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here is a function for: $f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$

what is the $f(x)$?

I calculate $f(x)$ as follows: $$\begin{align} x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\ f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\ 3f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6} \\ f(\frac{\pi}{3}) &=(1/3)\sin\frac{5\pi}{6} \end{align}$$

$f(x)=(1/3)\sin\frac{5x}{2}$

user123
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3 Answers3

3

Assuming $f$ is defined for all $x\in\mathbb{R}$. First, note that for any $x$, $$ f(x) = \sin\!\left(x+\frac{\pi}{2}\right)-2f\!\left(\frac{\pi}{3}\right) = \cos x -2f\!\left(\frac{\pi}{3}\right) $$ so it only remains to compute $f\!\left(\frac{\pi}{3}\right)$. From the expression above $$ f\!\left(\frac{\pi}{3}\right) = \cos \frac{\pi}{3} -2f\!\left(\frac{\pi}{3}\right) = \frac{1}{2} -2f\!\left(\frac{\pi}{3}\right) $$ and therefore rearranging the terms gives $f\!\left(\frac{\pi}{3}\right) = \frac{1}{6}$. Putting it all together, $$ \forall x\in \mathbb{R}, \quad f(x)=\cos x - \frac{1}{3}\;. $$

(It then only remains to check this expression satisfies the original functional equation, to be certain. It does; but even a quick sanity check for $x=0$, $x=\frac{\pi}{2}$ and $x=\pi$ will be enough to build confidence.)

Clement C.
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  • thanks, I understand your answer.but what is wrong with my answer that cause different equation? – user123 Sep 26 '14 at 17:11
  • The LHS in the third line is wrong. It should be $3f(\pi/3)$, not $-f(\pi/3)$. – Clement C. Sep 26 '14 at 17:15
  • I correct it, but last answer is different. – user123 Sep 26 '14 at 17:22
  • Once you have the right value of $f(\pi/3)$, your last step is wrong. Just replace $f(\pi/3)$ by its value in the first equation -- have you looked at the answers? What you are doing in your last step is roughly saying "if $h(e)=e^2$, then $h(x)=x^2$". This is not a valid reasoning. – Clement C. Sep 26 '14 at 18:30
  • Thanks, I looked at the answers and I know the answers are correct, but I don't know why such reasoning is incorrect. – user123 Sep 26 '14 at 18:45
  • Take $h$ defined as $h(x)=(x-e) + e^2$. $h(e)=e^2$, yet $h(x)\neq x^2$ in general. – Clement C. Sep 26 '14 at 18:46
  • OK, I understand it. In my answer, I can find other equation by using that solution. so it is wrong. – user123 Sep 26 '14 at 18:50
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$f(x)=f((x+\frac{1}{2}\pi)-\frac{1}{2}\pi)=\sin(x+\frac{1}{2}\pi)-2f(\frac{1}{3}\pi)=\cos(x)-2f(\frac{1}{3}\pi)$

To find $f(\frac{1}{3}\pi)$ we substitute $x=\frac{1}{3}\pi$:

$f(\frac{1}{3}\pi)=\cos(\frac{1}{3}\pi)-2f(\frac{1}{3}\pi)$

Then $f(\frac{1}{3}\pi)=\frac{1}{3}\cos(\frac{1}{3}\pi)=\frac{1}{6}$ so we end up with:

$$f(x)=\cos(x)-\frac{1}{3}$$

Vera
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1

Shift the argument by $\pi/2$: $$f(x)=\sin(x+\frac\pi2)-2f(\frac{\pi}3).$$ Then plug $\pi/3$: $$f(\frac{\pi}3)=\sin(\frac{5\pi}6)-2f(\frac{\pi}3).$$ Solve for $f(\pi/3)$: $$f(x)=\sin(x+\frac\pi2)-\frac23\sin(\frac{5\pi}6).$$