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Question Number 83425 by jagoll last updated on 02/Mar/20
((tan 3a)/(tan a)) = k show that ((sin 3a)/(sin a)) = ((2k)/(k−1))
$$\frac{\mathrm{tan}\:\mathrm{3a}}{\mathrm{tan}\:\mathrm{a}}\:=\:\mathrm{k} \\ $$$$\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{sin}\:\mathrm{3a}}{\mathrm{sin}\:\mathrm{a}}\:=\:\frac{\mathrm{2k}}{\mathrm{k}−\mathrm{1}} \\ $$
Commented by jagoll last updated on 02/Mar/20
tan 3a = k tan a ((3tan a−tan^3 a)/(1−3tan^2 a)) = k tan a 3−tan^2 a = k − 3k tan^2 a (3k−1)tan^2 a = k−3 tan^2 a = ((k−3)/(3k−1))⇒cot^2 a = ((3k−1)/(k−3)) csc^2 a = 1+ cot^2 a = 1+((3k−1)/(k−3)) csc^2 a = ((4k−4)/(k−3)) ⇒ sin^2 a = ((k−3)/(4k−4)) ⇒ ((sin 3a)/(sin a)) = ((3sin a−4sin^3 a)/(sin a)) = 3−4sin^2 a = 3 −4(((k−3)/(4k−4))) = 3−(((k−3)/(k−1))) = ((3k−3−k+3)/(k−1)) = ((2k)/(k−1))
$$\mathrm{tan}\:\mathrm{3a}\:=\:\mathrm{k}\:\mathrm{tan}\:\mathrm{a} \\ $$$$\frac{\mathrm{3tan}\:\mathrm{a}−\mathrm{tan}\:^{\mathrm{3}} \mathrm{a}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \mathrm{a}}\:=\:\mathrm{k}\:\mathrm{tan}\:\mathrm{a} \\ $$$$\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{a}\:=\:\mathrm{k}\:−\:\mathrm{3k}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{a} \\ $$$$\left(\mathrm{3k}−\mathrm{1}\right)\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}\:=\:\mathrm{k}−\mathrm{3} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}\:=\:\frac{\mathrm{k}−\mathrm{3}}{\mathrm{3k}−\mathrm{1}}\Rightarrow\mathrm{cot}\:^{\mathrm{2}} \mathrm{a}\:=\:\frac{\mathrm{3k}−\mathrm{1}}{\mathrm{k}−\mathrm{3}} \\ $$$$\mathrm{csc}^{\mathrm{2}} \:\mathrm{a}\:=\:\mathrm{1}+\:\mathrm{cot}^{\mathrm{2}} \:\mathrm{a}\:=\:\mathrm{1}+\frac{\mathrm{3k}−\mathrm{1}}{\mathrm{k}−\mathrm{3}} \\ $$$$\mathrm{csc}^{\mathrm{2}} \:\mathrm{a}\:=\:\frac{\mathrm{4k}−\mathrm{4}}{\mathrm{k}−\mathrm{3}}\:\Rightarrow\:\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{a}\:=\:\frac{\mathrm{k}−\mathrm{3}}{\mathrm{4k}−\mathrm{4}} \\ $$$$\Rightarrow\:\frac{\mathrm{sin}\:\mathrm{3a}}{\mathrm{sin}\:\mathrm{a}}\:=\:\frac{\mathrm{3sin}\:\mathrm{a}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{a}}{\mathrm{sin}\:\mathrm{a}} \\ $$$$=\:\mathrm{3}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{a}\:=\:\mathrm{3}\:−\mathrm{4}\left(\frac{\mathrm{k}−\mathrm{3}}{\mathrm{4k}−\mathrm{4}}\right) \\ $$$$=\:\mathrm{3}−\left(\frac{\mathrm{k}−\mathrm{3}}{\mathrm{k}−\mathrm{1}}\right)\:=\:\frac{\mathrm{3k}−\mathrm{3}−\mathrm{k}+\mathrm{3}}{\mathrm{k}−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2k}}{\mathrm{k}−\mathrm{1}} \\ $$

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