How do you simplify the expression #Sin(arctan(x)+arccos(x))#?

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Answer 1 · 2016-07-26T00:53:31

sin 2x

arctan x --> x
arccos x --> x
sin (arctan x + arccos x) = sin (x + x) = sin 2x

Answer 2 · 2016-07-26T04:08:42

.#=(1/sqrt(1+x^2))(sqrt(1-x^2)+-x^2)#, x in [-1, 1]#.

The negative sign is used, when #x in [-1, 0]#.

Let a = arc tan (x). The principal value of #a in [-pi/2, pi/2]#

Then x = tan a. sin a = #+-x/sqrt(1+x^2)# and cos a = #1/sqrt(1+x^2)#.

Let b = arc cos x. The principal value of #b in [0, pi]#

Then, x = cos b and sin b = #sqrt(1-x^2)#. Also, #x in [-1, 1]#.

The given expression =

#sin ( a + b )#

# = sin a cos b + cos a sin b#

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

#=(1/sqrt(1+x^2))(sqrt(1-x^2)+-x^2)#, x in [-1, 1]#.

The negative sign is used when #x in [-1, 0]#.