How do you find the exact value of Sin(arcsin(1/3)-arcsin(1/4))sin(arcsin(13)−arcsin(14))?

1 Answer
A. S. Adikesavan
Dec 22, 2016

1/12(sqrt15-2sqrt2)=0.0870112(√15−2√2)=0.0870, nearly.

Explanation:

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

Q_1Q1, 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)sinarcsin(13)cossinarcsin(14)−cosarcsin(13)sinarcarcsin(14)

=1/3cos arc cos sqrt(1-(1/4)^2)-1/4cos arc cossqrt(1-(1/3)^2)=13cosarccos√1−(14)2−14cosarccos√1−(13)2

=1/3sqrt(1-1/16)-1/4sqrt(1-1/9)=13√1−116−14√1−19

1/12(sqrt15-2sqrt2)112(√15−2√2).

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