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Very basic question

  1. Jan 7, 2009 #1
    Hello everyone! This is my first post here so pardon me if it's a little too simple... I just can't figure out where this equation came from (or rather how it got to that point):

    cos(2 theta) = 1 - 2sin^2(theta)

    Does it have something to do with the identity cos^2 + sin^2 = 1, and if so, how does it apply? (I just started my first year at college and I find myself wondering how I got here!)
  2. jcsd
  3. Jan 7, 2009 #2
    I would say first consider [tex]cos(2\theta) = cos(\theta + \theta)[/tex], then use the sum to product formula [tex]cos(\alpha + \beta) = (cos\alpha)(cos\beta) - (sin\alpha)(sin\beta)[/tex]. See if you can go from there.
    Last edited: Jan 7, 2009
  4. Jan 7, 2009 #3
    Remember that

    [tex] e^{i \theta} = cos(\theta) + i \cdot sin(\theta) [/tex]

    So let's square both sides to get

    [tex] e^{2i \theta} = cos^{2}(\theta) + 2i \cdot cos(\theta)sin(\theta) - sin^{2}(\theta) [/tex]

    But note also that

    [tex] e^{2i \theta} = cos(2 \theta) + i \cdot sin(\theta) [/tex]

    So matching up the real parts we get:

    [tex] cos(2 \theta) = cos^{2}(\theta) - sin^{2}(\theta) [/tex]

    and matching up the imaginary parts we get:

    [tex] sin(2 \theta) = 2 sin(\theta)cos(\theta) [/tex]
  5. Jan 8, 2009 #4
    Ok, so the question is actually

    [tex]\int sin^{2}(x)dx [/tex]

    Thanks to your help I got to

    [tex]\int sin^{2}(x)dx = \int 1/2 - cos(2\theta)/2 dx[/tex]

    Now, the answer says this integration equals

    [tex]1/2x - sin(2\theta)/4 + C [/tex]

    My question now is how does the cos bit integrate to the sin bit?
  6. Jan 8, 2009 #5


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    ?? you should have learned that d(sin x)/dx= cos(x) and so [itex]\int cos(x) dx= sin(x)+ C [/itex] before you start trying to integrate sin(2x)!
    Last edited by a moderator: Jan 26, 2009
  7. Jan 8, 2009 #6
    So the dividend is just integrated normally; how does the divisor go from 2 to 4?
  8. Jan 8, 2009 #7
    because [tex]\frac{d}{dx}\sin(2x)=2*\cos(2x) [/tex] and so [tex]\int \cos (2x) dx = \frac{\sin(2x)}{2}+C[/tex]

    So you have an extra 2 in the denominator, hence it becomes 2x2 which is 4...
  9. Jan 8, 2009 #8
    Just figured it out a minute ago! It's frustrating when I know it's something so simple... thanks everyone!
  10. Jan 26, 2009 #9
    lol guys if his asking a basic question like this i dont think it requires integration
    cos 2x = cos (x+x)
    cos (x+x) = cos x . cos x - sin x . sin x
    = cos^2 x - sin^2 x
    using the identity cos^2 x + sin^2 x = 1
    sin^2 x= 1-cos^2 x
    cos^2 x +cos^2 x - 1 = cos 2x
    2 cos^2 x -1=cos 2x
    There no integration or w/e these guys used xD
  11. Jan 26, 2009 #10

    Gib Z

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    Please, a) Don't revive old threads as you did with your other post, and b), read the entire thread before commenting.
  12. Jan 28, 2009 #11
    [tex]cos(A+B)=cosA \ cosB - sinA \ sinB[/tex]



    [tex]=cos\theta cos\theta-sin\theta sin\theta[/tex]


    remember that:


    for [tex]sin^2\theta=1-cos^2\theta[/tex]:







    for [tex]cos^2\theta=1-sin^2\theta[/tex]:







    [tex]\int \ sin^2x \ dx[/tex]

    [tex]=\int \ \frac{1-cos2x}{2} \ dx[/tex]

    [tex]=\frac{1}{2}\int \ 1-cos2x \ dx[/tex]

    let [tex]u=2x[/tex], then [tex]\frac{du}{dx}=2[/tex], [tex]du=2 \ dx[/tex]


    [tex]\frac{1}{2} \int \ 1-cos2x \ dx[/tex]

    [tex]=\frac{1}{2} \ \frac{1}{2} \int \ (1-cos2x)(2 \ dx)[/tex]

    [tex]=\frac{1}{2} \ \frac{1}{2} \int \ (1-cos \ u) \ du[/tex]

    [tex]=\frac{1}{4} \ \int \ (1-cos \ u) \ du[/tex]

    [tex]=\frac{1}{4} (u-sin \ u) + C[/tex]

    [tex]=\frac{1}{4} (2x-sin \ 2x) + C[/tex]

  13. Jan 28, 2009 #12

    Gib Z

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    That's correct, I dont see the problem.
  14. Jan 28, 2009 #13
    If you are unsure, differentiate your answer...
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