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How to prove that sin square x cos square x is identical

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    1) how to prove that sin square x cos square x is identical to 1/8 (1-
    cos 4x)?

    2) f(x) = x square +7x -6
    ----------------, show that when x is sufficiently small for
    (x-1)(x-2)(x+1)
    x^4 and higher powers to be neglected.


    2. Relevant equations
    1) cos (s + t) = cos s cos t – sin s sin t
    cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t
    sin 2t = 2 sin t cos t

    2) A Bx + C
    --------- + ------------
    factor quadractic


    3. The attempt at a solution
    1) from RHS:
    1- cos (2x +2x) 1-[ cos 2x cos 2x - sin 2x sin 2x]
    -------------- = ----------------------------------
    8 8
    1- cos^4 x + sin^4 x + 6 sin^2 x cos^2 x
    ----------------------------------------
    8
    * how to do the next step? is my solutions are correct so far?

    2) (x^2+7x-6)(x-1)^-1 ( x-2)^-1 (x+1)^-1
    expand (x-1)^-1 = 1+x+x^2+x^3
    (x-2)^-1 = 1/2 + x/4 + x^2/8 + x^3/16
    (1+x)^-1 = 1- x + x^2 - x^3
    multiply all : (x^2+7x-6)(1+x+x^2+x^3)( 1/2 + x/4 + x^2/8 + x^3/16)
    (1- x + x^2 - x^3)
    = 3x^2/2 - x^3/4 +2x -3
    * however , the answer I obtain is wrong. which part of my solution is
    wrong?
     
  2. jcsd
  3. Oct 20, 2011 #2
    Re: polynomials

    For the trig problem, try using the 2nd line of your relevant equations to get an expression for (cos x)^2 in terms of cos 2x. Then plug that expression into the Pythagorean identity to get an expression for (sin x)^2.

    Now, multiply those to get an expression for ((cos x)^2)*((sin x)^2) in terms of cos 2x. The next step should easy.


    I'm not sure what you're after in the rational function problem. Looks like if x is sufficiently small, you can ignore powers of x greater than 1 -- in the limit, the expression goes to -3.
     
  4. Oct 20, 2011 #3

    Mark44

    Staff: Mentor

    Re: polynomials

    You have not made it clear what you are trying to do, but I think you are trying to write a Maclaurin series representation for your rational function.

    I see a mistake in this line:
    It should be
    (x - 2)-1 = -1/(1 - x/2) = -(1 + x/2 + x2/4 + x3/8 + ...)
     
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