Question

If the given inverse trigonometric identity holds true that is ${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \dfrac{\pi }{6}$, then solve for x?

Answer 1

Hint – In this question use the inverse trigonometric identity of ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$, use this to put the value of ${\cos ^{ - 1}}x$ in terms of ${\sin ^{ - 1}}x$ in the given equation ${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \dfrac{\pi }{6}$.

Complete step-by-step solution -
Given trigonometric equation is
${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \dfrac{\pi }{6}$........................ (1)
Now as we know that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$
Therefore, ${\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x$
So substitute this value in equation (1) we have,
$\Rightarrow {\sin ^{ - 1}}x - \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right) = \dfrac{\pi }{6}$
Now simplify this equation we have,
$\Rightarrow {\sin ^{ - 1}}x - \dfrac{\pi }{2} + {\sin ^{ - 1}}x = \dfrac{\pi }{6}$
$\Rightarrow 2{\sin ^{ - 1}}x = \dfrac{\pi }{6} + \dfrac{\pi }{2} = \dfrac{{4\pi }}{6} = \dfrac{{2\pi }}{3}$
$\Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{3}$
$\Rightarrow x = \sin \dfrac{\pi }{3} = \sin {60^0} = \dfrac{{\sqrt 3 }}{2}$
So this is the required value of x.
So this is the required answer.

Note – Inverse trigonometric functions are the inverse of the trigonometric functions (with suitably restricted domains). Specifically they are the inverse of sine, cosine, tangent, cotangent, secant and cosecant functions and are used to obtain an angle from any of the angle’s trigonometric ratios. It is advised to remember the basic inverse trigonometric identities as it helps solving problems of this kind.