Question

Evaluate sin⁻¹ [$\frac{\pi}{3}$ - sin⁻¹($\frac{-1}{2}$)]

Answer 1

Given that

sin⁻¹ [π/3 - sin⁻¹ (-1/2) ]

                 ∵ If y = sin x , then x ∈ [ -π/2 , π/2 ]

So we have

= sin⁻¹ [π/3 - (-π/6) ]

= sin⁻¹ [π/3 + π/6 ]

= sin⁻¹ [(2π+π)/6]

= sin⁻¹ [3π/6]

= sin⁻¹ [π/2]

           ∵ π/2 = 1.5708

So it is not possible.

           ∵ Maximum value of Sine function is 1.

SO it has no Solution.

Answer 2

Answer:

$sin^{-1}[\frac{\pi}{3}-sin^{-1}(\frac{-1}{2})]=sin^{-1}[\frac{\pi}{2}]$

Step-by-step explanation:

To evaluate

$sin^{-1}[\frac{\pi}{3}-sin^{-1}(\frac{-1}{2})]\\\\as\:we\:know\:that\\\\sin^{-1}(-x)= -sin^{-1}x\\\\so\\\\sin^{-1}(\frac{-1}{2})=-sin^{-1}(\frac{1}{2})\\\\\\sin^{-1}[\frac{\pi}{3}+sin^{-1}(\frac{1}{2})]\\\\\\=sin^{-1}[\frac{\pi}{3}+sin^{-1}sin(\frac{\pi}{6})]\\\\=sin^{-1}[\frac{\pi}{3}+\frac{\pi}{6}]\\\\\\=sin^{-1}[\frac{\pi}{2}]$

it can not be solve furthur.

thus no real value exist.