# Thread: mei c3 jun 2006 question- its a toughie

1. ## mei c3 jun 2006 question- its a toughie

The question is:

Show that f(x + π) = (-e^(-1/5)* π)*f(x)

Background info:
f(x) = (e^(-1/5)* x)*sin x

I would really appreciate a walkthrough answer because I've been trying for hours to get their solution with no such luck.

Thanx

2. Originally Posted by hopingforhelp
The question is:

Show that f(x + π) = (-e^(-1/5)* π)*f(x)

Background info:
f(x) = (e^(-1/5)* x)*sin x

I would really appreciate a walkthrough answer because I've been trying for hours to get their solution with no such luck.

Thanx
The only thing I can think to suggest is just to do it!
Since f(x) is defined as $e^{-x/5} \sin (x)$, $f(x+ \pi) = e^{-(x+\pi)/5} \sin (x+\pi)$. Now you should know that $\sin (x+\pi)= -\sin (x)$ and, by the "laws of exponents", $e^{-(x+ \pi)/5}= e^{-x/5- \pi/5}= e^{-x/5}e^{-\pi/5}$. Putting those together,
$f(x+\pi)= e^{-x/5}e^{-\pi/5}(-\sin (x))= -e^{-\pi/5}e^{-x/5} \sin (x)= -e^{-\pi/5}f(x)$.

3. Thanks so much HallsofIvy, I had no idea that sin(x + π) = -sin x. Do rules like this come under a particular title so that I can search them up on google and learn them for my exam on monday?

4. Originally Posted by hopingforhelp
Thanks so much HallsofIvy, I had no idea that sin(x + π) = -sin x. Do rules like this come under a particular title so that I can search them up on google and learn them for my exam on monday?
sum formula for sine ...

$
\sin(x+\pi) = \sin{x}\cos(\pi) + \cos{x}\sin(\pi) = -\sin{x} + 0 = -sin{x}
$