If `x=2cost-cos2t,y=2sint-sin2t," then "(dy)/(dx)=`

Options

A. `tant`
B. `cot((3t)/(2))`
C. `tan((3t)/(2))`
D. `cott`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 2\cos t - \cos 2t\) and \(y = 2\sin t - \sin 2t\), we will follow these steps:

### Step 1: Differentiate \(y\) with respect to \(t\)
We start by differentiating \(y\) with respect to \(t\).

\[ y = 2\sin t - \sin 2t \]

Using the chain rule, we differentiate:

\[ \frac{dy}{dt} = 2\cos t - \frac{d}{dt}(\sin 2t) \]

Using the chain rule for \(\sin 2t\):

\[ \frac{d}{dt}(\sin 2t) = 2\cos 2t \]

So, we have:

\[ \frac{dy}{dt} = 2\cos t - 2\cos 2t \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Next, we differentiate \(x\) with respect to \(t\).

\[ x = 2\cos t - \cos 2t \]

Differentiating:

\[ \frac{dx}{dt} = -2\sin t + \frac{d}{dt}(\cos 2t) \]

Using the chain rule for \(\cos 2t\):

\[ \frac{d}{dt}(\cos 2t) = -2\sin 2t \]

So, we have:

\[ \frac{dx}{dt} = -2\sin t + 2\sin 2t \]

### Step 3: Find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the formula:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{2\cos t - 2\cos 2t}{-2\sin t + 2\sin 2t} \]

### Step 4: Simplify the expression
We can simplify this expression:

\[ \frac{dy}{dx} = \frac{2(\cos t - \cos 2t)}{2(-\sin t + \sin 2t)} \]

The \(2\) in the numerator and denominator cancels out:

\[ \frac{dy}{dx} = \frac{\cos t - \cos 2t}{-\sin t + \sin 2t} \]

### Final Result
Thus, the final result for \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t} \]

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