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Question Number 21260 by oyshi last updated on 17/Sep/17
if (θ−ϕ) subtle and sin θ+sin ϕ=(√(3(cos ϕ))  −cos θ)  so proof sin 3θ+sin 3ϕ=0
$${if}\:\left(\theta−\varphi\right)\:{subtle}\:{and}\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi=\sqrt{\mathrm{3}\left(\mathrm{cos}\:\varphi\right.} \\ $$$$\left.−\mathrm{cos}\:\theta\right) \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$
Answered by 951172235v last updated on 07/Feb/19
sin θ+sin ϕ =(√3) (cos ϕ−cos θ)  sin θ+(√3) cos θ = (√3) cos ϕ−sin ϕ  sin (θ+(Λ^− /3)) = sin ((Λ^− /3) −ϕ)  3sin (θ+(Λ^− /3))−4sin^3 (θ+(Λ^− /3)) = 3sin ((Λ^− /3)−ϕ)−4sin^3 ((Λ^− /3)−ϕ)  sin (3θ+Λ^− ) = sin (Λ^− −3ϕ)  −sin 3θ = sin 3ϕ  sin 3θ+sin 3ϕ =0
$$\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\:=\sqrt{\mathrm{3}}\:\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\varphi−\mathrm{sin}\:\varphi \\ $$$$\mathrm{sin}\:\left(\theta+\frac{\overset{−} {\Lambda}}{\mathrm{3}}\right)\:=\:\mathrm{sin}\:\left(\frac{\overset{−} {\Lambda}}{\mathrm{3}}\:−\varphi\right) \\ $$$$\mathrm{3sin}\:\left(\theta+\frac{\overset{−} {\Lambda}}{\mathrm{3}}\right)−\mathrm{4sin}\:^{\mathrm{3}} \left(\theta+\frac{\overset{−} {\Lambda}}{\mathrm{3}}\right)\:=\:\mathrm{3sin}\:\left(\frac{\overset{−} {\Lambda}}{\mathrm{3}}−\varphi\right)−\mathrm{4sin}\:^{\mathrm{3}} \left(\frac{\overset{−} {\Lambda}}{\mathrm{3}}−\varphi\right) \\ $$$$\mathrm{sin}\:\left(\mathrm{3}\theta+\overset{−} {\Lambda}\right)\:=\:\mathrm{sin}\:\left(\overset{−} {\Lambda}−\mathrm{3}\varphi\right) \\ $$$$−\mathrm{sin}\:\mathrm{3}\theta\:=\:\mathrm{sin}\:\mathrm{3}\varphi \\ $$$$\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi\:=\mathrm{0}\: \\ $$

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