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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 :)

Thomas Andrews
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Shubham
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3 Answers3

7

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 $y=1$.

Mark Bennet
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    @Mike - what's wrong, so I can correct it? – Mark Bennet Sep 26 '14 at 16:09
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    On closer inspection, it appears the mistake is mine. I apologize. – Mike Sep 26 '14 at 16:14
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    @Mike No worries - thanks for looking carefully. – Mark Bennet Sep 26 '14 at 16:15
  • @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! :) – Shubham Sep 26 '14 at 16:18
  • @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. – Shubham Sep 27 '14 at 12:25
  • @Shubham I'll change $y$ to $z$ to avoid the confusion. – Mark Bennet Sep 27 '14 at 16:14
  • Thank you for the reply. "It can clearly be made as large as we like by choosing z close to 1" By close to 1 we mean approaching 1 from right hand side like for example 1.01 which will make $ \frac 2{z-1} $ equal to 200 right? "z is as large as possible and will therefore be $ \frac {2}{3} $ ". How did we get z = 2/3 ? Again really thanks for replying :) – Shubham Sep 27 '14 at 17:17
  • @Shubham - OK on the high values. The least value comes when $z$ is at the top of the range of possible values - when $z=4$. Just plug it in. – Mark Bennet Sep 27 '14 at 17:24
  • Oh got it since $ 0 <= z <= 4 $ Now for $ z > 1 $ $ y = \frac 13 $ and

    for $ z < 1 $ $ y = 3 $ correct? But is there any explanation why did we leave $ (\frac 13, 3] $ ? I mean I get it that $y$ can't be equal to this region of values but how do we know for certain that none of the values in this region will ever be able to satisfy the original equation i.e. $ y = \frac {tan3x}{tanx} $ ?

     – Shubham Sep 27 '14 at 17:38
  • And sorry for asking stupid questions but this is really bugging me. :( – Shubham Sep 27 '14 at 17:40
  • @Shubham I have added another little paragraph to the answer. I hope it helps. – Mark Bennet Sep 27 '14 at 17:54
  • @MarkBennet Hey you mentioned above that because $ 0 \le z \le 4 $
    $ \therefore y \notin (1/3,3]$ We had the equation $ y = 1 - \frac 2{z-1}$ Now if we rewrite this in terms of $ z = \frac 2{1-y} + 1 $ and then put this equation in place of $z$ in $ 0 \le z \le 4 $ we get $ \frac13 \le y \le 3 $ Where am I wrong here? :(     – Shubham Sep 27 '14 at 18:46
5

$\tan x$ can take any value in $(-\infty,\infty)$. So you find the range for $\frac{3-x}{1-3x}$ for $x\in[0,+\infty)$.

QED
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  • Yes but how? This is the point where I got stuck. – Shubham Sep 26 '14 at 15:57
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    differentiate it with respect to $x$, to find the minima or maxima, and check the boundaries. – QED Sep 26 '14 at 16:03
  • 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! – Shubham Sep 26 '14 at 16:06
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    @AbishankaSaha Since it is possible for $x=\frac 13$ and the expression takes arbitrarily large positive and negative values, some care is required. – Mark Bennet Sep 26 '14 at 16:08
  • 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! – 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.

Dr. Sonnhard Graubner
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