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)

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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(52756) About Me  (Show Source):
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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) = -sqrt%281-sin%5E2%28u%29%29 = -sqrt%281-%285%2F13%29%5E2%29 = -sqrt%281+-+%2825%2F169%29%29 = -sqrt%28%28169-25%29%2F169%29 = -sqrt%28144%2F169%29 = -12%2F13.

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


2.  sin(v) = sqrt%281-cos%5E2%28v%29%29 = sqrt%281-%28-3%2F5%29%5E2%29%29 = sqrt%281-9%2F25%29%29 = sqrt%2816%2F9%29 = 4%2F5.

    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) = %28-12%2F13%29%2A%28-3%2F5%29+-+%285%2F13%29%2A%284%2F5%29 = 36%2F65+-+20%2F65 = 16%2F65.


Answer.  cos(u+v) = 16%2F65.  (u+v) is in QIV.