Without using a calculator find the value of t in 02pi that

Without using a calculator, find the value of t in [0,2pi) that corresponds to the following functions.

A) sin t= sqrt3/2; t in QII

B) tan t= -sqrt3; t in QII

Solution

A) sin t= sqrt3/2; t in QII

t = sin^-1(sqrt3/2)

t = pi/3

Now t is in Q-II So, t = pi-pi/3 = 2pi/3

Solution t = 2pi/3

B) tant = -sqrt3t in Q II

t = tan^-1(-sqrt3)

t = pi - pi/3 = 2pi/3 as t is in Q-II