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Math Help - Double Angle Identity

  1. #1
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    Double Angle Identity

    Stuck on this, don't know where to start

    2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}
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  2. #2
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    Quote Originally Posted by Beard View Post
    Stuck on this, don't know where to start

    2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}
    Sorry, I thought you were trying to prove an identity!
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  3. #3
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    Quote Originally Posted by Beard View Post
    Stuck on this, don't know where to start

    2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}
    I'll start you off:

    2\cos^{2}x = 2\sin(x)\cos(x) + 1 \implies 2\cos^{2}x -1 = 2\sin(x)\cos(x) \implies \cos{2x} = \sin{2x}

    When does the cosine of an angle equals the sine of an angle?
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  4. #4
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    Quote Originally Posted by Beard View Post
    Stuck on this, don't know where to start

    2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}
    First of all this is not an identity
    Put x= 0
    In case you need to solve the equation


    Put 1 = sin^2(x) + cos^2(x)

    EDIT: Far too late
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  5. #5
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    at 45
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  6. #6
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    I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.
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  7. #7
    Like a stone-audioslave ADARSH's Avatar
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    Read the initial lines
    This is an equation which has solutions and not identity
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  8. #8
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    Quote Originally Posted by Beard View Post
    at 45
    Sine and cosine are equal for 45 degrees and at 180+ 45= 225 degrees.

    That, of course, would be 2x. Now what values does x have?
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  9. #9
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    Quote Originally Posted by Chop Suey View Post
    I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.

    Sorry about this
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