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Finding an expression for (e.g. sin (3x)) in terms of (e.g. sin x) alone?

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Use double-angle and addition formulæ and other relations for trigonometrical functions to find an expression for sin(3x) in terms of sin x alone.

    My problem is I don't know what is meant by "find an expression for sin(3x) in terms of sin x alone.". I know the relevant formulae but do not know what is actually wanted of the question. I know I can 'split' it by going sin(2x + x) then using addition formulae... But I don't know why or what is expected as a final answer. An equation involving only sin and no cos?

    2. Relevant equations
    Trig identities, addition formulae


    3. The attempt at a solution
    No idea.


    PS. I found the answer online but had no idea why that's the answer - please don't just give me the answer :).
     
  2. jcsd
  3. Apr 4, 2010 #2

    danago

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    Basically, what the question is asking you is to find another way to write sin(3x) using only sin(x). So your final answer cannot have any sin(3x) or sin(2x), but only sin(x).

    I think your first idea is a good one:

    [tex]sin(3x) = sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)[/tex]

    However, this still has sin(2x), cos(2x) and cos(x) in it. How can you get rid of all these and be left with just a combination of sin(x)?
     
  4. Apr 4, 2010 #3
    1. Use your idea about addition formulae. Just apply it once.

    2. Then consider any formulas for [itex]\sin(2x)[/itex] and [itex]\cos(2x)[/itex]? In particular, you will want the identity for [itex]\cos(2x)[/itex] that involves only [itex]\sin x[/itex] as there are three identities for [itex]\cos(2x)[/itex]. And don't forget the most basic one: [itex](\sin x)^2 + (\cos x)^2 = 1[/itex].
     
  5. Apr 4, 2010 #4
    Doh! Looks like I type too slowly at this early hour... Danago beat me to the punch!
     
  6. Apr 5, 2010 #5
    Thanks a lot to both of you, I actually understand what is required now.

    I'm a bit lost with actually getting the solution, but I really do need to work on my math skills so I'll do that.

    Thanks again!
     
  7. Apr 5, 2010 #6

    danago

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    Give it a good shot, and if you get lost in the algebra and trig. identities, feel free to post back here and im sure someone will be able to help out :smile:

    All the best,
    Dan.
     
  8. Apr 5, 2010 #7

    danago

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    Gold Member

    Give it a good shot, and if you get lost in the algebra and trig. identities, feel free to post back here and im sure someone will be able to help out :smile:

    All the best,
    Dan.
     
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