Find-the-values-of-and-0-lt-lt-pi-2-satisfying-the-following-equation-cos-cos-cos-1-8-

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 by mrW1 last updated on 09/Jun/17

α = β = \frac{π}{3}

Commented by mrW1 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 \dots (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 \dots (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 by Tinkutara 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} .