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?


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

$If \underset{i = 1}{\sum^{n}} \sin (θ_{i}) = n then \cos (θ_{1}) + \cos (θ_{2}) + \cos (θ_{3}) + . . . + \cos (θ_{n}) = ?$ $i got 0 as answer . please who can correct ?$
Commented by mr W last updated on 08/Aug/20

$\underset{i = 1}{\sum^{n}} \sin (θ_{i}) ⩽ n (= o n l y f o r \sin θ_{i} = 1)$ $\underset{i = 1}{\sum^{n}} \sin (θ_{i}) = n \Rightarrow \sin θ_{i} = 1 \Rightarrow \cos θ_{i} = 0$ $\Rightarrow \underset{i = 1}{\sum^{n}} \cos θ_{i} = 0$
Commented by Rio Michael last updated on 08/Aug/20

$thanks to you both sirs$
	Answered by 1549442205PVT last updated on 08/Aug/20
	$Applying the inequality AM−GM We have n=Σ_(k=1) ^(n) cosθ_k =Σ_(k=1) ^(n) (√(1−sin^2 θ_k )) =1.(√(1−sin^2 θ_1 )) +1.(√(1−sin^2 θ_2 )) +...+1.(√(1−sin^2 θ_n )) ≤(√((1+1+..+1)(1−sin^2 θ_1 +1−sin^2 θ_2 +...+1−sin^2 θ_n ))) =(√(n(cos^2 θ_1 +cos^2 θ_2 +...+cos^2 θ_n ))) ≤(√(n.n))=n (since cosθ_k ^2 ≤1 ∀k=1...n^(−) ) This shows that the above inequality must ocurr equality sign which is equivalent to { (((1/(1−sin^2 θ_1 ))=(1/(1−sin^2 θ_2 ))=...=(1/(1−sin^2 θ_n )))),((sinθ_1 +sinθ_2 +...+sinθ_n =n)) :} ⇔ { ((sinθ_1 =sinθ_2 = ... =sinθ_n )),((sinθ_1 +sinθ_2 + ... +sinθ_n =n)) :} ⇔sinθ_1 =sinθ_2 = ... =sinθ_n =1 ⇔cosθ_1 =cosθ_2 = ... =cosθ_n =0 ⇒Σ_(k=1) ^(n) cos(θ_k )=0(q.e.d)$
	$Applying the inequality AM - GM$ $We have n = \underset{k = 1}{\overset{n}{Σ}} \cos θ_{k} = \underset{k = 1}{\overset{n}{Σ}} \sqrt{1 - \sin^{2} θ_{k}}$ $= 1 . \sqrt{1 - \sin^{2} θ_{1}} + 1 . \sqrt{1 - \sin^{2} θ_{2}} + . . . + 1 . \sqrt{1 - \sin^{2} θ_{n}}$ $⩽ \sqrt{(1 + 1 + . . + 1) (1 - \sin^{2} θ_{1} + 1 - \sin^{2} θ_{2} + . . . + 1 - \sin^{2} θ_{n})}$ $= \sqrt{n (\cos^{2} θ_{1} + \cos^{2} θ_{2} + . . . + \cos^{2} θ_{n})}$ $⩽ \sqrt{n . n} = n (since \cos θ_{k}^{2} ⩽ 1 \forall k = \overset{―}{1 . . . n})$ $This shows that the above inequality$ $must ocurr equality sign which is$ $equivalent to$ ${\begin{cases} \frac{1}{1 - \sin^{2} θ_{1}} = \frac{1}{1 - \sin^{2} θ_{2}} = . . . = \frac{1}{1 - \sin^{2} θ_{n}} \\ \sin θ_{1} + \sin θ_{2} + . . . + \sin θ_{n} = n \end{cases}$ $\Leftrightarrow {\begin{cases} \sin θ_{1} = \sin θ_{2} = . . . = \sin θ_{n} \\ \sin θ_{1} + \sin θ_{2} + . . . + \sin θ_{n} = n \end{cases}$ $\Leftrightarrow \sin θ_{1} = \sin θ_{2} = . . . = \sin θ_{n} = 1$ $\Leftrightarrow \cos θ_{1} = \cos θ_{2} = . . . = \cos θ_{n} = 0$ $\Rightarrow \underset{k = 1}{\overset{n}{Σ}} \cos (θ_{k}) = 0 (q . e . d)$
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