How do you simplify #cos[Arcsin(-2/5)-Arctan3]#?

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Answer 1 · 2016-06-29T17:38:40

#(sqrt 21-6)/(5sqrt 10)#, against principal values of arc sin and arc tan.
#(+-sqrt 2+-6)/(5sqrt 10)#, against general values..

Let a = arc sin (-2/5)3. Then, #sin a =(-2/5)<0#. So,

#cos a =sqrt 21/5# for principal a and #+-sqrt 21/5# for general a.

Let b = arc tan 3. Then, #tan b =3>0#. So,

#sin b =3/sqrt10#, for principal b and #+-3/sqrt10#, for general b..

#cos b =1/sqrt10#, for principal b and #+-1/sqrt10#, for general b..

Note that sin b and cos b are of the same sign , when tan b >0..

Now, the given expression is

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

#(sqrt 21-6)/(5sqrt 10)#,

against principal values of arc sin and arc tan.

#(+-sqrt 2+-6)/(5sqrt 10)#, against general values..