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..

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