If $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$, then $\frac{\mathrm{d}^2…
- $\frac{3}{2} \tan \frac{3 \theta}{2}$
- $\frac{3}{2} \sec \frac{3 \theta}{2} \tan \frac{3 \theta}{2}$
- $\frac{3}{2} \sec ^2 \frac{3 \theta}{2}$
- $\sec ^2 \frac{3 \theta}{2}$
Solution
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: $\frac{3}{2} \sec ^2 \frac{3 \theta}{2}$
Asked in: MHT CET 2024 (15 May Shift 1)