Question

Approximate value of ${\cos }^{-1}(-0.49)$ is _______

• A. $\frac{2\pi }{3}+\frac{1}{50\left(\sqrt{3}\right)}$
• B. $\frac{2\pi }{3}-\frac{1}{50\left(\sqrt{3}\right)}$
• C. $\frac{\pi }{3}-\frac{1}{50\left(\sqrt{3}\right)}$
• D. $\frac{\pi }{3}+\frac{1}{50\left(\sqrt{3}\right)}$

Answer 1

The correct option is B $\frac{2\pi }{3}-\frac{1}{50\left(\sqrt{3}\right)}$

$f\left(x\right)={\cos }^{-1}(-0.49)$
$f\left(x\right)=\pi -\left[{\cos }^{-1}(-0.49)\right]....\left(1\right)$
$g\left(x\right)={\cos }^{-1}(-0.49)$
Here $x+\Delta x=0.49$
$\therefore x+\Delta x=0.5+(-0.01)$
where $x=0.5=\frac{1}{2};\Delta x=-0.01$
Now $g\left(x\right)={\cos }^{-1}x$
$\therefore g\left(\frac{1}{2}\right)={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$
And $=-\frac{1}{\sqrt{1-x^2}}$
$\therefore g^{\prime }\left(\frac{1}{2}\right)=-\frac{1}{\sqrt{1-\frac{1}{4}}}=-\frac{2}{\sqrt{3}}$
$\therefore$ Approximate value of $g\left(x\right)$
$\cong g\left(x\right)+g^{\prime }\left(x\right)\Delta x\cong \frac{\pi }{3}+(-\frac{2}{\sqrt{3}})(-\frac{1}{100})\cong \frac{\pi }{3}+\frac{1}{50\sqrt{3}}$
(1) becomes as follows
$\therefore$ Approximate value of $f\left(x\right)$
$\cong \pi -(\frac{\pi }{3}+\frac{1}{50\sqrt{3}})\cong \frac{2\pi }{3}-\frac{1}{50\sqrt{3}}$