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Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Sin(A-B).Asked by Sade
Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Cos(A-B).
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Answered by
Reiny
Steve had already given you a good hint how to do these types in your previous trig question, this is the same type.
You have to know the definitions of the trig functions in terms of x, y, and r and you have to know the CAST rule
tanA = 5/1 = y/x , but we are in III, so y = -5, x = -1
so r^2 = (-5)^2 + (-1)^2 = 26
r = √26
then sinA = -5/√26 , cosA = -1/√26
similarely if sin B = 2/3, and B is in II
cos B = -√5/3
cos(A-B) = cosAcosB + sinAsinB
= (-1/√26)(-√5/3) + (-5/√26)(2/3)
= (√5 - 10)/(3√26)
You have to know the definitions of the trig functions in terms of x, y, and r and you have to know the CAST rule
tanA = 5/1 = y/x , but we are in III, so y = -5, x = -1
so r^2 = (-5)^2 + (-1)^2 = 26
r = √26
then sinA = -5/√26 , cosA = -1/√26
similarely if sin B = 2/3, and B is in II
cos B = -√5/3
cos(A-B) = cosAcosB + sinAsinB
= (-1/√26)(-√5/3) + (-5/√26)(2/3)
= (√5 - 10)/(3√26)
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