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Is the identity I came up with for sin(x) ^ 2 correct?

  1. Aug 18, 2009 #1
    I noticed that the graphs of sin(x) and sin(x) ^ 2 are very similar. So I offset sin(x) ^ 2 to exactly match sin(x):
    [tex]sin(x) = 2 sin^{2}\left(\frac{x}{2} +\frac{\pi}{4}\right) - 1[/tex]

    Is this right, or is it an illusion? I haven't been able to find any identity that this is based on.
    If it is right, then:
    [tex]sin^{2}(x) = \frac{sin(2x - \frac{\pi}{2}) + 1}{2}[/tex]

    Thanks,
    pro.in.vbdnf
     
    Last edited: Aug 18, 2009
  2. jcsd
  3. Aug 18, 2009 #2

    arildno

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    That is certainly true!

    Remember that [tex]sin(2x-\frac{\pi}{2})=-\cos(2x)[/tex]

    and therefore, your identity is a rewriting of the half-angle formula.
     
  4. Aug 18, 2009 #3
    You have indeed hit on something true. You are exploiting several trig identities. Namely:

    Cofunction (for cosine): [itex]\cos(x)=\sin\left( \frac{\pi}{2}-x \right)[/itex]

    Symmetry (for sine): [itex]\sin(-x) = \sin(x)[/itex]

    Double Angle (for cosine): [itex]\cos(2x)=1-2\sin^2(x)[/itex]

    So:

    [tex]\frac{\sin \left( 2x - \frac{\pi}{2} \right)}{2}= \frac{-\sin \left( \frac{\pi}{2} - 2x \right)}{2} = \frac{-\cos(2x)+1}{2} = \frac{-(1-2\sin^2x)+1}{2}= \frac{2\sin^2(x)}{2} = \sin^2(x)[/tex]​

    Incidently there is also the Half Angle (or Power Reducing) Identity

    [tex]\sin^2(x) = \frac{1-\cos(2x)}{2}[/tex]​

    which is closely related to what we have here.

    --Elucidus
     
  5. Aug 18, 2009 #4
    Thanks for your help! The identities I was missing putting together were [tex]sin^{2}(x) + cos^{2}(x) = 1[/tex] and [tex]cos(2x) = cos^{2}(x) - sin^{2}(x)[/tex].

    So [tex]cos^{2}(x) = \frac{cos(2x) + 1}{2}[/tex]. I'll have to remember that.
     
    Last edited: Aug 18, 2009
  6. Aug 18, 2009 #5
    Does that mean [tex]\sqrt{cos(x)}[/tex] has an identity in terms of cos(x)?
     
  7. Aug 18, 2009 #6
    I believe you meant [itex]\sin^2(x) + \cos^2(x) = 1[/itex]. The latter equation above is the Half Angle (or Power Reducing) Identity for cosine.

    I do not think so, but [tex]\sqrt{1 \pm \cos(x)}[/tex] does:

    [tex]\sqrt{1+ \cos(x)}=\sqrt{\frac{2(1+\cos(2x/2))}{2}}=\sqrt{2\cos^2(x/2)}=\sqrt{2}\left| \cos(x/2) \right|[/tex]

    [tex]\sqrt{1-\cos(x)}=\sqrt{2}\left| \sin(x/2) \right|[/tex]​

    Both of these are handy for certain types of methods in calculus.

    --Elucidus
     
  8. Aug 18, 2009 #7
    (Yes, I meant [itex]\sin^2(x) + \cos^2(x) = 1[/itex].)

    What are their names?

    By the way, is there a difference between the [tex] and [itex] tags? And is there a way to vertically center the LaTeX images with the text?
     
  9. Aug 18, 2009 #8

    arildno

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    1. The itex-tag can be used to generate Latex within the ordinary sentence structures, like the equation: [itex]2x-1=3[/itex], rather than [tex]2x-1=3[/tex]

    2. Note that the half-angle formula gets rid of a square root!!
    Thus, the right-hand side is easily integrated, the left-hand side would seem a hopeless target for integration without that nifty identity.
     
  10. Aug 18, 2009 #9
    [tex]\sqrt{cos(x)} = \sqrt{2 cos^{2}\left(\frac{x}{2}\right) - 1}[/tex]
    So there is no way to get rid of the right hand radical?
     
  11. Aug 18, 2009 #10

    arildno

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    Sure, by squaring both sides. :smile:
     
  12. Aug 18, 2009 #11
    ..while keeping the left radical? I wonder if there is an identity of the form [tex]\sqrt{cos(x)} = ... cos(x) ...[/tex]?
     
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