Expert: Ahmed Salami - 5/8/2009

Question
Can you show me how to solve these?

A) tan^-1 square root 3

B) sin^-1( -square root 3/2)

C) sin^-1(tan pi/4)

Answer
Hi Ashlee,
a)tan pi/3 = sqrt3
and so
tan^-1 sqrt3 = pi/3
but tan is also positive in the third quadrant and so pi/3 + pi is also a solution.
Therefore,
tan^-1 sqrt3 = pi/3  or  4pi/3

b)sin pi/3 = (sqrt3)/2
pi/3 is the basic angle here
but sin is negative in the third and fourth quadrant and so the true angles are pi/3 + pi and 2pi - pi/3
i.e
sin^-1 (-sqrt3)/2 = 4pi/3  or  5pi/3

c)tan pi/4 = 1
sin^-1(tan pi/4) = sin^-1 1
sin pi/2 = 1
and so
sin^-1 1 = pi/2
but sin is also positive in the second quadrant and so pi - pi/2 is also a solution.
Therefore,
sin^-1(tan pi/4) = sin^-1 1
                = pi/2 (twice i.e its a double solution)

It might be a little confusing so you can get back to me.

Regards