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    Solve: 2cos(2x)=1+cos(x)
    When 0<x<360
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    (Original post by JoshDes)
    Solve: 2cos(2x)=1+cos(x)
    When 0<x<360
     cos2x = 2 cos^2x-1

    Use this and solve quadratic
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    (Original post by DylanJ42)
     cos2x = 2 cos^2x-1

    Use this and solve quadratic
    Protip: whenever you're writing trigonometrical expressions using \LaTeX, put a \ in front of the trig function so that it typesets it correctly instead of writing down the three variables c, o and s mashed together. That is: \cos x v/s cos x. :yep:
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    (Original post by Zacken)
    Protip: whenever you're writing trigonometrical expressions using \LaTeX, put a \ in front of the trig function so that it typesets it correctly instead of writing down the three variables c, o and s mashed together. That is: \cos x v/s cos x. :yep:
    Noted, thank you (soon I will be a latex master) :curious:
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    (Original post by DylanJ42)
    Noted, thank you (soon I will be a latex master) :curious:
    You'll be teaching the young 'uns in no time. :yep:
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    (Original post by DylanJ42)
     cos2x = 2 cos^2x-1

    Use this and solve quadratic
    How did you get to this?
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    (Original post by Mitul106)
    How did you get to this?
     \cos^2x - \sin^2x = \cos2x ***

     \cos^2x - (1-\cos^2x) = \cos2x

     2\cos^2x - 1 = \cos2x

    *** This line can be obtained by expressing \cos(2x) as \cos(x+x) and expanding that

    However I could not tell you where \cos(x+y) = cosx.cosy - sinx.siny comes from, so that's as far back as I can go

    Zacken will probably know
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    (Original post by DylanJ42)
    Zacken will probably know
    There's an ugly geometric proof and a nicer one, but there's no way I'm writing it up - the person can google it. :lol:
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    e^i(x+y)=cos(x+y)+isin(x+y)

    (cosx+isinx)(cosy+isiny)=e^ix * e^iy

    cos(x+y)+isin(x+y)=(cosx+isinx)( cosy+isiny)

    Equate imaginary and real parts.
 
 
 
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