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If-lt-lt-lt-2pi-and-cos-x-cos-x-cos-x-0-for-all-x-R-then-is-2pi-3-




Question Number 16834 by sushmitak last updated on 26/Jun/17
If α<β<γ<2π and  cos (x+α)+cos (x+β)+cos (x+γ)=0  for all x∈R, then is  γ−α=((2π)/3)?
$$\mathrm{If}\:\alpha<\beta<\gamma<\mathrm{2}\pi\:\mathrm{and} \\ $$$$\mathrm{cos}\:\left({x}+\alpha\right)+\mathrm{cos}\:\left({x}+\beta\right)+\mathrm{cos}\:\left({x}+\gamma\right)=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R},\:\mathrm{then}\:\mathrm{is} \\ $$$$\gamma−\alpha=\frac{\mathrm{2}\pi}{\mathrm{3}}? \\ $$
Commented by ajfour last updated on 27/Jun/17
no, i believe   γ−α=((4π)/3) , then.
$$\mathrm{no},\:\mathrm{i}\:\mathrm{believe}\:\:\:\gamma−\alpha=\frac{\mathrm{4}\pi}{\mathrm{3}}\:,\:\mathrm{then}. \\ $$
Commented by ajfour last updated on 27/Jun/17
Commented by prakash jain last updated on 27/Jun/17
upon expansion  ∀x  cos (x)(cos α+cos β+cos γ)  −sin x(sin α+sin β+sin γ)=0  since above statement is true  for all x  cos α+cos β+cos γ=0  sin α+sin β+sin γ=0  cos β=−(cos α+cos γ)   ......A  sin β=−(sin α+sin γ)     ......B  square and add  1=2+2(cos αcos γ+sin αsin γ)  cos (γ−α)=−(1/2)  γ−α=2nπ±((2π)/3)  since γ<2π  γ−α=((2π)/3) or ((4π)/3)
$$\mathrm{upon}\:\mathrm{expansion} \\ $$$$\forall{x} \\ $$$$\mathrm{cos}\:\left({x}\right)\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma\right) \\ $$$$−\mathrm{sin}\:{x}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right)=\mathrm{0} \\ $$$${since}\:{above}\:{statement}\:{is}\:{true} \\ $$$${for}\:{all}\:{x} \\ $$$$\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma=\mathrm{0} \\ $$$$\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma=\mathrm{0} \\ $$$$\mathrm{cos}\:\beta=−\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\gamma\right)\:\:\:……{A} \\ $$$$\mathrm{sin}\:\beta=−\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\gamma\right)\:\:\:\:\:……{B} \\ $$$${square}\:{and}\:{add} \\ $$$$\mathrm{1}=\mathrm{2}+\mathrm{2}\left(\mathrm{cos}\:\alpha\mathrm{cos}\:\gamma+\mathrm{sin}\:\alpha\mathrm{sin}\:\gamma\right) \\ $$$$\mathrm{cos}\:\left(\gamma−\alpha\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\gamma−\alpha=\mathrm{2}{n}\pi\pm\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{since}\:\gamma<\mathrm{2}\pi \\ $$$$\gamma−\alpha=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\mathrm{or}\:\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$

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