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Math Help - trigo problem no. 14

  1. #1
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    trigo problem no. 14

    p216 q14 re.ex10
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  2. #2
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    Hello, afeasfaerw23231233!@

    Consider the equation: . \sin^8\!x + \cos^8\!x \;=\;\frac{17}{16}\cos^2\!2x\;\hdots\;(*)

    (a) Let c \:=\:\cos2x

    Show that (*) can be written as: . \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^4 \:=\:\frac{17}{16}c^2

    We have: . \left(\sin^2\!x\right)^4 +\left(\cos^2\!x\right)^4 \;=\;\frac{17}{16}\cos^2\!2x

    Then: . \left(\frac{1-\cos2x}{2}\right)^4 + \left(\frac{1+\cos2x}{2}\right)^4 \;=\;\frac{17}{16}\cos^2\!2x

    Let c = \cos2x and we have: . \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^2 \;=\;\frac{17}{16}c^2



    (b) Hence find the general solution of (*).

    We have: . \left(\frac{1-c}{2}\right)^4 + \left(\frac{1+c}{2}\right)^2 \;=\;\frac{17}{16}c^2

    Expand: . \frac{1 - 4c + 6c^2 - 4c^3 + c^4}{16} + \frac{1+4x + 6c^2 - 4c^3 + c^4}{16}\;=\;\frac{17}{16}c^2

    Simplify: . \frac{2 + 12c^2 + 2c^4}{16}\;=\;\frac{17}{16}c^2\quad\Rightarrow\qua  d 2 + 12c^2 + 2c^4 \;=\;17c^2


    And we have the quadratic: . 2c^2 - 5c^2 + 2 \:=\:0

    . . which factors: . (c^2-2)(2c^2-1) \:=\:0

    . . and has roots: . \begin{array}{ccccccc}c^2-2&=&0 & \to & c &=& \pm\sqrt{2} \\ 2c^2-1 &=&0 & \to & c &=&\pm\frac{1}{\sqrt{2}} \end{array}


    \cos2x \:=\:\pm\sqrt{2} has no real roots.


    \cos2x \:=\:\pm\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{4} + \frac{\pi}{2}n

    . . Therefore: . x \;=\;\frac{\pi}{8} + \frac{\pi}{4}n

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  3. #3
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    thanks. i made a calculating mistake and that's why my answer was wrong
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