range question

**ayushdadhwal** · March 8th 2011, 09:49 AM

range of
sin x/(cosx)^2
(i)without calculus
(ii)with calculus

**Soroban** · March 8th 2011, 12:49 PM

Hello, ayushdadhwal!

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

$\text{(i) without calculus}$

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

. . $\begin{array}{|c|c|c|} \hline \text{Quadrant} & \sin x,\,\cos^2\!x & \dfrac{\sin x}{\cos^2x} \\ \hline \hline 1 & \sin x\!: 0 \to 1 & \\ & \cos^2\!x\!: 1 \to 0 & 0 \to +\infty\\ \hline 2 & \sin x\!: 1 \to 0 & \\ &\cos^2\!x\!: 0 \to 1 & +\infty \to 0 \\ \hline 3 & \sin x\!: 0 \to \text{-}1 & \\ & \cos^2\!x\!: 1 \to 0 & 0 \to \text{-}\infty \\ \hline 4 & \sin x\!: \text{-}1 \to 0 & \\ & \cos^2\!x\!: 0 \to 1 & \text{-}\infty \to 0 \\ \hline \end{array}$

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

$\text}(ii) with calculus}$

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

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

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

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

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

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