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Trig identity

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data

    prove that cos^4(x)= 3/8 +1/2(cos2x)+1/8(cos4x)

    2. Relevant equations

    3. The attempt at a solution

    i attempted to change the cos2x to 2cos^2(x)-1.....then made all them over 8....and now im stuck......another question....if i multiply 2cos(4x) by 2, do i multiply the 2 and the 4x? or just he 4x?
  2. jcsd
  3. Sep 11, 2008 #2
    Well...Use the fact that cos^2(x) = (1 + cos(2x))/2 ...Then square both sides....you will then be pretty close...(that is..you will have cos^4(x) = some mess)..you will just need to use that above formula again to reduce the (cos^2(2x)) you get after squaring the right hand side of the equation........make sure you realize that when plugging (2x) into the left hand side...you will need to make the same plug in in the right...after that...the problem should unfold nicely... You just needed a different formula..
  4. Sep 11, 2008 #3
    do i isolate the cos(2x) in cos^2(x) = (1 + cos(2x))/2 and sun that in?
  5. Sep 11, 2008 #4
    .. I don't know what you mean.. With this method...that is starting with the formula "cos^2(x) = (1 + cos(2x))/2", after squaring and showing that the left hand side is indeed Cos^4(x), your goal is to git rid of every square you have on the right hand side of that...and it turns out, there is only 1...so you use the formula again, but make sure you keep track of that doubled variable...
  6. Sep 11, 2008 #5
    quick...and dumb question.....

    i get 3+cos2x+2cos2x all over 8....which id close except the cos2x should be a cos4x
  7. Sep 11, 2008 #6
    Ha! Try again...Your are making mistakes in your arithmetic.. Just take it nice and slow... The above isn't as close as you might think...because not only should a cos2x be a 4x...but 2cos2x divided by 8...isnt 1/2cos2x....
  8. Sep 11, 2008 #7
    lmao....il try it again slow
  9. Sep 11, 2008 #8
    ha...tell me what you get
  10. Sep 11, 2008 #9
    ahhhh i got it lol...when i was working out the cos^2(2x), i wrote it originally as 1+cos2x over 2, instead of 1+cos4x over 2....is that right now?
  11. Sep 11, 2008 #10
    I told you to watch out for that 2x! ><
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