+128407 sin ( arccos (2/3) - arctan (1/3) )
Let arccos (2/3) = A ...so ... cos A = 2/3 ......so sin A = sqrt (3^3 - 2^2) / 3 = sqrt (5) / 3
and arctan (1/3) ... so .. tan B = 1/3 ......so sin B = 1/ sqrt ( 1^2 + 3^2) = 1 / sqrt (10) and cos B = 3 / sqrt (10)
So we have
sin ( A - B ) = sin A cos B - sin B cos A
sin ( A - B) = sqrt (5) / 3 * 3 /sqrt (10) - 1/sqrt (10) * (2/3)
sin (A - B) = sqrt (5) / sqrt (10) - 2 / (3 sqrt (10)
sin (A - B) = 1/sqrt (2) - 2sqrt (10) / 30
sin ( A - B) = 1/sqrt (2) - sqrt (10) / 15
sin (A - B) = sqrt (2) / 2 - sqrt (10) /15
sin (A - B) = 15sqrt (2) - 2sqrt (10)
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30
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