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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 byajfour 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 byajfour last updated on 27/Jun/17

Commented byprakash 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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