The sum of cosines are given to be 'n'. The sum of its corresponding sines are to be found...

Question

I can neither make head nor tail of this. Could anyone please show me how to solve the following problem? "If $\sum_{i=1}^n cos \theta_i = n$ then what is the value of $\sum_{i=1}^n sin\theta_i$"?

Answer 1

If we have

$$\sum_{i=1}^n \cos \theta_i = n$$ then we also have

$$\frac{\sum_{i=1}^n \cos \theta_i}n = 1$$

which means that the average cosine of these angles is $1$.

If the average cosine of these angles is $1$, what is the average sine of these angles?

The average sine is $0$, since $\sin(\arccos 1) = 0$. Hence $$\frac{\sum_{i=1}^n \sin \theta_i}n = 0$$ which implies $\sum_{i=1}^n \sin \theta_i = 0$

Answer 2

Since

$$\cos \theta_i\leq1\implies\sum_{i=1}^n \cos \theta_i \leq n$$

and equality holds if and only if

$$\forall i \quad \cos \theta_i =1\implies \sin \theta_i=0$$