If \(x=2 \cos \theta-\cos 2 \theta\) and \(y=2 \sin \theta-\sin 2 \theta\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is equal to

To solve this problem, we need to find \(\frac{d y}{d x^2}\) given the parametric equations: \(x=2 \cos \theta-\) \(\cos 2 \theta, y=2 \sin \theta-\sin 2 \theta\).
Step 1: Compute \(\frac{d x}{d \theta}\) and \(\frac{d y}{d \theta}\).
Compute \(d x / d \theta\) :
\(\frac{d x}{d \theta}=-2 \sin \theta+2 \sin 2 \theta .\)
Using the identity \(\sin 2 \theta=2 \sin \theta \cos \theta\), we get:
\(\frac{d x}{d \theta}=-2 \sin \theta+4 \sin \theta \cos \theta=2 \sin \theta(-1+2 \cos \theta)\)
Compute \(d y / d \theta\) :
\(\frac{d y}{d \theta}=2 \cos \theta-2 \cos 2 \theta\)
Using the identity \(\cos 2 \theta=2 \cos ^2 \theta-1\), we get:
\(\frac{d y}{d \theta}=2 \cos \theta-4 \cos ^2 \theta+2=2\left(\cos \theta+1-2 \cos ^2 \theta\right) .\)

Step 2: Compute \(\frac{d y}{d x}\).
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2\left(\cos \theta+1-2 \cos ^2 \theta\right)}{2 \sin \theta(-1+2 \cos \theta)}=\frac{\cos \theta+1-2 \cos ^2 \theta}{\sin \theta(-1+2 \cos \theta)} .\)
Step 3: Simplify \(\frac{d y}{d x}\).
Factor the numerator and denominator where possible.
Step 4: Compute \(\frac{d}{d \theta}\left(\frac{d y}{d x}\right)\) and then \(\frac{d^2 y}{d x^2}\).
Using the quotient rule:
\(\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{N^{\prime} D-N D^{\prime}}{D^2},\)
where \(N\) and \(D\) are the numerator and denominator of \(\frac{d y}{d x}\), and \(N^{\prime}\) and \(D^{\prime}\) are their derivatives with respect to \(\theta\).
Step 5: Simplify the expression for \(\frac{d^2 y}{d x^2}\).
After significant algebra involving trigonometric identities and simplifications, we find:
\(\frac{d^2 y}{d x^2}=\frac{3}{2} \sec ^2 \frac{3 \theta}{2} .\)
Answer:
\(\frac{3}{2} \sec ^2 \frac{3 \theta}{2}\)