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Trig help

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    If sinW = 3/5 and cosT = 12/13, where both W and T are acute angles, then find the exact value of sin(W+T)

    2. Relevant equations





    3. The attempt at a solution

    I tried sin(3/5)cos(12/13) + cos(3/5)sin(12/13) = sin [(3/5) + (12/13)] but I doubt it's right.
     
  2. jcsd
  3. Nov 20, 2008 #2
    You were close in your attempt but you made a little mistake. The identity you are looking for is:
    sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
    You should try sinWcosT + cosWsinT. You are already given sinW and cosT. You should also be able to find cosW and sinT from those two values as well.
     
  4. Nov 20, 2008 #3
    Well I did do that w = x, t = y

    Sin(3/5)cos(12/13) + cos(3/5)sin(12/13)
    w t w t

    Or maybe I don't understand you, elaborate plaese
     
  5. Nov 20, 2008 #4
    sinW = 3/5 not sin(3/5)

    I think you are trying to use sin(sinW) instead of sinW.
     
  6. Nov 20, 2008 #5
    Okay, that's what I mean. sinW is not sin(3/5), sinW = 3/5.
     
  7. Nov 20, 2008 #6
    Oh ok thanks
     
  8. Nov 20, 2008 #7
    So, I got 72/65 for sin (w+t), can someone clarify please?
     
  9. Nov 20, 2008 #8
    Unless I fat fingered it:
    I got 56/65
    maybe double check...
     
  10. Nov 20, 2008 #9
    I recieved the same answer, 56/65.
     
  11. Nov 20, 2008 #10
    I must be doing something wrong then.

    (3/5)(12/13) + (3/5)(12/13) = 72/65
     
  12. Nov 20, 2008 #11
    No, you don't yet have cosW or SinT.

    You need to find them first from what you have been given.
     
  13. Nov 20, 2008 #12
    WEll i have been trying for awhile now to figure out cosw and sint. Any hints on how to get there?
     
  14. Nov 20, 2008 #13
    You could use another identity.

    sin^2(x) + cos^2(x)=1
     
  15. Nov 20, 2008 #14
    Erm.. what?
     
  16. Nov 20, 2008 #15
    Okay, here is another way that actually is really easy for this particular problem.
    [tex]sin W = \frac{o}{h}[/tex]
    You should know this, it is the definition of the sin function. Well, they tell you sin W = 3/5 so o = 3 and h = 5. You can then draw a right triangle with a hypotenuse of 5 and one leg of 3 and find the length of the other leg, a, using the Pythagorean theorem. Once you find a,
    [tex]cos W = \frac{a}{h}[/tex]
    you can plug it into there along with your h that you already know.

    You can use a similar method on T.
     
  17. Nov 20, 2008 #16
    Maybe you have seen it written sinx^2 + cosx^2 = 1

    it means the same thing.
     
  18. Nov 20, 2008 #17
    Yeah, Chaos' method does make it easier.
     
  19. Nov 20, 2008 #18
    He means
    [tex]sin^{2}x + cos^{2}x = 1[/tex]
     
  20. Nov 20, 2008 #19
    Thank you, I really should use latex. I am just so slow...
     
  21. Nov 21, 2008 #20
    sin(W+T)=sinWcosT+cosWsinT
    Since sinW=3/5 and cosT=12/13
    [tex]sin(W+T)=\frac{3}{5}*\frac{12}{13}+cosW*sinT[/tex]
    We don't know cosW and sinT, so we could find them by using:
    cosW=√1-sin2W
    cosW=√1-9/25
    cosW=4/5
    using the same method
    sinT=√1-cos2T
    sinT=√1-144/169
    sinT=√25/169
    sinT=5/13
    So we can plug them in the equation above:
    [tex]sin(W+T)=\frac{3}{5}*\frac{12}{13}+\frac{4}{5}*\frac{5}{13}[/tex]
    The final answer should be 56/65.
    Regards.
     
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