Question

Find the approximate value of ${\sin }^{-1}\left(0.51\right)$ given that $\sqrt{3}=1.7321$

• A. $\frac{\pi }{6}+0.011541$
• B. $\frac{\pi }{3}+0.011541$
• C. $\frac{\pi }{6}+0.011547$
• D. $\frac{\pi }{3}+0.011547$

Answer 1

The correct option is A $\frac{\pi }{6}+0.011541$

${\sin }^{-1}(0.5+0.01)={\sin }^1\left(0.51\right)$

$\therefore$ Approx value$=f\left(x\right)+h{f}^{\prime }\left(x\right)$

Here $x=0.5$

$h=0.01$

$f\left(u\right)={\sin }^{1}x$

$f‘\left(u\right)=\frac{1}{\sqrt{1-x^2}}$

$\therefore f\left(0.5\right)={\sin }^{-1}\left(0.5\right)=\frac{\pi }{6}$

$\therefore f^{\prime }\left(0.5\right)=\frac{1}{\sqrt{1-{\left(0.5\right)}^2}}$

$=\frac{1}{\sqrt{0.75}}$

$=\frac{2}{\sqrt{3}}$

$\therefore$ Approx value $=\frac{\pi }{6}+\frac{2}{\sqrt{3}}\times \left(\frac{1}{100}\right)$

$\frac{\pi }{6}+\frac{2}{100\sqrt{3}}$

$=\frac{\pi }{6}+0.011541$