SOLUTION: 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) Please help me solve this problem. It's from the homework given

Question 602898: 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)
Please help me solve this problem. It's from the homework given to us today, which I have no chance of understanding since I missed class per the doctor's orders, and my friend's notes aren't really that readable...
If you can, please show me the steps as well. It would really help me understand this lesson. Thank you!
Found 2 solutions by Alan3354, scott8148:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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)
--------------
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
-------------
cos(A + B) = cosA*cosB - sinA*sinB (from Wikipedia)
Find sin(A) and sin(B)
-------
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

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
cos(A + B) = [cos(A) * cos(B)] - [sin(A) * sin(B)]

you can use Pythagoras to find the missing functions
___ cos(A) = 40 / sqrt(40^2 + 9^2)
___ sin(A) = 9 / sqrt(40^2 + 9^2)

B is part of a 3-4-5 triangle , so sin(B) = -3/5

just figured out that 9-40-? is also a Pythagorean triple
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