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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 $\sum_{i=1}^n sin\theta_i$"?

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    $\begingroup$ Did you notice that it was a sum of exactly $n$ terms? $\endgroup$ – John Hughes Jan 9 '18 at 16:36
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    $\begingroup$ Please edit your question...clearly both sums can not be $n$. $\endgroup$ – lulu Jan 9 '18 at 16:37
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    $\begingroup$ Can you find an example of $\cos (\theta_1)+\cos (\theta_2) =2$? What then is $\sin (\theta_1)+\sin (\theta_2) $? How about $\cos (\theta_1)=1$ and $\sin (\theta_1)$? Can you see a pattern? $\endgroup$ – Henry Jan 9 '18 at 16:39
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    $\begingroup$ My point was that you have exactly $n$ cosines, and they sum up to $n$. You need to think about that. $\endgroup$ – John Hughes Jan 9 '18 at 16:47
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    $\begingroup$ When you have only $1$ equation and $n$ unknowns, you know something funny must be going on. Either the sum of sines can be gotten by rearranging the sum of cosines, or there is an inequality which forces all the cosines to a certain value, so the parameters can't actually vary. (The former turns out to not be true, so it must be the latter.) $\endgroup$ – 6005 Jan 9 '18 at 17:28
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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 $\sum_{i=1}^n \sin \theta_i = 0$

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Since

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

and equality holds if and only if

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

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