If $n \sum i = 1 cos (A_{i}) = n$ $sum_(i=1)^(n) cos(A_i) = n$ then find $n \sum i = 1 sin (A_{i})$ $sum_(i=1)^(n) sin(A_i)$ ?

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If $n \sum i = 1 cos (A_{i}) = n$ $sum_(i=1)^(n) cos(A_i) = n$ then find $n \sum i = 1 sin (A_{i})$ $sum_(i=1)^(n) sin(A_i)$ ?

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Answer 1 · 2018-07-02T00:48:18

0 (assuming all the $A_{i}$ $A_i$ s are real)

Explanation:

Since the maximum value of ${cos (A)}_{i}$ $cos(A)_i$ for real $A_{i}$ $A_i$ is 1, the sum of $n$ $n$ such numbers can be $n$ $n$ if and only if each one of them is equal to 1. Thus

$cos(A_i) = 1,qquad i=1,2,...,n implies$

$sin(A_i) = sqrt(1-cos^2(A_i))=0,qquad i=1,2,...,n$

Thus, the required sum is 0.

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If $n \sum i = 1 cos (A_{i}) = n$ $sum_(i=1)^(n) cos(A_i) = n$ then find $n \sum i = 1 sin (A_{i})$ $sum_(i=1)^(n) sin(A_i)$ ?

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