Question

Solve for x:
$(sin^{-1}x)^{3}+(cos^{-1}x)^{3}=\frac{\pi^{3}}{32}$

Answer 1

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

Identities Used :-

$\boxed{ \sf{ \: {sin}^{ - 1}x + {cos}^{ - 1}x = \frac{\pi}{2}}}$

$\boxed{ \sf{ \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

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

$\rm :\longmapsto\: {( {sin}^{ - 1}x)}^{3} + {({cos}^{ - 1}x)}^{3} = \dfrac{\pi^{3} }{32}$

can be rewritten as

$\rm :\longmapsto\: {\bigg( {sin}^{ - 1}x + {cos}^{ - 1}x\bigg) }^{3} - 3 {sin}^{ - 1}x{cos}^{ - 1}x( {sin}^{ - 1}x + {cos}^{ - 1}x) = \dfrac{\pi^{3} }{32}$

$\rm :\longmapsto\: {\bigg(\dfrac{\pi}{2}\bigg) }^{3} - 3 {sin}^{ - 1}x{cos}^{ - 1}x\bigg(\dfrac{\pi}{2}\bigg) = \dfrac{\pi^{3} }{32}$

$\rm :\longmapsto\:\dfrac{\pi^{3} }{8} - \dfrac{3\pi}{2} {sin}^{ - 1}x{cos}^{ - 1}x = \dfrac{\pi^{3} }{32}$

$\rm :\longmapsto\:\dfrac{\pi^{3} }{8} -\dfrac{\pi^{3} }{32} = \dfrac{3\pi}{2} {sin}^{ - 1}x{cos}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{4\pi^{3} - {\pi}^{3} }{32} = \dfrac{3\pi}{2} {sin}^{ - 1}x{cos}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{3\pi^{3} }{32} = \dfrac{3\pi}{2} {sin}^{ - 1}x{cos}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{\pi^{2} }{16} = {sin}^{ - 1}x{cos}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{\pi^{2} }{16} = {sin}^{ - 1}x \bigg(\dfrac{\pi}{2} - {sin}^{ - 1}x \bigg)$

$\rm :\longmapsto\:\dfrac{\pi^{2} }{16} = \dfrac{\pi}{2} {sin}^{ - 1}x - {( {sin}^{ - 1}x)}^{2}$

$\rm :\longmapsto\: {( {sin}^{ - 1}x)}^{2} - \dfrac{\pi}{2} {sin}^{ - 1}x + \dfrac{\pi^{2} }{16} = 0$

$\rm :\longmapsto\:16 {( {sin}^{ - 1}x)}^{2} - 8\pi \: {sin}^{ - 1}x + {\pi}^{2} = 0$

$\rm :\longmapsto\: {(4 {sin}^{ - 1}x - \pi)}^{2} = 0$

$\rm :\longmapsto\:4 {sin}^{ - 1}x - \pi= 0$

$\rm :\longmapsto\:4 {sin}^{ - 1}x - = \pi$

$\rm :\longmapsto\: {sin}^{ - 1}x = \dfrac{\pi}{4}$

$\rm :\longmapsto\:x = sin\dfrac{\pi}{4}$

$\bf\implies \:x = \dfrac{1}{ \sqrt{2} }$

Additional Information :-

$\boxed{ \sf{ \: {sec}^{ - 1}x + {tan}^{ - 1}x = \frac{\pi}{2}}}$

$\boxed{ \sf{ \: {cosec}^{ - 1}x + {cot}^{ - 1}x = \frac{\pi}{2}}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x - {cos}^{ - 1}y = {cos}^{ - 1}(xy + \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x + {cos}^{ - 1}y = {cos}^{ - 1}(xy - \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} }}}$