How do you simplify the expression #sin(arctan 2x - arccos x)#?

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Answer 1 · 2016-08-29T11:19:29

#"The Reqd. Value"=(2x^2-sqrt(1-x^2))/sqrt(4x^2+1)#.

Suppose that, #arc tan 2x=alpha, and, arc cos x=beta#

#rArr tan alpha=2x, and, cos beta=x#.

For the sake of brevity, we assume that #0<=x<=1#, so that,

#0<= alpha, beta <=pi/2 rArr "all Trigo. Ratios are" >0#.

Now, reqd. value #=sin (alpha-beta)#

#=sin alphacos beta- cos alphasin beta...........(star)#

#tan alpha=2x rArr sec^2 alpha=tan^2 alpha+1=4x^2+1#

#rArr cos alpha=1/(+sqrt(4x^2+1)).....(1)#.

#tan alpha=2x rArr sin alpha/cos alpha=2x#

#rArr sin alpha=2xcos alpha=(+2x)/sqrt(4x^2+1)..........(2)#

Also, #cos beta=x rArr sin beta=+sqrt(1-x^2)..........(3)#

Using #(1)-(3)# in #(star)#, bearing in mind that #cos beta=x#,

#"The Reqd. Value"=(2x^2)/sqrt(4x^2+1)-sqrt(1-x^2)/sqrt(4x^2+1)#

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

Enjoy Maths.!