How do we prove that the equality $$ \sin{x}+\sin{2x}+\sin{3x}=3 $$ does not hold for any real value of $x$ only using trigonometry ?


closed as off-topic by Namaste, Claude Leibovici, user26857, Davide Giraudo, Arnaldo May 7 '17 at 15:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Claude Leibovici, user26857, Davide Giraudo, Arnaldo
If this question can be reworded to fit the rules in the help center, please edit the question.


$\sin(x)=1$ implies that $x=\pi/2+2\pi k$, this implies that $\sin(2x)=0$, so it is not true. Since you must have $\sin(x)=\sin(2x)=\sin(3x)=1$ since $\sin(x)\in [-1,1]$.

  • $\begingroup$ how can we prove that the equality implies $\sin x =\sin 2x = \sin 3x = 1$ ? $\endgroup$ – ss1729 May 6 '17 at 19:36
  • 3
    $\begingroup$ because $sin(x)\leq 1$ for every $x$ so $sin(2x)\leq 1, sin(3x)\leq 1$. So if one of the numbers $sin(x),sin(2x), sin(3x)$ is strictly inferior to $1$, their sum will be strictly inferior to $3$. $\endgroup$ – Tsemo Aristide May 6 '17 at 19:37

That would only be possible if all of $\sin x$, $\sin 2x$ and $\sin 3x$ equalled $1$. But $\sin x=1$ implies $\sin 2x=0$.

A good follow-up: find the maximum of $\sin x+\sin 2x+\sin3x$ for real $x$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.