How do you simplify #Sin[cos^-1(5/13) - cos^-1(4/5)]#?

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Answer 1 · 2016-05-26T12:01:59

#+-33/65, +-63/65#.

Let #a = cos^(-1)(5/13#). Then #cos a = 5/13 > 0.# a is in the 1st

quadrant or in the 4th. So, sin a =+-12/13.

Let #b = cos^(-1)(4/5)#. Then #cos b = 4/5 > 0#. b is in the 1st

quadrant or in the 4th. So, #sin b =+-3/5#.

Now, the given expression is

#sin ( a - b ) = sin a cos b-cos a sin b#

#=(+-12/13)(4/5)-(5/13)(+-3/5)#

#=+-48/65# - or +#3/13#

#=+-63/65, +-33/65#.