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| Question 602906:  Hi, I need help with my homework. We are learning about the sum and difference of identities.
 Here is the problem:
 What is the value of sin(A-b) if tan A = 9/40 and A is in the first quadrant, while cos B = -4/5 and B is in the third quadrant?
 I really don't get this homework. What is up with A and B? Do we need to plot the graph as well? It's confusing me. I'm sorry if I'm asking stupid questions, I missed our class today, you see. I hope someone can explain this to me properly so I'll catch up with the lesson. Thank you!
 Found 2 solutions by  Alan3354, solver91311:
 Answer by Alan3354(69443)
      (Show Source): 
You can put this solution on YOUR website! Here is the problem: What is the value of sin(A-b) if tan A = 9/40 and A is in the first quadrant, while cos B = -4/5 and B is in the third quadrant?
 I really don't get this homework. What is up with A and B? Do we need to plot the graph as well?
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 Here's the one I just did with the same angles.
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 If tan A = 9/40, A is in the first quadrant, cos B = -4/5, B is in the third quadrant, find the value of cos(A+B)
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 Find the cos(A)
 Tan = y/x
 Plot the point (40,9)
 --> hyp = sqrt(40^2 + 9^2) = 41
 cos(A) = x/hyp = 9/41
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 cos(A + B) = cosA*cosB - sinA*sinB (from Wikipedia)
 Find sin(A) and sin(B)
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 sin(A) = y/hyp = 40/41
 sin(B) = -3/5 in Q3 (it's the 3, 4, 5 triangle)
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 cos(A + B) = cosA*cosB - sinA*sinB
 = (9/41)*(-4/5) - (40/41)*(-3/5)
 = -36/205 + 120/205
 = 84/205
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 Use some of the values found in the previous problem:
 sin(A-b) = sin(A)cos(B) - cos(A)sin(B)
 = (40/41)*(-4/5) - (9/41)*(-3/5)
 = -200/205 + 27/205
 = -173/205
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 What is up with A and B?
 Angles A & B are measured from the x-axis counter-clockwise.  All angles on the x-y plane are measured that way.
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 Do we need to plot the graph as well?  You don't have to if you don't want to.
 Not much to graph, just a point for each angle.
 For A draw a line from the Origin to (40,9)
 For B it's to (-3,4)
 
Answer by solver91311(24713)
      (Show Source): 
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