Question 1031119: sen(a)= 8/17, 0 < a < π/2 and cos(B) = 3/5, 0 < B < π/2, cos (a+b)= ??? Found 2 solutions by stanbon, ikleyn:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! sin(a)= 8/17, 0 < a < π/2 and cos(B) = 3/5, 0 < B < π/2,
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cos(a) = sqrt[17^2-8^2]/17 and sin(B) = sqrt[5^2-3^2]/5
cos (a+b)= cos(a)*cos(b)-sin(a)sin(b)
= (15/17)(3/5) - (8/17)(4/5)
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= 45/85 - 32/85
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= 13/85
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Cheers,
Stan H.
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You can put this solution on YOUR website! .
sin(a)= 8/17, 0 < a < pi/2 and cos(b) = 3/5, 0 < B < pi/2, cos (a+b)= ???
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1. We are going to use the formula
cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)
(regarding this formula, see the lesson Addition and subtraction formulas in this site).
For it, we need cos(a) and sin(b), in addition to the given sin(a) and cos(b).
2. cos(a) = = = = = = .
The sign "+" was chosen for the square root since the angle "a" is in Q1.
3. sin(b) = = = .
The sign "+" was chosen for the square root since the angle "b" is in Q1.
4. Now cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) = = = = .
(