Question

Prove that.
sin⁻¹(-1/2) + cos⁻¹(-√3/2) = cos⁻¹(-1/2)

Answer 1

sin⁻¹(-1/2) + cos⁻¹(-√3/2) = cos⁻¹(-1/2)

LHS

let
${sin}^{ - 1} ( \frac{ - 1}{2} ) = \alpha \\ \\ \sin( \alpha ) = \frac{ - 1}{2} = sin( \frac{ - \pi}{6} ) \: \: \: \\ \\ where \: \frac{ - \pi}{6} belongs \: to \: ( \frac{ - \pi}{2} \:, \frac{\pi}{2} )$
So, the principal value of sin⁻¹(-1/2) is -π/6

let cos⁻¹(-√3/2)

${cos}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) = \alpha \\ \\ \cos( \alpha ) = \frac{ - \sqrt{3} }{2} = cos( \frac{ - \pi}{6} ) \: \: \: \\ \\ where \: \frac{ - \pi}{6} \: does \: not \: belongs \: to \: ( 0, \: \pi ) \\ \\ so \: convert \: this \: into \: its \: principal \: value \: branch \\ \\ = \frac{5\pi}{6}$

$\frac{ - \pi}{6} + \frac{5\pi}{6} \\ \\ = \frac{4\pi}{6} \\ \\ = \frac{2\pi}{3}$
now solve RHS
we know that the range of principal value of Cos -1 is [0,π]

$let \\ \\ {cos}^{ - 1} ( \frac{ - 1}{2} ) = \beta \\ \\ cos \beta = \frac{ - 1}{2} = - cos \frac{\pi}{3} \\ \\ = \cos(\pi - \frac{\pi}{3} ) \\ \\ = \cos( \frac{2\pi}{3} ) \\ \\ \beta = \frac{2\pi}{3} \: belongs \: to \: (0,\pi)$

LHS = RHS

hence proved.