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Trig Identity

  1. Nov 6, 2007 #1
    [SOLVED] Trig Identity

    1. The problem statement, all variables and given/known data
    cos^4 (x) = (3/8) + (1/2)(cos(2x)) + (1/8)(cos(4x))


    2. Relevant equations
    cos2x = 2cos^2 x - 1

    cos^2 x = 1 - sin^2 x


    3. The attempt at a solution

    Can someone please give me hints? Thanks.
     
  2. jcsd
  3. Nov 6, 2007 #2

    rock.freak667

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    Try proving the RHS to the LHS

    and use the formula cos2A=cos[itex]^2[/itex]A-sin[itex]^2[/itex]A=2cos[itex]^2[/itex]A-1=1-2sin[itex]^2[/itex]A
     
  4. Nov 6, 2007 #3
    ok. Here is what I did:

    = (3/8) + (1/2)(cos(2x)) + (1/8)(cos(4x))
    = (3/8) + (2cos^2 x - 1) + (1/4cos^2 2x - 1)
    = (3/8) + (2cos^2 x - 1) + (1/2cos^4 x - 1 - 1)

    This is where I don't know what to do. Any help would be great. Thanks.
     
  5. Nov 6, 2007 #4

    Avodyne

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    You have one mistake going from the 1st line to the 2nd, and another going from the 2nd to the 3rd.
     
  6. Nov 6, 2007 #5
    I am sorry. I didn't realize that. Here is my new try:

    = (3/8) + (1/2)(cos(2(x)) + (1/8)(cos(4(x))
    = (3/8) + (1/2)(2cos^2 x - 1) + (1/8)(2cos^2 2x - 1)
    = (3/8) + (2cos^2 x) - (1/2) + (1/4cos^2 2x) - (1/8)

    Does this seem right? Please help. I am running out of time. Thanks.
     
  7. Nov 6, 2007 #6

    Avodyne

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    You still have a mistake going from the 2nd to the 3rd line.

    And, you eliminated some (cos 2x)'s, but you still have one in your last line.
     
  8. Nov 6, 2007 #7
    I am getting confused now. Can you elaborate a bit more please?

    Is this rite?

    = (3/8) + (((2cos^2 x)-1)/2) + (((2cos^2 2x)-1)/2)
    = (3/8) + (cos^2 x) - (1/2) + (cos ^2 2x)/4 - (1/8)

    I have been trying this for a long time. I think I am on the right track. Please help. Thanks.

    Side note: I read something about power reduction formulae on the internet, and our teacher hasn't taught us that. So we can only use trig identities to solve this question.
     
    Last edited: Nov 6, 2007
  9. Nov 6, 2007 #8

    Avodyne

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    Yes, that's right, and you are one the right track.

    Now, you still have a cos 2x in your next-to-last term in the last line. You want to write everything in terms of cos x.
     
  10. Nov 6, 2007 #9
    Thanks. Actually, that's the part which is most confusing.

    Here is what I make of it:

    (cos^2 2x)/4 = (2cos^4 x - 1) / 4

    Is this right? I know that cos2x = 2cos^2 x - 1. But I don't know whether the exponent would change or not. Please help. Thanks.
     
  11. Nov 6, 2007 #10

    Avodyne

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    No, it's not right. Suppose you have y^2/4, and I told you that y = 2 z^2 - 1; can you express y^2/4 in terms of z?
     
  12. Nov 6, 2007 #11
    I am so sorry. Let me try again:

    (cos^2 2x)/4 = ((2cos^2 x - 1)^2) / 4

    Is this right? I hope it is because this time I really thought about it.
     
  13. Nov 6, 2007 #12

    Avodyne

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    Yes, it's right!

    Now, with c = cos x, you have

    (3/8) + c^2 - (1/2) + ((2c^2 - 1)^2)/4 - (1/8)

    Expand out the square, and simplify as much as possible.
     
  14. Nov 6, 2007 #13
    Wow. Thanks very much Avodyne. Although it took me a long time to understand, it was still worth it.

    I am going to sleep now, but when I expanded the equation ((2c^2 - 1)^2)/4 , I got c^4-c^2+1/4 which was the key to solving this question. (And Thank God I got that, lol)

    I am very thankful to Avodyne(especially), and off course to rock.freak667 in helping me do this question and understand the concept. You are the best guys.
     
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