trig
posted by Ansley .
sin [sin 1 (7/25)  cos 1 (8/17)]

let A = sin^1 (7/25)
then cosA = 24/25 > sinA = 7/25
let B = cos^1 (8/17)
cosB = 8/17 , let's assume B is in II
sin B = 15/17
so sin [sin 1 (7/25)  cos 1 (8/17)]
= sin (A  B)
= sinAcosB  cosAsinB
= (7/25)(8/17)  (24/25)(15/17)
= (56  360)/425
= 416/425
check with calculator:
A = 16.26°
B = 118.07°
AB = 101.81°
sin(101.81°) = .9788
416/425 = .9788 , how about that?
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