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Assume that $f\in C[0,\pi]$ and $$ \int_0^\pi f(t)\sin (t)\ dt = \int_0^\pi f(t)\cos (t)\ dt , $$ then there exist $a,b\in (0,\pi)$ such that $a\neq b$ and $f(a)=f(b)$. Please, I want to solve this question, but don't know how. I would appreciate a hint or comment (not the whole solution). Thank you very much.

Senna
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    You forgot $ a \ne b$, otherwise the problem is trivial – Fred Feb 12 '21 at 10:38
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    Too many questions begin or end with "I don't even know how to begin with this problem". While this may be true ... it is still not a valid reason to limit your post to the statement of the problem without any mention of your own thoughts. – From Avoid "no clue" questions. – Martin R Feb 12 '21 at 10:39
  • I tried to apply mean value theorem and also a trigonometric change of variable, but I got nothing interesting. I am just asking for a hint. – Senna Feb 12 '21 at 10:48
  • Thanks Fred, I'll edit the post. – Senna Feb 12 '21 at 10:49
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    Unless I made an error, the statement is wrong: For $f(x) = x - \pi/2 - 1$ you have $\int_0^\pi f(t)\sin (t)\ dt = \int_0^\pi f(t)\cos (t)\ dt =-2$, but $f$ is injective. – Martin R Feb 12 '21 at 10:49
  • Wow, you're right Martin. Thank you very much for your comment... there must be something wrong with the problem. I had this in one of my exams... It was the only problem I couldn't solve... Now I see the teacher should be confused because the statement is false. – Senna Feb 12 '21 at 10:58
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    There is a similar question, where $\int_0^\pi f(t)\sin (t)\ dt = \int_0^\pi f(t)\cos (t)\ dt = 0$. In that case $f$ has two zeros in $(0, \pi)$: https://math.stackexchange.com/q/247385 – Martin R Feb 12 '21 at 11:02

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