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Math Help - Trig Functions

  1. #1
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    Trig Functions

    If sec 4x-sec2x = 2, find the value of cos^2x, 0<x<pi/2
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  2. #2
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    Hello, KingV15!

    We need two identities: . \begin{array}{ccc}<br />
\cos2\theta &=& 2\cos^2\!\theta - 1 \\ \cos4\theta &=& 8\cos^4\!\theta - 8\cos^2\!\theta + 1 \end{array}


    If \sec4x-\sec2x \:=\: 2, find the value of \cos^2\!x,\:\text{ for } 0 \leq x \leq \tfrac{\pi}{2}

    We have: . \sec4x - \sec2x \:=\:2 \quad \Rightarrow\quad \frac{1}{\cos4x} - \frac{1}{\cos2x} \:=\:2 \quad \Rightarrow \quad \cos2x - \cos4x \:=\:2\cos2x\cos4x

    . . (2\cos^2\!x-1) - (8\cos^4\!x - 8\cos^2\!x + 1) \;=\;2(2\cos^2\!x-1)(8\cos^4\!x - 8\cos^2\!x + 1)

    . . 32\cos^6\!x - 40\cos^4\!x + 10\cos^2\!x \:=\:0

    . . 2\cos^2\!x\left(16\cos^4\!x - 20\cos^2\!x + 5\right) \;=\;0


    Then:

    . . 2\cos^2\!x \:=\:0 \quad\Rightarrow\quad \boxed{\cos^2\!x \:=\:0}\quad [1]

    . . 16\cos^4\!x - 20\cos^2\!x + 5 \:=\:0 \quad\Rightarrow\quad\boxed{ \cos^2\!x \:=\:\frac{5\pm\sqrt{5}}{8}} \quad [2]


    From [1]: . x \:=\:\frac{\pi}{2}

    From [2]: . x \:=\:\frac{\pi}{10},\:\frac{3\pi}{10}

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