Question 923530
You don't need to worry about C in this problem. My guess is that it applies to another problem.


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First we need to find sin(A) based off of the given cos(A) = 1/3



{{{cos^2(A) + sin^2(A) = 1}}}



{{{(1/3)^2 + sin^2(A) = 1}}}



{{{1/9 + sin^2(A) = 1}}}



{{{sin^2(A) = 1-1/9}}}



{{{sin^2(A) = 8/9}}}



{{{sin(A) = sqrt(8/9)}}} A is in quadrant I, so sin(A) is positive



{{{sin(A) = sqrt(8)/sqrt(9)}}}



{{{sin(A) = sqrt(8)/3}}}



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Then we need to find cos(B) based on sin(B) = -1/2



{{{cos^2(B) + sin^2(B) = 1}}}



{{{cos^2(B) + (-1/2) = 1}}}



{{{cos^2(B) + 1/4 = 1}}}



{{{cos^2(B) = 1 - 1/4}}}



{{{cos^2(B) = 3/4}}}



{{{cos(B) = sqrt(3/4)}}} B is in quadrant IV, so cos(B) is positive



{{{cos(B) = sqrt(3)/sqrt(4)}}}



{{{cos(B) = sqrt(3)/2}}}


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Expand out sin(A-B) using an identity and then do substitutions.



{{{sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B)}}}



{{{sin(A-B) = (sqrt(8)/3)*(sqrt(3)/2) - (1/3)*(-1/2)}}}



{{{sin(A-B) = (sqrt(8)*sqrt(3))/6 + 1/6}}}



{{{sin(A-B) = (sqrt(8*3))/6 + 1/6}}}



{{{sin(A-B) = (sqrt(24))/6 + 1/6}}}



{{{sin(A-B) = (sqrt(4*6))/6 + 1/6}}}



{{{sin(A-B) = (sqrt(4)*sqrt(6))/6 + 1/6}}}



{{{sin(A-B) = (2*sqrt(6))/6 + 1/6}}}



{{{sin(A-B) = (2*sqrt(6) + 1)/6}}}



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Let me know if that helps or not. Thanks.


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