Question

If `sum_(i=1) ^(n)sin(theta_i)=n` then `cos(theta_1)+cos(theta_2)+...+cos(theta_n) =`?

Answer 1

`cos(theta_1)+cos(theta_2)+...+cos(theta_n)=0`

Explanation:

In the summation `sum_(i=1)^nsin(theta_i)=n`, we have `n` terms and they add up to `n`. However, the maximum possible value of any `sintheta_i` is `1`, when `theta_i=pi/2`.

Hence each term in `sum_(i=1)^nsin(theta_i)=n` is `1` and each `theta_i=pi/2`

therefore each `costheta_i=0` and

`cos(theta_1)+cos(theta_2)+...+cos(theta_n)=0`