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Math Help - Proving if it's true (Equality)

  1. #1
    Junior Member Cthul's Avatar
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    Proving if it's true (Equality)


    I'm no sure how to prove this equality true.
    (Oh and I used an image instead of latex, because I don't know how, someone also tell me how to make to the power and cos?)
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  2. #2
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    It doesn't. You need to use the identity

    \displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}.


    So that means

    \displaystyle \cos{(x + y)}\cdot \cos{(x - y)} = (\cos{x}\cos{y} - \sin{x}\sin{y})(\cos{x}\cos{y} + \sin{x}\sin{y})

    \displaystyle = \cos^2{x}\cos^2{y} - \sin^2{x}\sin^2{y}

    \displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{x})(1 - \cos^2{y})

    \displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{y} - \cos^2{x} + \cos^2{x}\cos^2{y})

    \displaystyle = \cos^2{y} + \cos^2{x} - 1.
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  3. #3
    Junior Member Cthul's Avatar
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    Thanks, that was helpful.
    (And I got how to do the latex now)
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    When you get to

    cos^2x\;cos^2y-sin^2x\;sin^2y=cos^2x-cos^2y

    you can combine some of the squares to get 1.

    cos^2x\;cos^2y-sin^2x\left(1-cos^2y\right)=cos^2x-cos^2y

    cos^2x\;cos^2y-sin^2x+sin^2x\;cos^2y=cos^2x-cos^2y

    cos^2y\left(cos^2x+sin^2x\right)-sin^2x=cos^2x-cos^2y

    cos^2y-sin^2x=cos^2x-cos^2y

    2cos^2y=cos^2x+sin^2x=1

    Note that there is a difference between an equation and an identity.
    An identity is true for all x.
    An equation is true for certain x for which we solve.
    Last edited by Archie Meade; December 6th 2010 at 03:21 PM. Reason: typo
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  5. #5
    Junior Member Cthul's Avatar
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    I have another that I'm not sure. I want to know how to prove its identity.
    \sin{\theta}\sin{2\theta}\cos{2\theta}=\frac{1}{4}  (\cos{3\theta}-\cos{5\theta})
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  6. #6
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    Start by noting that \displaystyle \cos{(5\theta)} = \cos{(3\theta + 2\theta)} and use the angle sum identity to simplify.

    Then note that \displaystyle \cos{(3\theta)} = \cos{(2\theta + \theta)} and again use the angle sum identity to simplify.
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  7. #7
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    Quote Originally Posted by Cthul View Post
    I have another that I'm not sure. I want to know how to prove its identity.
    \sin{\theta}\sin{2\theta}\cos{2\theta}=\frac{1}{4}  (\cos{3\theta}-\cos{5\theta})
    Alternatively, get everything in terms of "Sine", using the following identities

    cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]

    sinAcosB=\frac{1}{2}\left[sin(A+B)+sin(A-B)\right]

    Then...

    sin\theta\;sin2\theta\;cos2\theta=sin\theta\left[\frac{1}{2}\left(sin4\theta+sin0\right)\right]=\frac{1}{2}sin\theta\;sin4\theta

    \frac{1}{4}\left[cos3\theta-cos5\theta\right]=\frac{1}{4}\left[-2sin\left(\frac{8\theta}{2}\right)sin\left(-\frac{2\theta}{2}\right)\right]=-\frac{1}{2}\left[sin4\theta\left(-sin\theta\right)\right]
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  8. #8
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    Hello, Cthul!

    \cos(x+y)\cos(x-y) \:=\:\cos^2\!x - \cos^2\!y

    I'm no sure how to prove this equality true.

    I'm not surprised . . . The statement is not true.

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  9. #9
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    Quote Originally Posted by Cthul View Post
    Why not just let x = y = 0, then it gives:

    \Rightarrow \cos\left(0\right)\cdot\cos\left(0\right) = \cos^2\left(0\right)-\cos^2\left(0\right)

    \Rightarrow 1 = 0. A counterexample!
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  10. #10
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    Though if anyone got suckered into thinking it was an equation to be solved, then

    y=\frac{\pi}{4};\;\;x=\frac{\pi}{4} is a solution among others...

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  11. #11
    Junior Member Cthul's Avatar
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    Thanks for your response.
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