How do you simplify #Cos(arccos(2x) + arcsin(x))#?

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Answer 1 · 2016-07-24T03:58:07

cos 3x

arccos 2x --> 2x
arcsin x --> sin x
cos (arccos 2x + arcsin x) = cos (2x + x) = cos 3x

Answer 2 · 2016-07-24T04:48:51

#+-2 x sqrt (1 - x^2 ) +- x sqrt(1 - 4 x^2 ), -1/2<=x<=1/2#

arc cos 2x is the angle whose cosine is 2x.

Let #a = arc cos ( 2 x )#.

Then, #cos a = 2 x in [-1, 1 ] and sin a = +- sqrt (1 - 2 x^2 )#

Note that #x in [-1/2. 1/2]#.

Prefix negative sign for principal value #a > pi/4# (when x > 0)..

Let #b = arc sin x#.

Then, #sin b = x and cos b = +- sqrt (1 - x^2 )#.

Prefix negative sign for principal value #b < 0# (when x < 0).

Now, the given expression is

#cos ( a + b )#

#= cos a cos b - sin a sin b#

#=+-2 x sqrt( 1 - x^2) +- x sqrt ( 1 - 4 x^2 )#