0
$\begingroup$

Without using a computer but using pen and paper only, can anyone please help me calculate, simplify / evaluate the following?

$$ \frac{\tan(180^{\circ}/7)}{\tan(360^{\circ}/7)} - \frac{\tan(360^{\circ}/7)}{\tan(540^{\circ}/7)} - \frac{\tan(540^{\circ}/7)}{\tan(180^{\circ}/7)} $$

It evaluates to $-9$. This result is given for reference.

$\endgroup$
5
  • $\begingroup$ I assume you mean $180^\circ$ etc.? $\endgroup$ – mrf Feb 16 '15 at 8:38
  • $\begingroup$ @mrf: BTW, is your logo bi-polar coordinates map? $\endgroup$ – Narasimham Feb 16 '15 at 9:24
  • $\begingroup$ @Narasimham No, it's an excerpt of a plot of some rational function (color coding the argument). $\endgroup$ – mrf Feb 16 '15 at 9:28
  • $\begingroup$ @mrf sir do you have any solution without using computer? $\endgroup$ – Deddy Feb 16 '15 at 14:34
  • $\begingroup$ It may have to do with like complex $ z^7 + 1 =0 $ $\endgroup$ – Narasimham Feb 16 '15 at 18:26
1
$\begingroup$

EDIT1:

Among all arguments the highest common factor is $ (\pi/7) $ and $ \tan ( m \pi/7) $ can be expanded in terms of $ T_t =\tan (\pi/7), $ as m is an integer.

With obvious multiple argument notation:

$$ F(t) = \dfrac{T_t}{T_{2 t}} -\dfrac{T_{2t}}{T_{3 t}} - \dfrac{T_{3t}}{T_{ t}} $$

$$ T_t = t $$

$$ T_{2t}= \dfrac{2t}{1 - 2 t^2} $$

$$ T_{3t}= \dfrac{3 t- t^3 }{1 - 3 t^2} $$

and simplify it.

But it does simplify to the constant $ ( -9) $ you gave !

EDIT2: Honest, I used computer just to cross-verify that hand work cannot simplify fractions.

EDIT3:

$ F( \tan7 t) = 0 $. But using only paper/pencil such conclusion may not be so obvious to come to.

$\endgroup$
2
  • $\begingroup$ So is there any solution without computer help? $\endgroup$ – Deddy Feb 16 '15 at 11:39
  • $\begingroup$ I suppose it is there, but cumbersome and tedious. The computer has necessity wise made many of us lazy. $\endgroup$ – Narasimham Feb 17 '15 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.