Question

The greater of the two angles A = 2tan^-1(2√2 - 1) and B = 3sin^-1(1/3) + sin^-1(3/5) is ....

Answer 1

We have

A = 2tan^-1(2√2 - 1) > 2tan^-1(3) = 2/3

and B = 3tan^-1(1/3) + sin^-1(3/5)

= sin^-1(3(1/3) - 4(1/3)3) + sin^-1(3/5)

= sin^-1(23/27) + sin^-1(3/5) < sin^-1(√3/2) + sin^-1(√3/2)

= 2π/3

Thus, A > B