tanarcsin12 Can you explain how Im supposed to know to use

tan[arcsin(-1/2)] =

Can you explain how I\'m supposed to know to use the 11pi/6 value instead of the 7pi/6 based off of the unit circle?

Solution

tan[arcsin(-1/2)]

arcsin(-1/2) ; Range of arcsinx is [ -pi/2 , pi/2] which means Ist and Ivrt quadrants

So, arcsin(-1/2) = 2pi - pi/6 as sinx is negative in IVrth quadrant

7pi/6 is in IIIrd quadrant which is not in the range of arcsinx

So, arcsin(-1/2) = 2pi - pi/6 = 11pi/6

tan11pi/6) = - 1/sqrt3