Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=cos2theta+2costheta,y=sin2theta-2sintheta`.

Answer

To find \(\frac{dy}{dx}\) when \(x\) and \(y\) are connected parametrically by the equations \(x = \cos(2\theta) + 2\cos(\theta)\) and \(y = \sin(2\theta) - 2\sin(\theta)\), we will use the formula for parametric differentiation:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

### Step 1: Differentiate \(y\) with respect to \(\theta\)

Given:
\[ y = \sin(2\theta) - 2\sin(\theta) \]

Using the chain rule, we differentiate \(y\):
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin(2\theta)) - \frac{d}{d\theta}(2\sin(\theta)) \]

Calculating each derivative:
1. \(\frac{d}{d\theta}(\sin(2\theta)) = 2\cos(2\theta)\)
2. \(\frac{d}{d\theta}(2\sin(\theta)) = 2\cos(\theta)\)

Thus,
\[ \frac{dy}{d\theta} = 2\cos(2\theta) - 2\cos(\theta) \]

### Step 2: Differentiate \(x\) with respect to \(\theta\)

Given:
\[ x = \cos(2\theta) + 2\cos(\theta) \]

Using the chain rule, we differentiate \(x\):
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos(2\theta)) + \frac{d}{d\theta}(2\cos(\theta)) \]

Calculating each derivative:
1. \(\frac{d}{d\theta}(\cos(2\theta)) = -2\sin(2\theta)\)
2. \(\frac{d}{d\theta}(2\cos(\theta)) = -2\sin(\theta)\)

Thus,
\[ \frac{dx}{d\theta} = -2\sin(2\theta) - 2\sin(\theta) \]

### Step 3: Substitute into the formula for \(\frac{dy}{dx}\)

Now, substituting \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) into the formula:
\[ \frac{dy}{dx} = \frac{2\cos(2\theta) - 2\cos(\theta)}{-2\sin(2\theta) - 2\sin(\theta)} \]

### Step 4: Simplify the expression

We can factor out the common factor of \(2\) in both the numerator and the denominator:
\[ \frac{dy}{dx} = \frac{2(\cos(2\theta) - \cos(\theta))}{-2(\sin(2\theta) + \sin(\theta))} \]

This simplifies to:
\[ \frac{dy}{dx} = \frac{\cos(2\theta) - \cos(\theta)}{-(\sin(2\theta) + \sin(\theta))} \]

Thus, the final result is:
\[ \frac{dy}{dx} = -\frac{\cos(2\theta) - \cos(\theta)}{\sin(2\theta) + \sin(\theta)} \]