Question

The inequality $si{n}^{-1}x>co{s}^{-1}x$vholds for

• A. all values of x
• B. x $\in$ $(0,\frac{1}{\sqrt{2}})$
• C. x $\in$ $(\frac{1}{\sqrt{2}},1)$
• D. no value of x

Answer 1

Answer: (C)

x $\in$ $(\frac{1}{\sqrt{2}},1)$

The intersection point of sin(x) and cos(x) is $x=\frac{\pi }{4}$ between $[0,\frac{\pi }{2}]$.
Now cos(x) is a decreasing function in the above interval while sin(x) is an increasing function in the above interval.
Also.
$\cos \left(x\right)>\sin \left(x\right)$ between $[0,\frac{\pi }{4}]$ while
$\sin \left(x\right)>\cos \left(x\right)$ between $[\frac{\pi }{4},\frac{\pi }{2}]$
Hence
${\sin }^{-1}\left(x\right)>{\cos }^{-1}\left(x\right)$ between $[\frac{1}{\sqrt{2}},1]$.