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Math Help - Intercepts for two trig equations

  1. #1
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    Intercepts for two trig equations

    So I need a method to find the intercepts for two trig equations: y = sinx - cosx and y = 2cos(2x). I know that substitution is supposedly the easiest, but I can't seem to get anything usable. The segment I'm using is below. The actual intercepts are -1.1468, 1.3229 for the left and 0.7854, 0 for the right, but I need an algebraic/trigonometric/whatever way to do it.
    Cheers.
    Last edited by zapparage; July 30th 2008 at 01:45 PM.
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  2. #2
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    Quote Originally Posted by zapparage View Post
    So I need a method to find the intercepts for two trig equations: y = sinx - cosx and y = 2cos(2x). I know that substitution is supposedly the easiest, but I can't seem to get anything usable. The segment I'm using is below. The actual intercepts are -1.1468, 1.3229 for the left and 0.7854, 0 for the right, but I need an algebraic/trigonometric/whatever way to do it.
    Cheers.
    Solve sin x - cos x = 2 cos(2x):

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

    => sin x - cos x = 2 (cos x - sin x)(cos x + sin x)

    => sin x - cos x = 0 or 1 = -2 (cos x + sin x)

    => tan x = 1 or cos x + sin x = -1/2.

    To solve the second equation, note that cos x + sin x = sqrt{2} cos (x + pi/4).
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  3. #3
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    Thanks a lot for the help, but I'm completely lost.

    How'd you get:

    Code:
     sin x - cos x = 2 (cos^2 x - sin^2 x)
    and the 0 and 1 in:
    Code:
    => sin x - cos x = 0 or 1 = 2 (cos x + sin x)
    and also get to this:
    Code:
    => tan x = 1 or cos x + sin x = 1/2.
    Also, how do I get the exact intercept numbers once this is solved?

    You've probably guessed that I have no idea about the 'sqrt' - we haven't even gotten that far.
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  4. #4
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    Quote Originally Posted by zapparage View Post
    Thanks a lot for the help, but I'm completely lost.

    How'd you get:

    Code:
     sin x - cos x = 2 (cos^2 x - sin^2 x)
    Mr F asks: Have you been taught the double angle formulae?

    and the 0 and 1 in:
    Code:
    => sin x - cos x = 0 or 1 = 2 (cos x + sin x)
    Mr F says: Factorise and use the null factor law:

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

    => 2 (sin x - cos x)(cos x + sin x) + (sin x - cos x) = 0

    Common factor is sin x - cos x:

    => (sin x - cos x)(2(cos x + sin x) + 1) = 0

    sin x - cos x = 0 or 2(cos x + sin x) + 1 = 0

    etc.

    and also get to this:
    Code:
    => tan x = 1 or cos x + sin x = 1/2.
    Mr F says: sin x - cos x = 0 => sin x = cos x => sin x/cos x = 1.

    2(cos x + sin x) + 1 = 0 => 2(cos x + sin x) = -1 => cos x + sin x = -1/2.

    Also, how do I get the exact intercept numbers once this is solved?

    Mr F says: You solve the two trig equations to get the x-coordinates. Then you substitute the x-coordinates into either one of y = sinx - cosx and y = 2cos(2x) to get the y-coordinates.

    You've probably guessed that I have no idea about the 'sqrt' - we haven't even gotten that far.
    If you're required to get exact solutions for the x-intercepts, then it's expected that you:

    1. Are familiar with the double angle formula for cos(2x).
    2. Can re-arrange trig equations such as sin x - cos x = 0.
    3. Can solve simple trig equations.
    4. Can express cos x + sin x in the form I gave.
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