How do you find the exact value of #sin[arccos(-2/3)+arcsin(1/4)]#?

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Answer 1 · 2016-07-26T12:02:56

#sin(arccos(-2/3)+arcsin(1/4))=(5sqrt(3))/12-1/6#

Let #alpha = arccos(-2/3)# and #beta = arcsin(1/4)#

#alpha# is in Q2, so #sin(alpha) > 0# and we find:

#sin(alpha) = sqrt(1-cos^2(alpha)) = sqrt(1-(-2/3)^2) = sqrt(5/9) = sqrt(5)/3#

#beta# is in Q1, so #cos(beta) > 0# and we find:

#cos(beta) = sqrt(1-sin^2(beta))= sqrt(1-(1/4)^2) = sqrt(15/16) = sqrt(15)/4#

Then:

#sin(alpha+beta)#

#= sin alpha cos beta + sin beta cos alpha#

#=(sqrt(5)/3)(sqrt(15)/4) + (1/4)(-2/3)#

#=(5sqrt(3))/12-1/6#