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Let $v \in \mathbb{R}^3$ be a vector and let $\alpha$, $\beta$, and $\gamma$ be the angles between $v$ and coordinate axes, where $0 \leq \alpha, \beta, \gamma \leq\pi$.

Prove that

$$ \mathrm{abs}\left(\;\left| \begin{matrix} \cos\alpha & \cos\beta & \sin\gamma\\ \sin\alpha & \cos\beta & \cos\gamma\\ \cos\alpha & \sin\beta & \cos\gamma \end{matrix} \right|\;\right) \leq 2\sqrt{2}\sin\alpha\sin\beta\sin\gamma$$ where $\mathrm{abs}(a) = |a|$ is the absolute value of $a$.

My opinion is to consider 2 cases, since there is at most one of the three angles whose sine is $0$.

(1)If there is one angle whose sine is $0$, then it's trivial that LHS=RHS=$0$.

(2)If no such angle, then just divide $\sin\alpha$,$\sin\beta$, and $\sin\gamma$ on both sides, the the problem is reduced to

$$ \mathrm{abs}\left(\left| \begin{matrix} \cot\alpha & \cot\beta & 1\\ 1 & \cot\beta & \cot\gamma\\ \cot\alpha & 1 & \cot\gamma \end{matrix} \right|\right) \leq 2\sqrt{2}.$$

Then the LHS is just the volume of the parallelpiped formed by the three column vectors.

But I don't know how to do next? Any suggestion? Thanks very much!

share|cite|improve this question
See here for one approach. – David Mitra Mar 14 '12 at 16:48
yes,'s much easier than what i thought. – benben112358 Mar 14 '12 at 17:13

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