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If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$

Then prove that:

$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.

I have no idea how to go about this.

Please help.

user2369284
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  • Have you tried using the identity $\displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \tan (\alpha + \beta)$? Not sure if it'll work, but hey, worth a shot. – 2012ssohn Jan 01 '14 at 19:00
  • @2012ssohn I tried it but that makes the problem very lengthy and almost impossible to solve. – user2369284 Jan 01 '14 at 19:03
  • One method $$\frac{\sin \sin }{\cos \cos }=\frac{\cos \cdot\cos\frac\pi4}{\sin \cdot\sin\frac\pi4}$$ then apply http://en.wikipedia.org/wiki/Componendo_and_dividendo – lab bhattacharjee Jan 01 '14 at 19:03
  • I'm trying Let $A=\frac{-\alpha+\beta+\gamma}4$ etc,

    $\displaystyle \implies A+B+C=\frac{\alpha+\beta+\gamma}4=\frac\pi4$

    and $\displaystyle A+B=\frac\gamma2\iff\gamma=2(A+B)=2\left(\frac\pi4-C\right)=\frac\pi2-2C $ etc.

    $\cos\gamma=\cos(\frac\pi2-2C)=\sin2C$ then use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html. But things are shaping nicely, either I'm committing mistakes or

     – lab bhattacharjee Jan 01 '14 at 19:05
  • @labbhattacharjee Should I evaluate $1 + \sin2A + \sin2B + \sin2C$? – user2369284 Jan 01 '14 at 19:17
  • @user2369284, can you please follow the hint in my post – lab bhattacharjee Jan 01 '14 at 19:18
  • As an extension to @2012ssohn 's comment, maybe $\tan(\alpha + \beta + \gamma)=\dfrac{\tan \alpha + \tan \beta + \tan \gamma}{1-\tan \alpha \tan \beta - \tan \alpha \tan \gamma -\tan \beta \tan \gamma} $ will work. – K. Rmth Jan 01 '14 at 19:36

3 Answers3

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As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+\pi-\alpha}4=\frac\pi4-\frac\alpha2$

$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(\frac\pi4-\frac\alpha2\right)=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}=\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}$

$\displaystyle\implies \tan^2\left(\frac{-\alpha+\beta+\gamma}4\right)=\left(\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}\right)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$

$\displaystyle\implies \prod\left(\frac{1-\sin\alpha}{1+\sin\alpha}\right)=1$

$\displaystyle\implies\sum\sin\alpha+\prod\sin\alpha=0 $

Now set $\displaystyle\sin\alpha=2\sin\frac{\alpha}2\cos\frac{\alpha}2$ in $\displaystyle\prod\sin\alpha$

and use this for $\sum\sin\alpha$

Cancelling out $\displaystyle\prod\cos\frac{\alpha}2,$ (assuming $\displaystyle\prod\cos\frac{\alpha}2\ne0$ )

we get $\displaystyle4\prod \sin\frac{\alpha}2=-2$

Finally, using this, $\displaystyle\sum \cos A=1+4\prod\sin\frac{\alpha}2$

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lab bhattacharjee
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2

A little Generalization :

Let $\alpha+\beta+\gamma=4C$

and $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right) = \cot C$

As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+4C-\alpha}4=C-\frac\alpha2,$

$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(C-\frac\alpha2\right)=\frac{\sin\left(C-\frac\alpha2\right)}{\cos\left(C-\frac\alpha2\right)}$

So, the problem reduces to $\displaystyle\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)\sin\left(C-\frac\gamma2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)\cos\left(C-\frac\gamma2\right)}=\cot C$

$\displaystyle\implies\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)} =\frac{\cos\left(C-\frac\gamma2\right)\cos C}{\sin\left(C-\frac\gamma2\right)\sin C}$

Applying $\displaystyle2\sin A\sin B,2\cos A\cos B$ formula,

$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{4C-\beta-\alpha}2}{\cos\frac{\beta-\alpha}2+\cos\frac{4C-\beta-\alpha}2} =\frac{\cos\left(-\frac\gamma2\right)+\cos\frac{4C-\gamma}2}{\cos\left(-\frac\gamma2\right)-\cos\frac{4C-\gamma}2}$

