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

Find minimum value of : cos(ω−φ)+cos(φ−ϕ)+cos (ϕ−ω).

Findminimumvalueof:cos(ω−ϕ)+cos(ϕ−φ)+cos(φ−ω).

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

Ans:−(3/2)

Ans:−32

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

sinw=a_1 sin∅=a_2 sinϕ=a_3 cosw=b_1 cos∅=b_2 cosϕ=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 2k+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 2k+3=(a_1 +a_2 +a_3 )^2 +(b_1 +b_2 +b_3 )^2 k=(((a_1 +a_2 +a_3 )^2 +(b_1 +b_2 +b_3 )^2 −3)/2) value of k is minimum when (a_1 +a_2 +a_3 )^2 =0 and (b_1 +b_2 +b_3 )^2 =0 so k_(min) =((−3)/2) proved

sinw=a1sin∅=a2sinφ=a3cosw=b1cos∅=b2cosφ=b3k=b1b2+a1a2+b2b3+a2a3+b1b3+a1a32k+3=2(a1a2+a2a3+a1a3+b1b2+b2b3+b1b3)+a12+b12+a22+b22+a32+b322k+3=(a1+a2+a3)2+(b1+b2+b3)2k=(a1+a2+a3)2+(b1+b2+b3)2−32valueofkisminimumwhen(a1+a2+a3)2=0and(b1+b2+b3)2=0sokmin=−32proved

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

thank you sir!

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

let ω−φ=α, φ−ϕ=β ϕ−ω=−(ω−φ+φ−ϕ)=−(α+β) cos(ω−φ)+cos(φ−ϕ)+cos (ϕ−ω) =cos α+cos β+cos (α+β)=F(α,β) (∂F/∂α)=−sin α−sin (α+β)=0 ...(i) (∂F/∂β)=−sin β−sin (α+β)=0 ...(ii) ⇒α=β ⇒sin α=−sin 2α=−2sin α cos α ⇒sin α=0⇒F_(max) =3 ⇒cos α=−(1/2)⇒F_(min) =−(1/2)−(1/2)+2(−(1/2))^2 −1=−(3/2)

letω−ϕ=α,ϕ−φ=βφ−ω=−(ω−ϕ+ϕ−φ)=−(α+β)cos(ω−ϕ)+cos(ϕ−φ)+cos(φ−ω)=cosα+cosβ+cos(α+β)=F(α,β)∂F∂α=−sinα−sin(α+β)=0...(i)∂F∂β=−sinβ−sin(α+β)=0...(ii)⇒α=β⇒sinα=−sin2α=−2sinαcosα⇒sinα=0⇒Fmax=3⇒cosα=−12⇒Fmin=−12−12+2(−12)2−1=−32

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

Ausgezichnet! ����

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

Danke sehr!

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