If ##x## and ##y## are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.3. The Attempt at a Solution
do you mean?If ##x## and ##y## are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.
No, I meant:do you mean?
Ok I can write ##\frac{y}{a} =1-2sin^2(\omega t) ##, also to ## \frac{x^2}{a^2}\ = sin^2(\omega t) ## and then insert ## sin^2(\omega t) ##No, I meant:
##\frac{y}{a} = \cos(2 \omega t) =## something to do with ##\sin(\omega t) =## something to do with ##\frac{x}{a}##
You're terribly over-complicating this.
You seem to have a mental block about replacing ##\sin(\omega t) ## by ##\frac{x}{a}##.Ok I can write ##\frac{y}{a} =1-2sin^2(\omega t) ##, also to ## \frac{x^2}{a^2}\ = sin^2(\omega t) ## and then insert ## sin^2(\omega t) ##
to the equation ##\frac{y}{a} =1-2sin^2(\omega t) ## . would it be correct?
first of all thanks for your help it's not taken for grantedYou seem to have a mental block about replacing ##\sin(\omega t) ## by ##\frac{x}{a}##.
It is!first of all thanks for your help it's not taken for granted
Last try:
in the end I will get : ## \frac{2x^2}{a^2}+ \frac{y}{a}=1 ## then : ## y= - \frac{2}{a}x^2+a ## and thats parabola figure . right ?