Find the value of sin pi by 3 minus sin inverse minus 1 by 2

'sin(pi/3- sin^-1(-1/2))' || find the value of 'sin(pi/3- sin^-1(-1/2))' ||

Evaluating Inverse Trigonometric Functions

evaluate sin pi by 3 minus sin inverse minus one by two | find sin pi/3-sin^-1(-1/2) |

`Sin[pi/3-sin^(-1)(-1/2)]`

Evaluate : sin [π/3 - sin⁻¹(-1/2)]

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