Question 602906
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)