If x=sit t nad y=cos 2t prove that d^2y/dx^2+4=0
To prove that \(\frac{{d^2y}}{{dx^2}} + 4 = 0\) given \(x = \sin(t)\) and \(y = \cos(2t)\), we need to find the second derivative of \(y\) with respect to \(x\), and then substitute the given values for \(x\) and \(y\).
Let's start by finding the first derivative of \(y\) with respect to \(t\):
\(\frac{{dy}}{{dt}} = -2\sin(2t)\)
Next, we need to find \(\frac{{dx}}{{dt}}\) using the given equation \(x = \sin(t)\):
\(\frac{{dx}}{{dt}} = \cos(t)\)
Now, we can find the derivative of \(y\) with respect to \(x\) by applying the chain rule:
\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = -2\sin(2t) \cdot \frac{{1}}{{\cos(t)}} = -2\sin(2t) \cdot \sec(t)\)
To find the second derivative of \(y\) with respect to \(x\), we differentiate \(\frac{{dy}}{{dx}}\) with respect to \(t\) again:
\(\frac{{d^2y}}{{dx^2}} = \frac{{d}}{{dt}}\left(-2\sin(2t) \cdot \sec(t)\right)\)
To simplify this expression, we need to use some trigonometric identities. We know that \(\sec(t) = \frac{{1}}{{\cos(t)}}\) and \(\sin(2t) = 2\sin(t)\cos(t)\).
Substituting these identities into the previous expression, we get:
\(\frac{{d^2y}}{{dx^2}} = \frac{{d}}{{dt}}\left(-2 \cdot 2\sin(t) \cos(t) \cdot \frac{{1}}{{\cos(t)}}\right) = \frac{{d}}{{dt}}(-4\sin(t))\)
Taking the derivative of \(-4\sin(t)\) with respect to \(t\), we get:
\(\frac{{d^2y}}{{dx^2}} = -4\cos(t)\)
Finally, let's substitute \(x = \sin(t)\) and \(y = \cos(2t)\) into \(-4\cos(t)\):
\(-4\cos(t) = -4\cos(\arcsin(x)) = -4\sqrt{1 - x^2}\)
So, we have proven that \(\frac{{d^2y}}{{dx^2}} + 4 = 0\) given \(x = \sin(t)\) and \(y = \cos(2t)\).