Question

drawing the graph of inverse trigonometric functions sin inverse X and cos inverse X and solving the inequality cos inverse X less than sin inverse X

Answer 1

The value of x is given by

$\displaystyle \sf{ x \in \: \bigg[ - 1, \frac{1}{ \sqrt{2} } \bigg) }$

Given :

$\sf {sin}^{ - 1} x \: \: and \: \: {cos}^{ - 1} x$

To find :

1. Draw the graph of the inverse Trigonometric function

2. To solve the inequality

$\sf {cos}^{ - 1} x \: > \: {sin}^{ - 1} x$

Solution :

Step 1 of 2 :

Draw the graph of the inverse Trigonometric functions

The graph of the inverse Trigonometric function is referred to the attachment

$\sf Green \: line \: : {sin}^{ - 1} x$

$\sf Red \: line : \: {cos}^{ - 1} x$

Domain of each function is [ - 1 , 1 ]

Step 2 of 2 :

Solve the inequality

$\sf {cos}^{ - 1} x \: > \: {sin}^{ - 1} x$

$\displaystyle \sf\implies 2{cos}^{ - 1} x > {sin}^{ - 1} x +{cos}^{ - 1} x \:$

$\displaystyle \sf\implies 2{cos}^{ - 1} x > \frac{\pi}{2}$

$\displaystyle \sf\implies {cos}^{ - 1} x > \frac{\pi}{4}$

$\displaystyle \sf\implies x < cos \frac{\pi}{4} \: \: \bigg( \because \: {cos}^{ - 1} x \: is \: decreasing \: )$

$\displaystyle \sf\implies x < \frac{1}{ \sqrt{2} }$

$\because \: \: \displaystyle \sf{ Domain \: of \: {cos}^{ - 1} x \: \: is \: \: \bigg[ - 1, 1 \bigg]}$

Hence the required solution set is

$\displaystyle \sf{ x \in \: \bigg[ - 1, \frac{1}{ \sqrt{2} } \bigg) }$

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