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Question Number 15300 by Tinkutara last updated on 09/Jun/17

$Find the values of α and β, 0 < α, β < \frac{π}{2}$ $satisfying the following equation$ $\cos α \cos β \cos (α + β) = - \frac{1}{8} .$
Commented bymrW1 last updated on 09/Jun/17

$α = β = \frac{π}{3}$
Commented bymrW1 last updated on 10/Jun/17

$F (α, β) = \cos α \cos β \cos (α + β)$ ${let}^{'} s find the minimum of this function$ $\frac{\partial F}{\partial α} = - \sin α \cos β \cos (α + β) - \cos α \cos β \sin (α + β)$ $= - \cos β [\sin α \cos (α + β) + \cos α \sin (α + β)]$ $= - \cos β \sin (2 α + β) = 0$ $\Rightarrow \sin (2 α + β) = 0$ $\Rightarrow 2 α + β = π . . . (i)$ $\frac{\partial F}{\partial β} = - \cos α \sin β \cos (α + β) - \cos α \cos β \sin (α + β)$ $= - \cos α [\sin β \cos (α + β) + \cos β \sin (α + β)]$ $= - \cos α \sin (α + 2 β) = 0$ $\Rightarrow \sin (α + 2 β) = 0$ $\Rightarrow α + 2 β = π . . . (ii)$ $from (i) and (ii)$ $\Rightarrow α = β = \frac{π}{3}$ $minimum = \cos \frac{π}{3} \cos \frac{π}{3} \cos \frac{2 π}{3} = - \frac{1}{8}$ $∴ for \cos α \cos β \cos (α + β) = - \frac{1}{8}$ $there is only one solution :$ $α = β = \frac{π}{3}$
Commented byTinkutara last updated on 10/Jun/17

$Thanks Sir!$
	Answered by Tinkutara last updated on 09/Jul/17

	$8 \cos α \cos β \cos (α + β) + 1 = 0$ $4 \cos (α + β) [\cos (α + β) + \cos (α - β)] + 1 = 0$ $4 \cos^{2} (α + β) + 4 \cos (α + β) \cos (α - β)$ $+ \cos^{2} (α - β) + \sin^{2} (α - β) = 0$ ${[2 \cos (α + β) + \cos (α - β)]}^{2} + \sin^{2} (α - β) = 0$ $\sin (α - β) = 0 and 2 \cos (α + β) + \cos (α - β) = 0$ $\Rightarrow α = β and this gives$ $2 \cos 2 α + 1 = 0 \Rightarrow α = \frac{π}{3}, as 0 < α, β < \frac{π}{2} .$ $So α = β = \frac{π}{3} .$
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