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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).

$${Find}\:{minimum}\:{value}\:{of}\:: \\ $$$${cos}\left(\omega−\phi\right)+\mathrm{cos}\left(\phi−\varphi\right)+\mathrm{cos}\:\left(\varphi−\omega\right). \\ $$

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

Ans:−(3/2)

$${Ans}:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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

$${sinw}={a}_{\mathrm{1}} \:\:\:{sin}\emptyset={a}_{\mathrm{2}} \:\:\:\:{sin}\varphi={a}_{\mathrm{3}} \\ $$$${cosw}={b}_{\mathrm{1}} \:\:\:{cos}\emptyset={b}_{\mathrm{2}} \:\:\:\:{cos}\varphi={b}_{\mathrm{3}} \\ $$$${k}={b}_{\mathrm{1}} {b}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{2}} {b}_{\mathrm{3}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} +{b}_{\mathrm{1}} {b}_{\mathrm{3}} +{a}_{\mathrm{1}} {a}_{\mathrm{3}} \\ $$$$\mathrm{2}{k}+\mathrm{3}=\mathrm{2}\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} +{a}_{\mathrm{1}} {a}_{\mathrm{3}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} +{b}_{\mathrm{2}} {b}_{\mathrm{3}} +{b}_{\mathrm{1}} {b}_{\mathrm{3}} \right)+{a}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} +{b}_{\mathrm{3}} ^{\mathrm{2}} \\ $$$$\mathrm{2}{k}+\mathrm{3}=\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$${k}=\frac{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{3}}{\mathrm{2}} \\ $$$${value}\:{of}\:{k}\:{is}\:{minimum}\:{when}\:\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${and}\:\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${so}\:{k}_{{min}} =\frac{−\mathrm{3}}{\mathrm{2}}\:\:{proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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)

$${let}\:\omega−\phi=\alpha,\:\phi−\varphi=\beta \\ $$$$\varphi−\omega=−\left(\omega−\phi+\phi−\varphi\right)=−\left(\alpha+\beta\right) \\ $$$$ \\ $$$${cos}\left(\omega−\phi\right)+\mathrm{cos}\left(\phi−\varphi\right)+\mathrm{cos}\:\left(\varphi−\omega\right) \\ $$$$=\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\left(\alpha+\beta\right)={F}\left(\alpha,\beta\right) \\ $$$$ \\ $$$$\frac{\partial{F}}{\partial\alpha}=−\mathrm{sin}\:\alpha−\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{0}\:\:\:...\left({i}\right) \\ $$$$\frac{\partial{F}}{\partial\beta}=−\mathrm{sin}\:\beta−\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow\alpha=\beta \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=−\mathrm{sin}\:\mathrm{2}\alpha=−\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\mathrm{0}\Rightarrow{F}_{{max}} =\mathrm{3} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{F}_{{min}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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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