Prove: $\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma =4\mathrm{cos}\frac{\alpha}{2}\mathrm{cos}\frac{\beta}{2}\mathrm{cos}\frac{\gamma}{2}$ when $\alpha +\beta +\gamma =\pi $

Ximena Skinner
2022-07-08
Answered

Prove: $\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma =4\mathrm{cos}\frac{\alpha}{2}\mathrm{cos}\frac{\beta}{2}\mathrm{cos}\frac{\gamma}{2}$ when $\alpha +\beta +\gamma =\pi $

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Prove that $\frac{\mathrm{cos}3A+\mathrm{cos}3B}{2\mathrm{cos}(A-B)-1}=(\mathrm{cos}A+\mathrm{cos}B)\mathrm{cos}(A+B)-(\mathrm{sin}A+\mathrm{sin}B)\mathrm{sin}(A+B)$

I used$\mathrm{cos}3A=4{\mathrm{cos}}^{3}A-3\mathrm{cos}A$ , but it is getting more and more complicated.

I used

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Evaluating $(\mathrm{cot}\frac{\pi}{18}-3\mathrm{cot}\frac{\pi}{6})\cdot (\mathrm{csc}\frac{\pi}{9}+2\mathrm{cot}\frac{\pi}{9})$

Try:

$\mathrm{cot}\frac{\pi}{18}\mathrm{csc}\frac{\pi}{9}-3\sqrt{3}\mathrm{csc}\frac{\pi}{9}+2\mathrm{cot}\frac{\pi}{18}\mathrm{cot}\frac{\pi}{9}-6\sqrt{3}\mathrm{cot}\frac{\pi}{9}$

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In any triangle is $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C=\frac{3\sqrt{3}}{2}$ always
Well, I came with an interesting proof. But I just want to verify it
Applied at function $y=\mathrm{sin}x$
We have, $\mathrm{sin}(\frac{A+B+C}{2})\ge \frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}$
From here we will get
$\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\le \frac{3\sqrt{3}}{2}$
Also by A.M. $\ge $ G.M. in an acute angled triangle
$\frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}\ge \sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}$
$\Rightarrow \mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3(\sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C})$
$\Rightarrow \mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3(\frac{\sqrt{3}}{2})=\frac{3\sqrt{3}}{2}>2$
and from this I get
$\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge \frac{3\sqrt{3}}{2}$

asked 2022-01-14

Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Using$x=2\mathrm{cos}\left(\theta \right)$ we get $2\mathrm{cos}\left(\theta \right)=\sqrt{2+\sqrt{2-\sqrt{2+2\mathrm{cos}\theta}}}$

$2\mathrm{cos}\theta =\sqrt{2+\sqrt{2-2\mathrm{cos}\frac{\theta}{2}}}$

$2\mathrm{cos}\theta =\sqrt{2+2\mathrm{sin}\frac{\theta}{4}}$

$4{\mathrm{cos}}^{2}\theta =2+2\mathrm{sin}\frac{\theta}{4}$

After this step I am not able to solve it

Using

After this step I am not able to solve it