If $\frac{\tan 8°}{1-3\tan^{2}8°}+\frac{3\tan 24°}{1-3\tan^{2}24°}+\frac{9\tan 72°}{1-3\tan^{2}72°}+\frac{27\tan 216°}{1-3\tan^{2}216°}=x\tan 108°+y\tan 8°$, find x and y. I am unable to simplify the first and third terms. I am getting power 4 expressions. Thanks.
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What did you do to get four degree terms? – Hiten May 13 '15 at 17:49
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Combined the first and third terms – May 13 '15 at 17:50
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Try expanding $$\tan 3x = \frac{\tan x}{1 - 3\tan^2 x}(3 - \tan^2 x)$$ – uranix May 13 '15 at 17:51
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1Yeah, I did that. But it doesn't help much. – May 13 '15 at 17:58
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Isn't this a line? So there are infinitely many solutions $(x,y)$. – ET93 May 13 '15 at 18:16
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@uranix could you please tell me what to do after writing all the terms in the form of tan3x – May 14 '15 at 05:53
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@You-know-me Sorry, I don't know. I just saw similar expression and thought that was a right track – uranix May 14 '15 at 06:49
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@ET93 I suppose that The problem asks about simple (rational or so) expressions for $x$, $y$. I.e. simplify to the givenform and find $x, y$. – uranix May 14 '15 at 07:53
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Yeah, I know.. But the question is simplification.. – May 14 '15 at 07:54
1 Answers
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HINT:
$$\tan(3\cdot8^\circ)=\dfrac{3\tan 8^\circ-\tan^38^\circ}{1-3\tan^28^\circ}$$
Now, $$\frac{\tan 8^\circ}{1-3\tan^28^\circ}-y\tan8^\circ=\dfrac{(1-y)\tan 8^\circ-(-3y)\tan^38^\circ}{1-3\tan^28^\circ}$$ which will be a multiple of $\tan(3\cdot8^\circ)$ if $$\dfrac{1-y}{-3y}=\dfrac31\iff y=-\dfrac18$$
$$\implies\frac{\tan A}{1-3\tan^2A}-\left(-\dfrac18\right)\tan A=\dfrac38\tan3A$$
and I should leave it here.
lab bhattacharjee
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Who said anything of a multiple? I mean, why should it be a multiple of tan24°? – May 14 '15 at 06:42
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@You-know-me, Observe the angles $8,24,72,216:$ Each is three times of the previous – lab bhattacharjee May 14 '15 at 08:04
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@You-know-me, Put $A=8^\circ,24^\circ,72^\circ,216^\circ$ one by one – lab bhattacharjee May 14 '15 at 16:17