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-   -   Sin2X - sinXcosX = cosX (https://www.askmehelpdesk.com/showthread.php?t=527012)

  • Nov 18, 2010, 08:10 AM
    cptn13
    sin2X - sinXcosX = cosX
  • Nov 18, 2010, 09:38 AM
    ebaines

    Sorry - not true.

    sin2X = 2sinXcosX, so sin2X - sinXcosX = sinXcosX, which is not the same thing as cosX.
  • Nov 18, 2010, 07:04 PM
    harum

    sin(2x) = 2*sin(x)*cos(x), therefore:

    sin(2x) - cos(x)*sin(x) = sin(x)*cos(x)

    What you get is a simple equation:

    sin(x)*cos(x) = cos(x) or

    cos(x)*(sin(x)-1)=0

    This equation holds if:

    (a) cos(x)=0 => x = pi/2 + pi*n; n is any whole number, or
    (b) sin(x)=1 => x = pi/2 + 2*pi*k; k is any whole number

    The answer (b) is a subset of and contained in the answer (a), therefore just answer (a) would be sufficient. Hope this helps.
  • Nov 19, 2010, 04:15 AM
    Unknown008

    Yes, the answer would depend on what the question asks.

    If it's a proof the identify is false, and if it's an equation then the post of harum is the way to go.

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