How do you write the trigonometric expression #sin(arctan2x-arccosx)# an an algebraic expression?

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Answer 1 · 2018-01-23T13:09:55

The answer is #=(2x^2-sqrt(1-x^2))/(sqrt(1+4x^2))#

#"Reminder"#

#sin(a-b)=sinacosb-sinbcosa#

#sin^2a+cos^2a=1#

Let #y=arctan(2x)#, #=>#, #tany=2x#

Let #z=arccos(x)#, #=>#, #cosz=x#

Therefore,

#sin(arctan(2x)-arccos(x))=sin(y-z)#

#=sinycosz-sinzcosy#

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

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