How do you find the exact value of #Sin(arcsin(1/3)-arcsin(1/4))#?

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Answer 1 · 2016-12-22T05:06:54

#1/12(sqrt15-2sqrt2)=0.0870#, nearly.

Use #sin (a-b)=sin a cos b - cos a sin b and f f^(-1)(y) = y.# and, for x in

#Q_1#, cos x = sqrt(1-sin^2 x) and, likewise, sin x = sqrt( 1-cos^2x)#

The given expression becomes

#sin arcsin(1/3)cos sin arc sin (1/4)-cos arc sin (1/3) sin arc arc sin (1/4)#

#=1/3cos arc cos sqrt(1-(1/4)^2)-1/4cos arc cossqrt(1-(1/3)^2)#

#=1/3sqrt(1-1/16)-1/4sqrt(1-1/9)#

#1/12(sqrt15-2sqrt2)#.