Now as $\displaystyle\alpha+\beta+\gamma=4C,$ this becomes

$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{\gamma}2}{\cos\frac{\beta-\alpha}2+\cos\frac{\gamma}2} =\frac{\cos\frac\gamma2+\cos\frac{\alpha+\beta}2}{\cos\frac\gamma2-\cos\frac{\alpha+\beta}2}$

Applying Componendo and dividendo,

$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2}{-\cos\frac{\gamma}2} =\frac{\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha+\beta}2}$

$\displaystyle\implies\cos\frac{\beta-\alpha}2\cos\frac{\beta+\alpha}2=-\cos^2\frac{\gamma}2 $

Applying $2\cos A\cos B,\cos2x=2\cos^2x-1,$ $\displaystyle\frac{\cos\alpha+\cos\beta}2=-\frac{1+\cos\gamma}2$

Here $\displaystyle C=\frac\pi4\implies \alpha+\beta+\gamma=4C=\pi$

If $\displaystyle C=-\frac\pi4\implies \alpha+\beta+\gamma=4C=-\pi\equiv\pi\pmod{2\pi}$

So, $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right)=-1$ and $\alpha+\beta+\gamma=-\pi$ or $\pi$ will satisfy the required identity

lab bhattacharjee
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1

Let's introduce the shorthand notations $$ \begin{align} a &= \frac{-\alpha+\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\alpha}{2},\\ b &= \frac{\alpha-\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\beta}{2},\\ c &= \frac{\alpha+\beta-\gamma}{4} = \frac{\pi}{4} - \frac{\gamma}{2}, \end{align} $$ then $$ a+b = \frac{\gamma}{2},\quad a+c = \frac{\beta}{2},\quad b+c = \frac{\alpha}{2}, $$ $$ a+b+c = \frac{\pi}{4}, $$ $$ \sin a\sin b \sin c = \cos a\cos b\cos c.\tag{1} $$ Now $$ \cos(a+b+c) = \cos(a+b)\cos(c) - \sin(a+b)\sin c= \frac{\sqrt{2}}{2}, $$ so that $$ \cos(a+b)\cos(c) - \sin a\sin c\cos b - \sin b\sin c\cos a = \frac{\sqrt{2}}{2} $$ add $(2\cos a\cos b\cos c)$ to both sides, and we get $$ \cos(a+b)\cos(c) + \cos(a+c)\cos b+ \cos(b+c)\cos(a)= \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c, $$ which reduces to $$ \begin{multline} \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\ \cos\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c. \tag{2} \end{multline} $$ On the other hand, $$ \sin(a+b+c) = \sin(a+b)\cos c + \cos(a+b)\sin c = \frac{\sqrt{2}}{2}, $$ so that, $$ \sin(a+b)\cos c + \cos a\cos b\sin c = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{3} $$ Similarly, if we take $\sin(a+b+c) = \sin((a+c)+b)$, we get $$ \sin(a+c)\cos b + \cos a\cos c\sin b = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{4} $$ If we add eqs. $(3)$ and $(4)$, we get $$ \sin(a+b)\cos c + \sin(a+c)\cos b + \sin(b+c)\cos a = \sqrt{2} + 2\sin a\sin b\sin c, $$ which reduces to (using also $(1)$) $$ \begin{multline} \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\ \sin\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \sqrt{2} + 2\cos a\cos b\cos c. \tag{5} \end{multline} $$ Finally, we subtract eq $(5)$ from eq $(2)$: $$ \begin{multline} \left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right) \cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \ldots = - \frac{\sqrt{2}}{2}, \end{multline} $$ which is $$ \begin{multline} \frac{\sqrt{2}}{2}\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)\left(\cos\left(\frac{\alpha}{2}\right)+\sin\left(\frac{\alpha}{2}\right)\right) + \ldots + \frac{\sqrt{2}}{2} =0, \end{multline} $$ so that finally $$ \cos\alpha +\cos\beta+\cos\gamma + 1 = 0. $$

Pulsar
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