Find-minimum-value-of-cos-cos-cos-

Question Number 56954 by rahul 19 last updated on 27/Mar/19

F i n d m i n i m u m v a l u e o f :

c o s (ω - ϕ) + \cos (ϕ - φ) + \cos (φ - ω) .

Commented by rahul 19 last updated on 27/Mar/19

A n s : - \frac{3}{2}

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19

s i n w = a_{1} s i n \emptyset = a_{2} s i n φ = a_{3}

c o s w = b_{1} c o s \emptyset = b_{2} c o s φ = b_{3}

k = b_{1} b_{2} + a_{1} a_{2} + b_{2} b_{3} + a_{2} a_{3} + b_{1} b_{3} + a_{1} a_{3}

2 k + 3 = 2 (a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{3} + b_{1} b_{2} + b_{2} b_{3} + b_{1} b_{3}) + a_{1}^{2} + b_{1}^{2} + a_{2}^{2} + b_{2}^{2} + a_{3}^{2} + b_{3}^{2}

2 k + 3 = {(a_{1} + a_{2} + a_{3})}^{2} + {(b_{1} + b_{2} + b_{3})}^{2}

k = \frac{{(a_{1} + a_{2} + a_{3})}^{2} + {(b_{1} + b_{2} + b_{3})}^{2} - 3}{2}

v a l u e o f k i s m i n i m u m w h e n {(a_{1} + a_{2} + a_{3})}^{2} = 0

a n d {(b_{1} + b_{2} + b_{3})}^{2} = 0

s o k_{m i n} = \frac{- 3}{2} p r o v e d

Commented by rahul 19 last updated on 27/Mar/19

thank you sir!

Answered by mr W last updated on 27/Mar/19

l e t ω - ϕ = α, ϕ - φ = β

φ - ω = - (ω - ϕ + ϕ - φ) = - (α + β)

c o s (ω - ϕ) + \cos (ϕ - φ) + \cos (φ - ω)

= \cos α + \cos β + \cos (α + β) = F (α, β)

\frac{\partial F}{\partial α} = - \sin α - \sin (α + β) = 0 \dots (i)

\frac{\partial F}{\partial β} = - \sin β - \sin (α + β) = 0 \dots (i i)

\Rightarrow α = β

\Rightarrow \sin α = - \sin 2 α = - 2 \sin α \cos α

\Rightarrow \sin α = 0 \Rightarrow F_{m a x} = 3

\Rightarrow \cos α = - \frac{1}{2} \Rightarrow F_{m i n} = - \frac{1}{2} - \frac{1}{2} + 2 {(- \frac{1}{2})}^{2} - 1 = - \frac{3}{2}

Commented by rahul 19 last updated on 27/Mar/19

Ausgezichnet! ��

Commented by mr W last updated on 27/Mar/19

Danke sehr!