Question

The Taylor expansion of sin $x$ about $x=\frac{\pi }{6}$ by

• A. $\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})-\frac{1}{4}{(x-\frac{\pi }{6})}^2-\frac{\sqrt{3}}{12}{(x-\frac{\pi }{6})}^3+....$
• B. $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$
• C. $(x-\frac{\Pi }{6})-\frac{{(x-\frac{\Pi }{6})}^3}{3!}+\frac{{(x-\frac{\Pi }{6})}^5}{5!}-\frac{{(x-\frac{\Pi }{6})}^7}{7!}+...$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})-\frac{1}{4}{(x-\frac{\pi }{6})}^2-\frac{\sqrt{3}}{12}{(x-\frac{\pi }{6})}^3+....$

$f\left(x\right)=sinx,f\left(\frac{\pi }{6}\right)=\frac{1}{2}$
$f^{\prime }\left(x\right)=cosx,f^{\prime }\left(\frac{\pi }{6}\right)=+\frac{\sqrt{3}}{2}$
$f^{\prime \prime }\left(x\right)=-sinx,f^{\prime \prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}$
$f^{\prime \prime \prime }\left(x\right)=-cosx,f^{\prime \prime \prime }\left(\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{2}$
$f\left(x\right)=f\left(\frac{\pi }{6}\right)+(x-\frac{\pi }{6})f^{\prime }\left(\frac{\pi }{6}\right)+\frac{1}{2!}{(x-\frac{\pi }{6})}^2{f}^{\prime \prime }\left(\frac{\pi }{6}\right)+....$
$sinx=\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})-\frac{1}{4}{(x-\frac{\pi }{6})}^2-\frac{\sqrt{3}}{12}{(x-\frac{\pi }{6})}^3$