Question

If $x={\sin }^{-1}\left(\sin 10\right)$ and $y={\cos }^{-1}\left(\cos 10\right)$, then $y-x$ is equal to:

• A. $\pi$
• B. $7\pi$
• C. $0$
• D. $10$

Answer 1

The correct option is A $\pi$

$y={\cos }^{-1}\left(\cos 10\right)$ and $x={\sin }^{-1}\left(\sin 10\right)$

answer is not zero , it means here $10$ is in radian not in degree.

so, $10=\frac{10\pi }{\pi }=\frac{10\pi }{\left(22/7\right)}=\frac{70\pi }{22}=\frac{35\pi }{11}=\frac{(33\pi +2\pi )}{11}=3\pi +\frac{2\pi }{11}$

we know, $y={\cos }^{-1}\left(\cos A\right)=A$ if $0<A<\pi$

see graph of ${\cos }^{-1}\left(\cos x\right)$

here it is clear that, graph between $3\pi$ to $4\pi$

so, equation of line between $3\pi$ to $4\pi$ :

$Y-\pi =\frac{(0-\pi )}{(4\pi -3\pi )}(X-3\pi )$

$Y-\pi +(X-3\pi )=0$

$X+Y-4\pi =0$

so, $Y=4\pi -X$

so, $y={\cos }^{-1}\left(\cos 10\right)=4\pi -10$

again, for $x={\sin }^{-1}\left(\sin 10\right)$ see graph, between $3\pi$ to $\frac{7\pi }{2}$ [ because $3\pi +\frac{2\pi }{11}$ lies between them ]

here, equation of line lies between $3\pi$ to $\frac{7\pi }{2}$

points are $(\frac{5\pi }{2},\frac{\pi }{2})$ and $(\frac{7\pi }{2},\frac{-\pi }{2})$ is

$Y+\frac{\pi }{2}=\frac{(\frac{-\pi }{2}-\frac{\pi }{2})}{(\frac{7\pi }{2}-\frac{5\pi }{2})}(X-\frac{7\pi }{2})$

$Y+\frac{\pi }{2}+(X-\frac{7\pi }{2})=0$

$Y+X-3\pi =0$

$Y=3\pi -X$

hence, $x={\sin }^{-1}\left(\sin 10\right)=3\pi -10$

now, $y-x=4\pi -10-(3\pi -10)=\pi$

hence, answer should be $\pi$