solve for 1 period sin (α/2) =sin 2α sin (β/3) =sin 3β cos (γ/2) =cos 2γ cos (δ/3) =cos 3δ tan (ε/2) =tan 2ε tan (ζ/3) =tan 3ζ


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Question Number 40285 by MJS last updated on 18/Jul/18

$solve for 1 period$ $\sin \frac{α}{2} = \sin 2 α$ $\sin \frac{β}{3} = \sin 3 β$ $\cos \frac{γ}{2} = \cos 2 γ$ $\cos \frac{δ}{3} = \cos 3 δ$ $\tan \frac{ϵ}{2} = \tan 2 ϵ$ $\tan \frac{ζ}{3} = \tan 3 ζ$
	Answered by MrW3 last updated on 19/Jul/18

	$\sin \frac{α}{2} = \sin 2 α$ $\Rightarrow 2 α + 2 m π = \frac{α}{2} + 2 n π \Rightarrow α = \frac{4 k π}{3}$ $o r$ $\Rightarrow 2 α + 2 m π = (2 n + 1) π - \frac{α}{2} \Rightarrow α = \frac{2 (2 k + 1) π}{5}$ $\Rightarrow s o l u t i o n i s :$ $α = \frac{4 k π}{3} a n d α = \frac{2 (2 k + 1) π}{5} w i t h k = 0, \pm 1, \pm 2, . . .$ $o n e p e r i o d : 0 ⩽ α ⩽ 4 π$ $\frac{4 k π}{3} ⩽ 4 π \Rightarrow k ⩽ 3$ $\frac{2 (2 k + 1) π}{5} ⩽ 4 π \Rightarrow k ⩽ 4.5$ $\Rightarrow α = 0, \frac{2 π}{5}, \frac{6 π}{5}, \frac{4 π}{3}, 2 π, \frac{8 π}{3}, \frac{14 π}{5}, \frac{18 π}{5}, 4 π$
	Answered by MrW3 last updated on 19/Jul/18

	$\tan \frac{ϵ}{2} = \tan 2 ϵ$ $\frac{ϵ}{2} + m π = 2 ϵ + n π \Rightarrow ϵ = \frac{2 k π}{3}$ $\Rightarrow s o l u t i o n i s :$ $ϵ = \frac{2 k π}{3}$
	Answered by MrW3 last updated on 19/Jul/18

	$\cos \frac{γ}{2} = \cos 2 γ$ $\frac{γ}{2} + 2 m π = 2 n π \pm 2 γ \Rightarrow {\begin{cases} γ = \frac{4 k π}{3} \\ γ = \frac{4 k π}{5} \end{cases}$ $\Rightarrow s o l u t i o n i s :$ $γ = \frac{4 k π}{3} a n d γ = \frac{4 k π}{5}$
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