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Question Number 106944 by Rio Michael last updated on 08/Aug/20

If Σ_(i= 1) ^n sin (θ_i ) = n then cos (θ_1 ) + cos (θ_2 ) + cos (θ_3 ) + ... + cos (θ_n ) = ? i got 0 as answer. please who can correct?

$$\mathrm{If}\:\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{sin}\:\left(\theta_{{i}} \right)\:=\:{n}\:\mathrm{then}\:\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \right)\:+\:\mathrm{cos}\:\left(\theta_{\mathrm{2}} \right)\:+\:\mathrm{cos}\:\left(\theta_{\mathrm{3}} \right)\:+\:...\:+\:\mathrm{cos}\:\left(\theta_{{n}} \right)\:=\:? \\ $$$$\mathrm{i}\:\mathrm{got}\:\mathrm{0}\:\mathrm{as}\:\mathrm{answer}.\:\mathrm{please}\:\mathrm{who}\:\mathrm{can}\:\mathrm{correct}? \\ $$

Commented by mr W last updated on 08/Aug/20

Σ_(i= 1) ^n sin (θ_i )≤n (= only for sin θ_i =1) Σ_(i= 1) ^n sin (θ_i )=n ⇒ sin θ_i =1⇒cos θ_i =0 ⇒Σ_(i=1) ^n cos θ_i =0

$$\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{sin}\:\left(\theta_{{i}} \right)\leqslant{n}\:\left(=\:{only}\:{for}\:\mathrm{sin}\:\theta_{{i}} =\mathrm{1}\right) \\ $$$$\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{sin}\:\left(\theta_{{i}} \right)={n}\:\Rightarrow\:\mathrm{sin}\:\theta_{{i}} =\mathrm{1}\Rightarrow\mathrm{cos}\:\theta_{{i}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\:\theta_{{i}} =\mathrm{0} \\ $$

Commented by Rio Michael last updated on 08/Aug/20

thanks to you both sirs

$$\mathrm{thanks}\:\mathrm{to}\:\mathrm{you}\:\mathrm{both}\:\mathrm{sirs} \\ $$

Answered by 1549442205PVT last updated on 08/Aug/20

Applying the inequality AM−GM We have n=Σ_(k=1) ^(n) cosθ_k =Σ_(k=1) ^(n) (√(1−sin^2 θ_k )) =1.(√(1−sin^2 θ_1 )) +1.(√(1−sin^2 θ_2 )) +...+1.(√(1−sin^2 θ_n )) ≤(√((1+1+..+1)(1−sin^2 θ_1 +1−sin^2 θ_2 +...+1−sin^2 θ_n ))) =(√(n(cos^2 θ_1 +cos^2 θ_2 +...+cos^2 θ_n ))) ≤(√(n.n))=n (since cosθ_k ^2 ≤1 ∀k=1...n^(−) ) This shows that the above inequality must ocurr equality sign which is equivalent to { (((1/(1−sin^2 θ_1 ))=(1/(1−sin^2 θ_2 ))=...=(1/(1−sin^2 θ_n )))),((sinθ_1 +sinθ_2 +...+sinθ_n =n)) :} ⇔ { ((sinθ_1 =sinθ_2 = ... =sinθ_n )),((sinθ_1 +sinθ_2 + ... +sinθ_n =n)) :} ⇔sinθ_1 =sinθ_2 = ... =sinθ_n =1 ⇔cosθ_1 =cosθ_2 = ... =cosθ_n =0 ⇒Σ_(k=1) ^(n) cos(θ_k )=0(q.e.d)

$$\mathrm{Applying}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{AM}−\mathrm{GM} \\ $$$$\mathrm{We}\:\mathrm{have}\:\:\:\:\mathrm{n}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\mathrm{cos}\theta_{\mathrm{k}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{k}} } \\ $$$$=\mathrm{1}.\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{1}} }\:+\mathrm{1}.\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{2}} }\:+...+\mathrm{1}.\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{n}} \:} \\ $$$$\leqslant\sqrt{\left(\mathrm{1}+\mathrm{1}+..+\mathrm{1}\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{1}} +\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{2}} +...+\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{n}} \right)} \\ $$$$=\sqrt{\mathrm{n}\left(\mathrm{cos}^{\mathrm{2}} \theta_{\mathrm{1}} +\mathrm{cos}^{\mathrm{2}} \theta_{\mathrm{2}} +...+\mathrm{cos}^{\mathrm{2}} \theta_{\mathrm{n}} \right)} \\ $$$$\leqslant\sqrt{\mathrm{n}.\mathrm{n}}=\mathrm{n}\:\left(\mathrm{since}\:\mathrm{cos}\theta_{\mathrm{k}} ^{\mathrm{2}} \:\leqslant\mathrm{1}\:\forall\mathrm{k}=\overline {\mathrm{1}...\mathrm{n}}\right) \\ $$$$\mathrm{This}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{above}\:\mathrm{inequality} \\ $$$$\mathrm{must}\:\:\mathrm{ocurr}\:\:\mathrm{equality}\:\mathrm{sign}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{equivalent}\:\mathrm{to} \\ $$$$\begin{cases}{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{2}} }=...=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta_{\mathrm{n}} }}\\{\mathrm{sin}\theta_{\mathrm{1}} +\mathrm{sin}\theta_{\mathrm{2}} +...+\mathrm{sin}\theta_{\mathrm{n}} =\mathrm{n}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{sin}\theta_{\mathrm{1}} =\mathrm{sin}\theta_{\mathrm{2}} =\:\:\:...\:\:\:\:\:\:\:\:=\mathrm{sin}\theta_{\mathrm{n}} }\\{\mathrm{sin}\theta_{\mathrm{1}} +\mathrm{sin}\theta_{\mathrm{2}} +\:\:\:\:...\:\:\:\:+\mathrm{sin}\theta_{\mathrm{n}} =\mathrm{n}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{sin}\theta_{\mathrm{1}} =\mathrm{sin}\theta_{\mathrm{2}} =\:\:\:\:...\:\:\:\:\:=\mathrm{sin}\theta_{\mathrm{n}} =\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{cos}\theta_{\mathrm{1}} =\mathrm{cos}\theta_{\mathrm{2}} =\:\:\:\:...\:\:\:\:=\mathrm{cos}\theta_{\mathrm{n}} =\mathrm{0} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\mathrm{cos}\left(\theta_{\mathrm{k}} \right)=\mathrm{0}\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right)\: \\ $$

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