}x^{27}=-\frac{256}{3}x^{27}$. When $n = 2$: $\frac{(-1)^{2}}{(2\times2+1)!}(8x^{9})^{2\times2 + 1}=\frac{1}{5!}\times8^{5}x^{45}=\frac{32768}{120}x^{45}=\frac{4096}{15}x^{45}$. When $n = 3$: $\frac{(-1)^{3}}{(2\times3+1)!}(8x^{9})^{2\times3 + 1}=\frac{-1}{7!}\times8^{7}x^{63}=-\frac{2097152}{5040}x^{63}=-\frac{131072}{315}x^{63}$.

Answer:

$8x^{9}-\frac{256}{3}x^{27}+\frac{4096}{15}x^{45}-\frac{131072}{315}x^{63}$