Answer
$\sin (1)=1-\dfrac{1}{3!}+\dfrac{1}{5!}-\dfrac{1}{7!}+....$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as:
$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+....$
Plug $x=1$ in the above equation to obtain:
$\sin (1)=1-\dfrac{1^3}{3!}+\dfrac{1^5}{5!}-\dfrac{1^7}{7!}+...,$
Therefore, the first $4$ non-zero terms of an infinite series is equal to:
$\sin (1)=1-\dfrac{1}{3!}+\dfrac{1}{5!}-\dfrac{1}{7!}+....$
