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Identies: Sum And Difference, Double Angle hurry

  1. Mar 4, 2008 #1
    I need some help with these kind of problems. I don't clearly understand how to solve them. Here is one...PLEASE HELP!!!!

    sin 3x = (sin x)(3-4sin^2 x) PROVE THEY ARE EQUAL
     
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2
    You can use identities sin(x+y) = cos(x)sin(y) + sin(x)cos(y) and sin(2x) = 2sin(x)cos(x) for this. Merely write 3x slightly differently so that the use of these identities becomes more evident.
     
  4. Mar 4, 2008 #3
    thank you......alright this one i am stuck on.....i have been working on it for 10 minutes and can't come up with the answer.
    2csc2x=(csc^2x)(tanx)
     
  5. Mar 4, 2008 #4
    would the correct answer for my first problem be:
    change sin3x to (sinx)(sin2x)
    then change sin2x to 2sinxcosx
    then change that to (2sinx)(2-2sinx)
    then....i don't know.....but am I on the right track??
     
  6. Mar 4, 2008 #5
    This is incorrect. Put sin(3x) = sin(x + 2x) instead.
     
  7. Mar 4, 2008 #6
    okay thank you I will try that
     
  8. Mar 4, 2008 #7
    alright as my next step I put (sinx)(cos2x)+(sin2x)(cosx)...is this correct?
     
  9. Mar 4, 2008 #8
    do I then change the cos's to sin's?
     
  10. Mar 4, 2008 #9

    cristo

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    Yes. Perhaps you should attempt the entire problem and then post your solution instead of posting line by line.
     
  11. Mar 4, 2008 #10
    sorry I just need to know where I'm messing up.
     
  12. Mar 4, 2008 #11
    Okey, my first advice wasn't flawless because I didn't notice that you need the formula for cos(2x) too. You cannot switch sines to cosines arbitrarily.
     
  13. Mar 4, 2008 #12
    so by using cos2x and sin2x formulas I should get (sinx)(1-2sin^2 x)+(2sinxcosx)(cosx)...right??
     
  14. Mar 4, 2008 #13

    cristo

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    Correct.
     
  15. Mar 4, 2008 #14
    then do I multiply the (2sinxcosx)(cosx) ???
     
  16. Mar 4, 2008 #15
    Rewrite what your problem is and what you want to solve starting from the beginning please.
     
  17. Mar 4, 2008 #16
    the problem is:
    sin3x = (sinx)(3-4sin^2 x) ...PROVE THAT THEY ARE EQUAL
     
  18. Mar 4, 2008 #17
    Ok, show some work now.

    (Im going to bed in 10 min, so type fast if you want help).
     
    Last edited: Mar 4, 2008
  19. Mar 5, 2008 #18
    ok......
    first i split sin3x making sin(x+2x)
    than i used the sum identity making (sinx)(cos2x)+(sin2x)(cosx)
    then i used double angle identity making (sinx)(1-2sin^2 x)+(2sinxcosx)(cosx)
    dunno what to do next
     
  20. Mar 5, 2008 #19
    Factor out a sin(x).
     
  21. Mar 5, 2008 #20
    i got it :) thanks!!!
    new homework... will you help me with it if I have trouble?
     
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