Question

State True or False:
${\sin ^{ - 1}}\sqrt {2 - x} = {\cos ^{ - 1}}\sqrt {x - 1} $ holds for all real x.
a) True
b) False

Answer 1

Hint: In order to solve the problem convert both of the inverse trigonometric functions into one function by using inverse trigonometric identities. Then simplify the terms in order to convert the functions in some comparable constants.

Complete step-by-step answer:

Given equation is: ${\sin ^{ - 1}}\sqrt {2 - x} = {\cos ^{ - 1}}\sqrt {x - 1} $

First we will convert ${\cos ^{ - 1}}{\text{ into }}{\sin ^{ - 1}}$

As we know that:

$

  {\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2} \\

   \Rightarrow {\cos ^{ - 1}}\theta = \dfrac{\pi }{2} - {\sin ^{ - 1}}\theta \\

 $

So using the above formula to modify the given equation

$

   \Rightarrow {\sin ^{ - 1}}\sqrt {2 - x} = {\cos ^{ - 1}}\sqrt {x - 1} \\

   \Rightarrow {\sin ^{ - 1}}\sqrt {2 - x} = \dfrac{\pi }{2} - {\sin ^{ - 1}}\sqrt {x - 1} \\

   \Rightarrow {\sin ^{ - 1}}\sqrt {2 - x} + {\sin ^{ - 1}}\sqrt {x - 1} = \dfrac{\pi }{2} \\

 $

Now let us use inverse trigonometric identity for sum of inverse sine angle which is given as:

$ \Rightarrow {\sin ^{ - 1}}A + {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^2}} + B\sqrt {1 - {A^2}} } \right)$

Using the above formula we get

$

  \because {\sin ^{ - 1}}\left( {\sqrt {2 - x} } \right) + {\sin ^{ - 1}}\left( {\sqrt {x - 1} } \right) = \dfrac{\pi }{2} \\

   \Rightarrow {\sin ^{ - 1}}\left( {\left( {\sqrt {2 - x} } \right)\sqrt {1 - {{\left( {\sqrt {x - 1} } \right)}^2}} + \left( {\sqrt {x - 1} } \right)\sqrt {1 - {{\left( {\sqrt {2 - x} } \right)}^2}} } \right) = \dfrac{\pi }{2} \\

 $

Now let us bring the ${\sin ^{ - 1}}$ from LHS to RHS in the form of $\sin $ to find the constant value of LHS

$ \Rightarrow \left( {\left( {\sqrt {2 - x} } \right)\sqrt {1 - {{\left( {\sqrt {x - 1} } \right)}^2}} + \left( {\sqrt {x - 1} } \right)\sqrt {1 - {{\left( {\sqrt {2 - x} } \right)}^2}} } \right) = \sin \left( {\dfrac{\pi }{2}} \right)$

Let us simplify further to find the constant value of LHS by putting the value of RHS.

\[

   \Rightarrow \left( {\left( {\sqrt {2 - x} } \right)\sqrt {1 - \left( {x - 1} \right)} + \left( {\sqrt {x - 1} } \right)\sqrt {1 - \left( {2 - x} \right)} } \right) = 1{\text{ }}\left[ {\because \sin \left( {\dfrac{\pi }{2}} \right) = 1} \right] \\

   \Rightarrow \left( {\left( {\sqrt {2 - x} } \right)\left( {\sqrt {2 - x} } \right) + \left( {\sqrt {x - 1} } \right)\left( {\sqrt {x - 1} } \right)} \right) = 1 \\

   \Rightarrow \left( {\left( {2 - x} \right) + \left( {x - 1} \right)} \right) = 1 \\

   \Rightarrow \left( {2 - x + x - 1} \right) = 1 \\

   \Rightarrow \left( {2 - x + x - 1} \right) = 1 \\

   \Rightarrow 1 = 1 \\

 \]

As by solving the LHS and the RHS we have got the final value of LHS equal to RHS. So the equation always holds true.

Hence, the equation ${\sin ^{ - 1}}\sqrt {2 - x} = {\cos ^{ - 1}}\sqrt {x - 1} $ holds true for all the real value of x.

So, option A is the correct option.

Note: In order to solve such types of problems students must remember the identities for inverse trigonometry. For such problems where we need to prove the relation exists for all the real values the best way to proceed is to prove the LHS and RHS as constant and equal. The values of trigonometric terms for some common angles must be remembered.