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

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Question 1060448: Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sin (u + v)
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52756)   (Show Source): You can put this solution on YOUR website!
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Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sin (u + v)
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I am going to use the formula

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

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


Answer.  sin(u+v) = .  


Answer by MathTherapy(10549)   (Show Source): You can put this solution on YOUR website!

Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sin (u + v)
Problem # 1060297 "asked" for sec (v - u). That answer was given to you. Why can't you follow the same steps to obtain sin (u + v)?
Why do you want someone to do everything for you? You have to make an effort, and give someone else a chance to get their problem solved.
Don't you think you're being UNFAIR?
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