cos(sin^-1(2/3)+cos^-1(-5/13))
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Use Pythagorean theorem,
2^2 + 5 = 3^2
5^2 + 12^2 = 13^2
So, cos(sin^-1(2/3)+cos^-1(-5/13)) =
cos(sin^-1(2/3)) (-5/13) - (2/3) sin(cos^-1(-5/13)) =
(sqrt(5)/3)(-5/13) - (2/3)(12/13)
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Attn: I assumed major branch for inverse trig. Otherwise, the answer can be +/-(sqrt(5)/3)(-5/13) +/- (2/3)(12/13)
2^2 + 5 = 3^2
5^2 + 12^2 = 13^2
So, cos(sin^-1(2/3)+cos^-1(-5/13)) =
cos(sin^-1(2/3)) (-5/13) - (2/3) sin(cos^-1(-5/13)) =
(sqrt(5)/3)(-5/13) - (2/3)(12/13)
---------
Attn: I assumed major branch for inverse trig. Otherwise, the answer can be +/-(sqrt(5)/3)(-5/13) +/- (2/3)(12/13)