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Thread: range question

  1. #1
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    range question

    range of
    sin x/(cosx)^2
    (i)without calculus
    (ii)with calculus
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  2. #2
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    Hello, ayushdadhwal!

    $\displaystyle \text{Range of: }\,\dfrac{\sin x}{\cos^2x}$

    $\displaystyle \text{(i) without calculus}$

    I substituted values: .$\displaystyle x \:=\:0,\,\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\p i}{4},\,\pi,\,\frac{5\pi}{4},\,\hdots$
    . . and found the following behavior:


    . . $\displaystyle \begin{array}{|c|c|c|} \hline
    \text{Quadrant} & \sin x,\,\cos^2\!x & \dfrac{\sin x}{\cos^2x} \\ \hline \hline
    1 & \sin x\!: 0 \to 1 & \\
    & \cos^2\!x\!: 1 \to 0 & 0 \to +\infty\\ \hline

    2 & \sin x\!: 1 \to 0 & \\
    &\cos^2\!x\!: 0 \to 1 & +\infty \to 0 \\ \hline

    3 & \sin x\!: 0 \to \text{-}1 & \\
    & \cos^2\!x\!: 1 \to 0 & 0 \to \text{-}\infty \\ \hline

    4 & \sin x\!: \text{-}1 \to 0 & \\
    & \cos^2\!x\!: 0 \to 1 & \text{-}\infty \to 0 \\ \hline
    \end{array}$


    $\displaystyle \text{Therefore, the range of }\,\dfrac{\sin x}{\cos^2\!x}\,\text{ is: }\,(\text{-}\infty,\,+\infty)$




    $\displaystyle \text}(ii) with calculus}$

    $\displaystyle \text{We note that: }\:\dfrac{\sin x}{\cos^2\!x} \:=\:\sec x\tan x$


    $\displaystyle \text{This is the }derivative\text{ of: }\:f(x) \,=\,\sec x$

    $\displaystyle \text{The graph of }\sec x\text{ looks like this:}\begin{array}{c} \cup\;\; \cup \\ \quad\;\cap\;\; \cap \end{array}$

    $\displaystyle \text{And its }slope\text{ ranges from }\text{-}\infty \text{ to }+\infty.$

    $\displaystyle \text{Therefore, the range of }\dfrac{\sin x}{\cos^2\!x}\text{ is: }\,(\text{-}\infty,\,+\infty)$

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