SOLUTION: What is the exact value of the trigonometric function given that sin u= 5/13 and cos v= -3/5 (Both are in Quadrant II.) cos= (u+v)

Question 1060117: What is the exact value of the trigonometric function given that sin u= 5/13 and cos v= -3/5 (Both are in Quadrant II.) cos= (u+v)
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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What is the exact value of the trigonometric function given that sin u= 5/13 and cos v= -3/5 (Both are in Quadrant II.) cos= (u+v)
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I am going to use the formula

cos(u+v) = cos(u)*cos(v) - sin(u)*sin(v).     (*)


For it, in addition to the given values  sin(u) = 5/13 and cos(v) = -3/5  I need to know  cos(u)  and  sin(v).


1.   cos(u) =  =  =  =  =  = .

    The sign "-" is at the sqrt since cosine is negative in QII.


2.  sin(v) =  =  =  =  = .

    The sign "+" is at the sqrt since sine is positive in QII.


3.  Now you have everything to use the formula (*). Substitute all given and found values into (*). You will get

    cos(u+v) = cos(u)*cos(v) - sin(u)*sin(v) =  =  = .


Answer.  cos(u+v) = .  (u+v) is in QIV.

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