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Question-14592




Question Number 14592 by Tinkutara last updated on 02/Jun/17
Commented by mrW1 last updated on 02/Jun/17
∣cos x∣=(q/p)≤1  ∣cos y∣=(u/v)≤1  ∣cos z∣=(1/((qu)/(pv)))=((pv)/(qu))≥1≤1  that means the absolute value of  cos x, cos y and cos z must be 1.  there are 4 possibilities:  (x,y,z)=(0,0,0),(0,π,π),(π,0,π),(π,π,0)
$$\mid\mathrm{cos}\:{x}\mid=\frac{{q}}{{p}}\leqslant\mathrm{1} \\ $$$$\mid\mathrm{cos}\:{y}\mid=\frac{{u}}{{v}}\leqslant\mathrm{1} \\ $$$$\mid\mathrm{cos}\:{z}\mid=\frac{\mathrm{1}}{\frac{{qu}}{{pv}}}=\frac{{pv}}{{qu}}\geqslant\mathrm{1}\leqslant\mathrm{1} \\ $$$${that}\:{means}\:{the}\:{absolute}\:{value}\:{of} \\ $$$$\mathrm{cos}\:{x},\:\mathrm{cos}\:{y}\:{and}\:\mathrm{cos}\:{z}\:{must}\:{be}\:\mathrm{1}. \\ $$$${there}\:{are}\:\mathrm{4}\:{possibilities}: \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\pi,\pi\right),\left(\pi,\mathrm{0},\pi\right),\left(\pi,\pi,\mathrm{0}\right) \\ $$
Commented by Tinkutara last updated on 03/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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