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Thread: 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: .$\displaystyle \sin^8\!x + \cos^8\!x \;=\;\frac{17}{16}\cos^2\!2x\;\hdots\;(*)$

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

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

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

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

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



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

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

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

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


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

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

    . . and has roots: .$\displaystyle \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} $


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


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

    . . Therefore: .$\displaystyle 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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