Question

Evalute : Sin [π/3 - Sin-'(-1/2)]​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:sin\bigg(\dfrac{\pi}{3} - {sin}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) \bigg)$

$\: \: \: \: \: \: \: {\bigg \{ \because \: \sf \: {sin}^{ - 1}( - x) = - {sin}^{ - 1}x \bigg \}}$

$\rm \rm \: = \: \: \:\:sin\bigg(\dfrac{\pi}{3} + {sin}^{ - 1}\bigg(\dfrac{1}{2} \bigg) \bigg)$

$\rm \rm \: = \: \: \:\:sin\bigg(\dfrac{\pi}{3} + {sin}^{ - 1}\bigg(sin\dfrac{\pi}{6} \bigg) \bigg)$

${\bigg \{ \because \: \sf \: {sin}^{ - 1}(sinx) = x \: \: if \: x \in \: \bigg[ - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg] \bigg \}}$

$\rm \: = \: \: \:sin\bigg(\dfrac{\pi}{3} + \dfrac{\pi}{6} \bigg)$

$\rm \: = \: \: \:sin\bigg(\dfrac{2\pi + \pi}{6} \bigg)$

$\rm \: = \: \: \:sin\bigg(\dfrac{3\pi}{6} \bigg)$

$\rm \: = \: \: \:sin\bigg(\dfrac{\pi}{2} \bigg)$

$\rm \: = \: \: \:1$

Hence,

$\bf :\longmapsto\:sin\bigg(\dfrac{\pi}{3} - {sin}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) \bigg) = 1$

Additional information :-

$\rm :\longmapsto\: {cos}^{ - 1}(cosx) = x \: \: if \: x \: \in \: \bigg[0, \: \pi\bigg]$

$\rm :\longmapsto\: {tan}^{ - 1}(tanx) = x \: \: if \: x \: \in \: \bigg( - \dfrac{\pi}{2} , \dfrac{\pi}{2}\bigg)$

$\rm :\longmapsto\: {sin}^{-1}x+{sin}^{ - 1}y ={sin}^{-1}\bigg(x \sqrt{1-{y}^{2}} + y\sqrt{1-{x}^{2} }\bigg)$

$\rm :\longmapsto\: {sin}^{-1}x - {sin}^{ - 1}y ={sin}^{-1}\bigg(x \sqrt{1-{y}^{2}} - y\sqrt{1-{x}^{2} }\bigg)$

$\rm :\longmapsto\: {cos}^{-1}x - {cos}^{ - 1}y ={cos}^{-1}\bigg(xy + \sqrt{1-{y}^{2}} \sqrt{1-{x}^{2} }\bigg)$

$\rm :\longmapsto\: {cos}^{-1}x + {cos}^{ - 1}y ={cos}^{-1}\bigg(xy - \sqrt{1-{y}^{2}} \sqrt{1-{x}^{2} }\bigg)$