Trigonometry

por **stuart clark** » Seg Mai 30, 2011 06:26

If $sin\;\alpha+sin\;\beta+sin\;\gamma = cos\;\alpha+cos\;\beta+cos\;\gamma = 0$

Then prove that $cos\;\left(\alpha+\beta\right)+cos\;\left(\beta+\gamma\right)+cos\;\left(\gamma+\alpha\right)=0$

por **FilipeCaceres** » Seg Mai 30, 2011 12:01

hello, are $\alpha,\beta,\gamma$ angles of a triangle?

por **stuart clark** » Seg Mai 30, 2011 15:32

Yes $\alpha,\beta$ and $\gamma$ are angle of a $\triangle$

por **FilipeCaceres** » Seg Mai 30, 2011 17:44

stuart clark escreveu:Yes $\alpha,\beta$ and $\gamma$ are angle of a $\triangle$

Then $\alpha+\beta+\gamma=\pi$ so
$\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)= \cos(\pi-\gamma)+\cos(\pi-\beta)+\cos(\pi-\alpha)=-\cos(\alpha)-\cos(\beta)-\cos(\gamma)=0$

:y:

por **FilipeCaceres** » Seg Mai 30, 2011 22:58

$\left\{\begin{array}{c}cos\alpha+cos\beta+cos\gamma = 0\\\ sin\alpha+sin\beta+sin\gamma=0\end{array}\right\implies\left\{\begin{array}{c}cos\;\left(\alpha+\beta\right)+cos\;\left(\beta+\gamma\right)+cos\;\left(\gamma+\alpha\right)=0\\\ sin\;\left(\alpha+\beta\right)+sin\;\left(\beta+\gamma\right)+sin\;\left(\gamma+\alpha\right)=0\end{array}\right$

Proof. $\left\{\begin{array}{c}u=\cos\alpha+i\cdot\sin\alpha\\\ v=\cos\beta+i\cdot\sin\beta\\\ w=\cos\gamma+i\cdot\sin\gamma\end{array} \implies |u|=|v|=|w|=1$ and $\overline{u+v+w}= \overline u+\overline v+\overline w= \frac {1}{u}+\frac {1}{v}+\frac {1}{w}=\frac{uv+vw+wu}{uvw}\ (*)$

Therefore,
$\left\{\begin{array}{c}cos\alpha+cos\beta+cos\gamma = 0\\\sin\alpha+sin\beta+sin\gamma=0\end{array}\right$
$u+v+w=0\iff \overline{u+v+w}=0\stackrel{(*)}{\iff}vw+wu+uv=0$

Thus,
$\left\{\begin{array}{c}cos\;\left(\alpha+\beta\right)+cos\;\left(\beta+\gamma\right)+cos\;\left(\gamma+\alpha\right)=0\\\ sin\;\left(\alpha+\beta\right)+sin\;\left(\beta+\gamma\right)+sin\;\left(\gamma+\alpha\right)=0\end{array}\right$

credits: Nicula

por **stuart clark** » Ter Mai 31, 2011 14:21

Thanks FilipeCaceres.

why Things not strike in my mind earilier.

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