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(52754) (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)
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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) = .
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?