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Question-219279

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Question Number 219279 by Mingma last updated on 22/Apr/25
Answered by som(math1967) last updated on 22/Apr/25
((2sin((β+α)/2)cos((α−β)/2))/(2sin ((α+β)/2)sin((α−β)/2)))=(√3) ⇒ cot((α−β)/2)=cot(π/6) ⇒ α−β=(π/3) sin 3α+sin β =2sin ((3(α+β))/2)×cos ((3(α−β))/2) =2sin ((3(α+β))/2)×cos(π/2) =0
$$\:\frac{\mathrm{2}{sin}\frac{\beta+\alpha}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}}{\mathrm{2sin}\:\frac{\alpha+\beta}{\mathrm{2}}{sin}\frac{\alpha−\beta}{\mathrm{2}}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:{cot}\frac{\alpha−\beta}{\mathrm{2}}={cot}\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\:\alpha−\beta=\frac{\pi}{\mathrm{3}} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin}\:\beta \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}×\mathrm{cos}\:\frac{\mathrm{3}\left(\alpha−\beta\right)}{\mathrm{2}} \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}×{cos}\frac{\pi}{\mathrm{2}} \\ $$$$=\mathrm{0} \\ $$
Answered by Nicholas666 last updated on 22/Apr/25
((2sin(((α+β)/2))cos(((α−β)/2)))/(2sin(((α+β)/2))sin(((α−β)/2)))) =(√3) cot(((α−β)/2))=(√3) ⇒((α−β)/2) = (π/6)+kπ α−β=(π/3)+2kπ sin (3α)+sin(3β)=2sin(((3(α+β))/2))cos(((3(α−β))/2)) sin(3α)+sin(3β)=2sin(((3(α+β))/2))cos(((3((π/3)+2kπ))/2)) sin(3α)+sin(3β)=2sin(((3(α+β))/2))cos((π/2)+kπ) cos ((π/2)+3kπ)=0 sin(3α)+sin(3β)=2sin(((3(α+β))/2)×0=0 final ans; sin(3α) + sin(3β) = 0
$$\:\:\frac{\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}\:=\sqrt{\mathrm{3}} \\ $$$${cot}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)=\sqrt{\mathrm{3}}\:\Rightarrow\frac{\alpha−\beta}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{6}}+{k}\pi \\ $$$$\:\:\alpha−\beta=\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi \\ $$$$\:\:{sin}\:\left(\mathrm{3}\alpha\right)+{sin}\left(\mathrm{3}\beta\right)=\mathrm{2}{sin}\left(\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{3}\left(\alpha−\beta\right)}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:{sin}\left(\mathrm{3}\alpha\right)+{sin}\left(\mathrm{3}\beta\right)=\mathrm{2}{sin}\left(\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{3}\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\right)}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:{sin}\left(\mathrm{3}\alpha\right)+{sin}\left(\mathrm{3}\beta\right)=\mathrm{2}{sin}\left(\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\mathrm{2}}+{k}\pi\right) \\ $$$$\:\:\:{cos}\:\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{3}{k}\pi\right)=\mathrm{0} \\ $$$$\:\:\:\:{sin}\left(\mathrm{3}\alpha\right)+{sin}\left(\mathrm{3}\beta\right)=\mathrm{2}{sin}\left(\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}×\mathrm{0}=\mathrm{0}\right. \\ $$$${final}\:{ans};\:\:{sin}\left(\mathrm{3}\alpha\right)\:+\:{sin}\left(\mathrm{3}\beta\right)\:=\:\mathrm{0} \\ $$$$ \\ $$

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