Question

Prove that.
sin⁻¹(1/√2) - 3sin⁻¹(√3/2) = -3π/4

Answer 1

We know that the principal value branch of
${sin}^{ - 1} \: is \: \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]$
let
${sin}^{ - 1} ( \frac{1}{ \sqrt{2} } ) = \alpha \\ \\ then \\ \\ \sin( \alpha ) = \frac{1}{ \sqrt{2} } = sin( \frac{\pi}{4} ) \\ \\ where \: \frac{\pi}{4} \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]$

Now

${sin}^{ - 1} ( \frac{ \sqrt{3} }{ 2} ) = \beta \\ \\ then \\ \\ \sin( \beta ) = \frac{ \sqrt{3} }{ 2} = sin( \frac{\pi}{3} ) \\ \\ where \: \frac{\pi}{3} \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]$
now put both the values in equation

$= \frac{\pi}{4} - 3 \times \frac{\pi}{3} \\ \\ = \frac{\pi}{4} - \pi \\ \\ = \frac{\pi - 4\pi}{4} \\ \\ = \frac{ - 3\pi}{4}$
= RHS

HENCE PROVED