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Sin^4Ө =3/8-3/8cos(2Ө) Prove the following trigonometric identity

  1. Aug 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove the following trigonometric identity. The question is sin^4Ө =3/8-3/8cos(2Ө)

    2. Relevant equations
    I think I'm supposed to use the power reducing formulas for trigonometric identities which are
    sin^2(u)= (1- cos(2u))/2
    cos^2(u)=(1+cos(2u))/2
    *Let u represent any integer/value*

    3. The attempt at a solution
    To expand the equation I separated the equation into (1- cos^2Ө)(1- cos^2Ө) = 1- 2cos^2Ө + (cos^2Ө)^2. I reduced it because sin^2Ө = 1 – cos^2Ө and after this I'm confused on what the next steps are.
     
  2. jcsd
  3. Aug 13, 2015 #2

    BvU

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    So, a much better post than the first try. Good !

    Your other entry is from the righthand side: what have you got for ##\cos(2\theta)## that might be useful here ?

    [edit] ah: your second relevant equation !
     
  4. Aug 13, 2015 #3

    jedishrfu

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    So you applied the sin^2(u) to the lefthand side of the equation and got (1/4) * (1 - cos(2u) )^2 right?

    Next expand the

    (1/4)*(1 - cos(2u) )^2 = (1/4) * ( 1 - 2*cos(2u) - cos(2u)^2 )

    next apply the cos^2(u) rule to the last term and see what you get.
     
  5. Aug 13, 2015 #4
    As well as the power reducing formulas, you'll want to glance at the double angle formulas.
     
  6. Aug 13, 2015 #5

    BvU

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    But now I see a problem coming up:
    let ##\theta = \pi/2\ \ ## then ##(sin\theta)^4 = 1\ \ ## and ##\ \ 3/8-3/8\cos(2\theta)=3/4## !?
    So no identity at all !
    Or did I read the original thingy in the wrong way ?
     
  7. Aug 13, 2015 #6

    haruspex

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    I believe the identity ought to read ##\sin^4(\theta)=\frac 38-\frac 12\cos(2\theta)+\frac 18\cos(4\theta)##.
    Looks like someone turned that ##4\theta## into another ##2\theta##.
     
  8. Aug 14, 2015 #7

    HOI

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    Why would u have to be an integer?
     
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