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$$ y = \frac {\tan 3x}{\tan x} $$

Now this is what I did :

$$ y = \frac {3\tan x - \tan^3 x}{\tan x(1 - 3\tan^2 x)} $$

$$ y = \frac {3 - \tan^2x}{1 - 3\tan^2x} $$

but I was not able to do further than this, I tried converting it into $sine$ & $cosine$ form but got stuck. I think it will involve using concepts of limit and continuity but I have not been taught that yet. Any help will be appreciated :)

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If you multiply your numerator and denominator by $\cos^2 x$ and use $\sin^2 x=1-\cos^2 x$, the expression becomes $$\frac {3-4\cos^2 x}{1-4\cos^2 x}$$

Now set $z=4\cos^2 x$ and note that $0\le z \le 4$ and the function becomes $$\frac {z-3}{z-1}=1-\frac 2{z-1}$$

You should be able to sketch that to see what is happening - the proof will then be easy.


If $z\gt 1$ then $\frac 2{z-1}\gt 0$ and is decreasing with increasing $z$. For this without calculus suppose $w\gt z\gt 1$ then $\frac 2{z-1}-\frac 2{w-1}=\frac {2(w-z)}{(z-1)(w-1)}\gt 0$. It can clearly be made as large as we like by choosing $z$ close to $1$, and for a decreasing function the minimum value will occur when $z$ is as large as possible and will therefore be $\frac 23$.

Since we are subtracting this term from $1$ the signs are reversed and for $z\gt 1$ the range of values is $(-\infty, \frac 13]$.

When $z\lt 1$ a similar analysis shows the range to be $[3,\infty)$


Note therefore that your original $y$ cannot take values in the open interval $(\frac 13, 3)$. This is because $0\le 4\cos^2 x\le 4$ - if the value of $z=4\cos^2 x$ were unconstrained you would be able to obtain every real value except $y=1$.

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    $\begingroup$ @Mike - what's wrong, so I can correct it? $\endgroup$ – Mark Bennet Sep 26 '14 at 16:09
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    $\begingroup$ On closer inspection, it appears the mistake is mine. I apologize. $\endgroup$ – Mike Sep 26 '14 at 16:14
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    $\begingroup$ @Mike No worries - thanks for looking carefully. $\endgroup$ – Mark Bennet Sep 26 '14 at 16:15
  • $\begingroup$ @Mark Bennet If by sketching you mean drawing a graph then mate Im sorry but I can't understand you. I haven't covered that part of my curriculum yet. But really thanks for helping! :) $\endgroup$ – Shubham Sep 26 '14 at 16:18
  • $\begingroup$ @MarkBennet is it possible to do this further without drawing a graph? And in the place of $y$ shouldn't there be any other variable since $f(x) = y$ in the original question? Please respond. $\endgroup$ – Shubham Sep 27 '14 at 12:25
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$\tan x$ can take any value in $(-\infty,\infty)$. So you find the range for $\frac{3-x}{1-3x}$ for $x\in[0,+\infty)$.

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  • $\begingroup$ Yes but how? This is the point where I got stuck. $\endgroup$ – Shubham Sep 26 '14 at 15:57
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    $\begingroup$ differentiate it with respect to $x$, to find the minima or maxima, and check the boundaries. $\endgroup$ – Abishanka Saha Sep 26 '14 at 16:03
  • $\begingroup$ I can do that but the place where I found this question relates to this so I was hoping if I could get an answer similar to this. If you know-how please post here :) Differentiation is fine too though :) Thanks! $\endgroup$ – Shubham Sep 26 '14 at 16:06
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    $\begingroup$ @AbishankaSaha Since it is possible for $x=\frac 13$ and the expression takes arbitrarily large positive and negative values, some care is required. $\endgroup$ – Mark Bennet Sep 26 '14 at 16:08
  • $\begingroup$ Dammit I am sorry for being ignorant. Your answer is right @Abishanka Saha the answer would be $ y \in (-\infty, 1/3 ) \cup (3, \infty) $ Thanks! $\endgroup$ – Shubham Sep 26 '14 at 16:15
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i have found that $\frac{\tan(3x)}{\tan(x)}=\frac{1+2\cos(2x)}{-1+2\cos(2x)}$ and now you can set $t=\cos(2x)$ with $|t|\le 1$. Now you can apply calculus.

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  • $\begingroup$ You mean apply the principle of maxima and minima? $\endgroup$ – Shubham Sep 27 '14 at 12:20

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