6
$\begingroup$

Calculate $x$, if $$\tan(x)=\tan9\tan69\tan33$$

(Using sexagesimal degrees) Since $\tan3x=\tan(60-x)\tan x \tan(60+x)$: \begin{align*} \tan27&=\tan69\tan9\tan51\\ \implies\tan27\tan39&=\tan69\tan9 \end{align*}

So the problem is equivalent to calculating $x$ in $$\tan(x)=\tan27\tan33\tan39$$ But thats all my progress so far. Interestingly enough, the answer is $x=15$. Is there some way tu constructively solve the equation? If not, a straightforward proof of $\tan(15)=\tan9\tan69\tan33$ would be nice too.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ Hint: $27=60-33$. $\endgroup$ – Lucian Jan 8 '15 at 21:24
  • $\begingroup$ another hint: $33 = 30 + 3$ $\endgroup$ – John Joy Jan 8 '15 at 21:36
  • $\begingroup$ @Lucian So $\tan99=\tan27\tan33\tan93\implies \tan81\tan3=\tan27\tan33$. I don't see what to do next. $\endgroup$ – chubakueno Jan 8 '15 at 22:26
  • 1
    $\begingroup$ upload.wikimedia.org/math/e/0/5/… This may help $\endgroup$ – Teoc Jan 11 '15 at 2:35
  • 1
    $\begingroup$ With tangent addition formulae you can compute the closed forms of the tangents. $\endgroup$ – Teoc Jan 11 '15 at 2:36
4
+50
$\begingroup$

$$ \tan(x)=\tan(9)\tan(69)\tan(33) $$ $$ \tan(x) = \frac{\tan(39)\tan(3)}{\tan(9)} $$ Thus, $$ \tan^2(x) = \tan(3)\tan(33)\tan(39)\tan(69) $$ $$ \tan^2(x) = \tan(3)\tan(3-36)\tan(3-72)\tan(3+36) $$ $$ \tan^2(x)\tan(75) = \tan(3-72)\tan(3-36)\tan(3)\tan(3+36)\tan(3+72) $$

I will now show that ,

$$ \tan(5x)= \tan(x-72)\tan(x-36)\tan(x)\tan(x+36)\tan(x+72) $$

Let $z=\cos(x)+i\sin(x)$, and $ \omega = \cos(36)+i\sin(36) $ Then, $$ \tan(x) = -\frac{i(z^2-1)}{2(z^2+1)} $$ Similarly we can get $\tan(x-36),\tan(x-72),\tan(x+36),\tan(x+72) $ and then multiply . We very easily get the above identity. So, we have $$ \tan^2(x)=\tan^2(15) \implies x=15 $$

Improve this answer
$\endgroup$
  • $\begingroup$ Where did you use $\omega$? $\endgroup$ – user84413 Jan 18 '15 at 2:10
  • $\begingroup$ to write, tan(x-36),tan(x+36)... in terms of complex number (for example for tan(x-36), replace z by z/w $\endgroup$ – Shivang jindal Jan 18 '15 at 15:59

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.