How do you simplify #sin (2 * arcsin (x))#?

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Answer 1 · 2016-11-08T20:11:19

The answer is #=2xsqrt(1-x^2) #

Let #y=arcsinx#, then #x=siny#

#sin(2arcsinx)=sin2y=2sinycosy#

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

#cos^2y=1-x^2##=>##cosy=sqrt(1-x^2)#

#:.sin(2arcsinx)=2xsqrt(1-x^2) #

Answer 2 · 2018-06-15T23:32:19

If we interpret #arcsin a# as all the solutions to #sin x = a# then

#sin(2 arcsin x) = 2 (sin arcsin x)(cos arcsin x) = 2 x cos arcsin (x/1) = pm 2x sqrt{1-x^2}#

#arcsin (x/1)# refers to a right triangle, opposite #x#, hypotenuse #1# so adjacent #sqrt{1-x^2}#. The sign is ambiguous so we prepend #pm#.