# Found an odd relationship! Could someone help me to prove or debunk it?

I finished up in hospital which typically means that one has A LOT of spare time to kill and after using electronic devices so much that it makes you sorry I flinched into doodling and and light-headedly playing with a calculator.

It happend than that I bounced upon one odd thing:

For every right-angled triangle I sketched it seemed to be true that $$\frac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{\cos^2{\alpha}+\cos^2{\beta}+\cos^2{\gamma}}=2$$

Because my medicament dispenser is kinda slacking my brain I can't really concentrate on finding a way to prove this relation and a counterexample is also not in sight. So can someone please help me out? This open question is always on my mind and starts to become annonying :)

yours, Levix

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Now that it has been proved that the statement is true for every right-angled triangle. It will be more interesting to see if it is true for ANY triangle. –  Mick Aug 19 '14 at 3:52
@Mick: It isn't. For an equilateral triangle, e.g., the fraction yields 3. –  Dennis Aug 19 '14 at 4:48
@Dennis Thanks for pointing out a counterexample. It is just a thought that tries to explore the original to another extension, if there is one. –  Mick Aug 19 '14 at 14:55

L.H.S. $$\large\frac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{\cos^2{\alpha}+\cos^2{\beta}+\cos^2{\gamma}}$$

Since,

$\alpha+\beta+\gamma$=$180^\circ$

Since, the triangle is a right angled triangle (assuming it is at $\gamma$), that means

$\alpha+\beta=90^\circ$

Or,

$\beta=90^\circ-\alpha$

So,

$\sin \beta$=$\cos \alpha$

and

$\cos \beta$=$\sin \alpha$

The equation then changes to :

$$\large\frac{\sin^2 \alpha+\cos^2 \alpha+1}{\cos^2 \alpha+\sin^2 \alpha+ 0}$$

$$\large\frac{1+1}{1 + 0}=2=\text {R.H.S}$$

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WLOG $\displaystyle\alpha=90^\circ\implies\beta+\gamma=90^\circ\iff\gamma=90^\circ-\beta$

$$\sin^2\beta+\sin^2\gamma=\sin^2\beta+\sin^2(90^\circ-\beta)=\sin^2\beta+\cos^2\beta=?$$

$$\cos^2\beta+\cos^2\gamma=?$$

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Wish I know the mistake here –  lab bhattacharjee Aug 19 '14 at 5:04

In a right triangle, exactly one of $\alpha$, $\beta$, $\gamma$ must equal $\pi/2$, so let this be $\gamma = \pi/2$. Then $\sin \gamma = 1$ and $\cos \gamma = 0$. We must also have $\alpha + \beta = \pi/2$, hence $\beta = \pi/2 - \alpha$ and it immediately follows that $\sin \beta = \sin (\pi/2 - \alpha) = \cos \alpha$, and $\cos \beta = \sin \alpha$. The rest is simple substitution and the circular identity $\sin^2 \alpha + \cos^2 \alpha = 1$.

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$cos^2(\alpha)+cos^2(\beta)+cos^2(\gamma)=t =>$$sin^2(\alpha)+sin^2(\beta)+sin^2(\gamma)=1-cos^2(\alpha)+1-cos^2(\beta)+1-cos^2(\gamma)=3-t$
where $-3<=t<=3$
then $\frac{3-t}{t}=2 => 2t=3-t => t=1$

that means $cos^2(\frac\pi2)+cos^2(\beta)+cos^2(\frac\pi2-\beta)=1$

that means $0+cos^2(\beta)+sin^2(\beta)=1$

and it is always true(does not matter which angle is $\frac\pi2$ and $cos(\alpha)=sin(\frac\pi2-\alpha) is always true$ )

or directly going $\frac{1+sin^2(\alpha)+cos^2(\alpha)}{0+sin^2(\alpha)+cos^2(\alpha)}= \frac{1+1}{0+1}=2$

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