Calculus

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If √2cos(x)-1 = (1+√3)/2 and √2cos(x)+1 = (1-√3)/2, find the value of cos4x.

  • Calculus -

    let's add the two given equations
    √2cos(x)-1 = (1+√3)/2
    √2cos(x)+1 = (1-√3)/2

    2√2cosx = 1
    cosx = 1/(2√2) or 1/√8
    Then by Pythagoras , sinx = √7/√8

    using some identities
    cos 4x = cos^2 2x - sin^2 2x
    = (cos 2x)(cos 2x) - (sin 2x)(sin 2x)
    = (cos^2 x - sin^2 x)^2 - (2sinxcosx)^2
    = (1/8 - 7/8)^2 - (2√7/8)^2
    = 36/64 - 28/64
    = 8/64
    = 1/8

  • Calculus -

    I don't "get" it.

    If √2cos(x)-1 = (1+√3)/2,
    you can solve for x directly, and then calculate 4x, and its cosine. There may be more than one answer. The second equation may have a different set of answers with some agreeing with the first set.

    You need to clarify what is "under" the square root sign, using parentheses. Is it [sqrt2*(cosx)] -1 or sqrt(2cosx-1) ?

  • ???? (something wrong) HELP Calculus -

    Just looking over this question again
    especially the first equation

    √2cos(x)-1 = (1+√3)/2
    √2cos(x) = (1+√3)/2 + 1
    √2cosx = (3 + √3)/2
    cosx = (3+√3)/(2√2) = 1.673

    but the cosine of any angle is between -1 and +1,
    so it would be undefined.

    check the typing of the question please.

  • Calculus -

    I'm pretty sure I typed the question correctly, the answer at the back says 1/2.

  • Calculus -

    Retyped with brackets, it would be
    If √2[cosx)-1 = (1+√3)/2 and √2(cosx)+1 = (1-√3)/2, find the value of cos4x.

  • Transcription ambiguity -

    In books where equations are typeset, the square-root sign includes a horizontal bar to indicate the extent of the square-root. When the equation is transcribed to a single line, the horizontal line should be replaced with parentheses.
    For example:
    √2cos(x)-1 should be written as √(2)cos(x)-1 if the horizontal line is short, and √(2cos(x))-1 if the horizontal line extends over the cos(x), and √(2cos(x)-1) if the square-root includes the "-1".

    As they are, there is ambiguity in the given equations.

  • Calculus -

    find the positive roots of the equation
    2x^3-5x^2+6=0
    and evaluate the function y=500x^6 at each root. round you answers to the nearest integer, but only in the final step.

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