SOLUTION: sen(a)= 8/17, 0 < a < &#960;/2 and cos(B) = 3/5, 0 < B < &#960;/2, cos (a+b)= ???

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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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Answer by ikleyn(52792)   (Show Source): 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)= ???
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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) =  =  =  = .


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