Question 365266
Let A = 2sin^1(4/5).  Then A/2 = sin^1(4/5), or {{{sin(A/2) = 4/5}}}.  We have to find cosA.  But from the identity {{{cos A = 1-2sin^2(A/2)}}} we get {{{cosA = 1-2*(4/5)^2 = 1-2*(16/25) = 1-32/25 = -7/25}}}