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Math Help - inequalities of trigo

  1. #1
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    inequalities of trigo

    inequalities of trigo
    p124 q12 re.ex5
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  2. #2
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    Hello, afeasfaerw23231233!


    We have to "talk" our way through these . .


    Find the values of \theta in the range  0 \leq \theta < 2\pi

    (a)\;\;2\sin^2\!\theta \:\leq \:1

    We have: . \sin^2\!\theta \;\leq \;\frac{1}{2}\quad\Rightarrow\quad |\sin\theta| \;\leq \;\frac{1}{\sqrt{2}} \quad\Rightarrow\quad -\frac{1}{\sqrt{2}}\;\leq \;\sin\theta \;\leq \;\frac{1}{\sqrt{2}}

    Clearly, we are dealing with multiples of 45: . \frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\:  \frac{7\pi}{4}

    When is \sin\theta between -\frac{1}{\sqrt{2}} \text{ and } +\!\frac{1}{\sqrt{2}} ?


    Examine the unit circle . . .

    Answer: . {\color{blue}\left[0,\:\frac{\pi}{4}\right],\;\left[\frac{3\pi}{4},\:\frac{5\pi}{4}\right], \;\left[\frac{7\pi}{4},\:2\pi\right)}




    (b)\;\sin2\theta + \cos\theta \:\leq \:0
    This one takes a bit more "talking" . . .


    We have: . 2\sin\theta\cos\theta + \cos\theta \:\leq \:0 \quad\Rightarrow\quad \cos\theta(2\sin\theta + 1) \:\leq\:0

    The product of two factors is negative if the factors have opposite parity.
    Hence, there are two cases to examine . . .


    [1]\;\begin{Bmatrix} \cos\theta & \leq & 0 \\ 2\sin\theta + 1 & \geq & 0 \end{Bmatrix}

    . . \cos\theta is negative in quadrants 2 and 3: . \left[\frac{\pi}{2},\:\frac{3\pi}{2}\right]

    . . \sin\theta \:\geq \:-\frac{1}{2} \quad\Rightarrow\quad \left[0,\:\frac{7\pi}{6}\right] \cup \left[\frac{11\pi}{6},\:2\pi\right)

    The intersection of the two sets is: . {\color{blue}\left[\frac{\pi}{2},\:\frac{7\pi}{6}\right]}


    [2]\;\begin{Bmatrix}\cos\theta & \geq & 0 \\ 2\sin\theta + 1 & \leq 0 \end{Bmatrix}

    . . \cos\theta is positive in quadrants 1 and 4: . \left[0,\:\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},\:2\pi\right)

    . . \sin\theta \:\leq\:-\frac{1}{2} \quad\Rightarrow\quad \left[\frac{7\pi}{6},\:\frac{11\pi}{6}\right]

    The intersection of the two sets is: . {\color{blue}\left[\frac{7\pi}{6},\:\frac{3\pi}{2}\right]}


    The solution is: . \left[\frac{\pi}{2},\:\frac{7\pi}{6}\right] \cup \left[\frac{7\pi}{6}\:\frac{3\pi}{2}\right] \;=\;{\bf{\color{blue}\left[\frac{\pi}{2},\:\frac{3\pi}{2}\right]}} . . . Quadrants 2 and 3

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