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Trig Identity proof

  1. Aug 22, 2012 #1
    Show that:

    [itex]4(\sin^4x+\cos^4x) \equiv \cos4x + 3[/itex].

    Really stuck with this, no idea how to go ahead with it. The book gives a hint: [itex]\sin ^4 x = (\sin ^2 x)^2[/itex] and use [itex]\cos 2x = 1 - 2\sin ^2 x[/itex]

    But I don't even understand the hint, where did they get
    [itex]\cos 2x = 1 - 2\sin ^2 x[/itex] from?
     
  2. jcsd
  3. Aug 22, 2012 #2

    Mark44

    Staff: Mentor

    This is one of three double-angle identities for cos(2x).

    The other two are
    cos(2x) = cos2(x) - sin2(x)
    cos(2x) = 2cos2(x) - 1

    The 2nd one above and the one you're asking about can be obtained from the first one I showed by using the identity sin2(x) + cos2(x) = 1.
     
  4. Aug 22, 2012 #3

    CAF123

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    Gold Member

    [tex] cos (x + x) = cos^2x - sin^2x = (1 - sin^2x) -sin^2x = 1 - 2sin^2x, [/tex] using the identity [itex] sin^2x +cos^2x =1 [/itex]
     
  5. Aug 23, 2012 #4

    CAF123

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    To start you off, write the given equation as, [tex] 4((\sin^2x)^2 + (\cos^2x)^2), [/tex] and use the double angle identities to get this solely in terms of [itex] \cos.[/itex]
     
  6. Aug 23, 2012 #5
    You can start by using this:
    a2+b2=(a+b)2-2ab
    This would mean that:
    [tex]4(\sin^4x+\cos^4x)=4((sin^2x+cos^2x)^2-2sin^2xcos^2x)[/tex]
    Now rest should be easy to solve.
     
  7. Aug 24, 2012 #6
    working :

    [itex]4((sin^2x)^2 + (cos^4x) [/itex]
    [itex]4((1-cos^2x)^2 + cos^4x))[/itex]
    [itex]4(1-2cos^2x + cos^4x + cos^4x) [/itex]
    [itex]4(1-2cos^2x + 2cos^4x) [/itex]

    Pretty much stuck here.
    continuing from your working: [itex]4(1 - 2sin^2xcos^2x)[/itex], but I don't really know any identities to go from here...
     
  8. Aug 24, 2012 #7

    Bacle2

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    How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?
     
  9. Aug 24, 2012 #8
    He doesn't have [itex] cos(4x) [/itex], he has [itex] cos^{4}(x) [/itex], or [itex] (cos(x))^{4} [/itex]
     
  10. Aug 24, 2012 #9

    ehild

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    Use the following double-angle identities:
    sin2(x)=0.5(1-cos(2x))
    cos2(x)=0.5(1+cos(2x))

    ehild
     
    Last edited: Aug 25, 2012
  11. Aug 24, 2012 #10

    Mark44

    Staff: Mentor

    Bacle2 explicitly was referring to the right side of the original identity, which has a cos(4x) term in it.

    See below.

     
  12. Aug 24, 2012 #11

    Bacle2

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    Look at the RH side of the original post, it has a Cos(4x)+3.


    Edit: Sorry, I did not see Mark44's post, and now I cannot see the 'delete' option.
     
    Last edited: Aug 25, 2012
  13. Aug 25, 2012 #12
    Try using these identities:
    i)sin(2x)=2sin(x)cos(x)
    ii)cos(2x)=1-2sin2x

    If you square both the sides of i identity, you may see a way through.
     
  14. Aug 25, 2012 #13

    CAF123

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    Use what ehild suggested if attempting the way I started (which is just one way to tackle the problem - there are many others).
    We have, as suggested in my last post,
    [itex] 4((\sin^2x)^2 + (\cos^2x)^2) [/itex]

    Now, use the double angle identities that ehild suggested, specifically [itex] \sin^2x = -\frac{1}{2}(\cos2x - 1) [/itex] and [itex] \cos^2x = \frac{1}{2}(cos2x +1) [/itex]

    Substitute these into the eqn, expand the brackets and simplify...
     
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