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Trig - can't see how to simplify

  1. May 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Its all part of a show that question from an example paper that I'm doing. Apart from this part I have the answer. Just wondered if anyone could show me the substitution that I would have to make.

    Starting from [tex](\frac{sin3\gamma}{sin\gamma})^2[/tex], and ending up with [tex](1+2cos2\gamma)^2[/tex]

    Other than this, I have the question right, and if I got to this stage in an exam, and couldn't do it after spending some time on it, then I'd just go to the final answer, and hope that its a logical step that I just haven't seen. But as I say I have the rest of the answer, its just this little part that needs manipulating.

    Thanks

    Brewer
     
    Last edited: May 14, 2007
  2. jcsd
  3. May 14, 2007 #2

    malawi_glenn

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    hints:

    sin(3y) = sin (2y + y) = .. ?

    sin(2y) = .. ?

    (cosy)^2 = .. ?

    :)
     
  4. May 14, 2007 #3

    Hootenanny

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    Since you've done the rest of the question, I see no harm is showing you this little bit. So, from the triple angle indentity [itex]\sin3\theta = 3\sin\theta - 4\sin^3\theta[/itex] we can write;

    [tex]\left(\frac{\sin3\gamma}{\sin\gammer}\right)^2 = \left(\frac{3\sin\gamma - 4\sin^3\gamma}{\sin\gamma}\right)^2[/tex]

    [tex]= \left(3-4\sin^2\gamma\right)^2[/tex]

    Recalling that [itex]\sin^2\theta = \left(1-\cos2\theta\right)/2[/itex] gives;

    [tex]\left(3-4\sin^2\gamma\right)^2 = \left(3 - 4\frac{1-\cos2\gamma}{2}\right)^2[/tex]

    [tex]=\left(3 -\left(2 - 2\cos2\gamma\right)\right)^2[/tex]

    [tex]=\left(1+2\cos2\gamma\right)^2[/tex]

    As required. I hope that was helpful :smile:
     
    Last edited: May 15, 2007
  5. May 14, 2007 #4

    malawi_glenn

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    my "solution" is simpler i guess..

    i dont think this guy is at the level were one uses the triple angle indentity..
     
  6. May 14, 2007 #5
    It was actually. Both versions. I would have tried the first method, but a maths student I was talking to mentioned the triple angle identity (but I'm sure I've not heard of it/done it yet), but I didn't follow it. But I see how its done now. Thanks a lot, both of you.
     
  7. May 14, 2007 #6

    Hootenanny

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    An undergraduate doing a mathematics related course doesn't know the triple angle identity? :rolleyes:

    Of course the same result can be arrived using the angle sum and double angle indentities, but this way is quicker...:biggrin:
     
    Last edited: May 14, 2007
  8. May 14, 2007 #7

    cristo

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    The triple angle identity (which I admit, I don't remember explicitly) is easily derived by expanding sin(2x+x) and then using the double angle identities.

    Guilty as charged! :blushing:
     
    Last edited: May 14, 2007
  9. May 14, 2007 #8
    As far as I can remember, I haven't been taught it yet and I'm in my second year of a Physics degree. In fact I'm pretty definate that I haven't been taught it. I wasn't taught it last year, and my maths course this year focused mainly on Fourier transforms and series.
     
  10. May 14, 2007 #9

    malawi_glenn

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    depends on how much you must/want remember.. i think its easier to remember a few simple ones that have broad area of usage, than very very specific ones.
     
  11. May 14, 2007 #10

    Office_Shredder

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    If you've memorized triple angle identities, you're a loser

    All statements made here reflect the opinions of the author and in no way reflect the views of normal people
     
  12. May 14, 2007 #11

    malawi_glenn

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    naw that wasn't very nice =(
     
  13. May 14, 2007 #12

    Hootenanny

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    Guys, chill out a little! Obviously sarcasm doesn't work well over forums...

    I wouldn't have known the triple angle forumlae either if I hadn't been using them alot recently.

    Office: Gee Thanks :mad:
     
    Last edited: May 14, 2007
  14. May 14, 2007 #13

    cristo

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    Ahh.. so it was sarcastic! No, it doesn't work too well. There should be some kind of sarcastic smily on here! Anyway, I for one am totally chilled out! :biggrin:

    You're right about the memorising thing though-- if you use something enough, you're bound to remember it!
     
  15. May 14, 2007 #14

    Office_Shredder

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    I was being sarcastic too (hence the italics)...

    let's just pretend this conversation never happened
     
  16. May 14, 2007 #15

    Integral

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    The triple angle identity is readily available in any decent table of trig identies. Once you get out of trig class, that is a perfectly valid reference which every student of math or physics should have at hand. You do not have to be "shown" each and every identity before you can use it. Learn to use references. My CRC Standard Math Tables is perhaps my most used reference.
     
  17. May 15, 2007 #16

    Curious3141

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    The neatest way, IMHO, is to use factor formulae.

    Forget about the squares on the LHS and RHS, and let's call gamma x since it saves typing.

    [tex]\frac{\sin 3x}{\sin x} = \frac{2\sin 3x \cos x}{2 \sin x \cos x} = \frac{\sin 4x + \sin 2x}{\sin 2x} = \frac{\sin 2x(2\cos 2x + 1)}{\sin 2x} = 2\cos 2x + 1[/tex]

    BTW, Hootenanny, there's a little error in your working, and the final sign is wrong. :smile:
     
  18. May 15, 2007 #17

    Hootenanny

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    :blushing: Duely corrected in the post, thanks Curious :smile:

    This line;

    [tex]=\left(3 -2 - 2\cos2\gamma\right)^2[/tex]

    Should read;

    [tex]=\left(3 -\left(2 - 2\cos2\gamma\right)\right)^2[/tex]
     
  19. May 15, 2007 #18

    Curious3141

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    Much better! :approve:
     
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