What is the function $f$ verifying $f\left(\frac{x}{2}+\frac{x}{2}\cos\left(\frac{v\pi}{x}\right)\right)=\frac{x}{2}\sin\left(\frac{v\pi}{x}\right)$?

Question

What are the solutions to the functional equality (for a constant $v$): $$ \forall\, x > 0, \ \ \ \ f\left(\frac{x}{2}+\frac{x}{2}\cos\left(\frac{v\pi}{x}\right)\right)=\frac{x}{2}\sin\left(\frac{v\pi}{x}\right)\text.$$

Thanks

Answer 1

There is no such function $ f $. To see that, let $ a = \frac { v \pi } 2 $ and use double-angle formulas to rewrite the functional equation as $$ f \left( x \left( \cos \frac a x \right) ^ 2 \right) = x \left( \cos \frac a x \right) \left( \sin \frac a x \right) \text . $$ Now note that $ x \mapsto x \left( \cos \frac a x \right) ^ 2 $ is a continuous function the graph of which oscillates between the lines $ y = 0 $ and $ y = x $. That means you can choose $ x $ and $ x ' $ such that $ 0 < x < x ' $ and $ x \left( \cos \frac a x \right) ^ 2 = x ' \left( \cos \frac a { x ' } \right) ^ 2 \ne 0 $. As $ f $ is a function, we must have $ f \left( x \left( \cos \frac a x \right) ^ 2 \right) = f \left( x ' \left( \cos \frac a { x ' } \right) ^ 2 \right) $, or equivalently $ x \left( \cos \frac a x \right) \left( \sin \frac a x \right) = x ' \left( \cos \frac a { x ' } \right) \left( \sin \frac a { x ' } \right) $. Dividing by $ x \left( \cos \frac a x \right) ^ 2 = x ' \left( \cos \frac a { x ' } \right) ^ 2 $ we get $ \tan \frac a x = \tan \frac a { x ' } $, which means there is an integer $ n $ such that $ \frac a x - \frac a { x ' } = n \pi $. This shows that $ \cos \frac a x = ( - 1 ) ^ n \cos \frac a { x ' } $, and therefore $ \left( \cos \frac a x \right) ^ 2 = \left( \cos \frac a { x ' } \right) ^ 2 $. Thus we must have $ x = x ' $, which leads to a contradiction.