The angle of intersection of the curves `y=2sin^2x` and y=cos2x at `x=pi/6` is:

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. None of these

Answer

To find the angle of intersection of the curves \( y = 2\sin^2 x \) and \( y = \cos 2x \) at \( x = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Differentiate the curves to find the slopes
We need to find the derivatives of both functions to get the slopes of the tangents at the point of intersection.

1. **For the first curve \( y = 2\sin^2 x \)**:
\[ \frac{dy}{dx} = 2 \cdot 2\sin x \cdot \cos x = 2\sin(2x) \]

2. **For the second curve \( y = \cos 2x \)**:
\[ \frac{dy}{dx} = -\sin(2x) \cdot 2 = -2\sin(2x) \]

### Step 2: Evaluate the derivatives at \( x = \frac{\pi}{6} \)
Now we will substitute \( x = \frac{\pi}{6} \) into the derivatives to find the slopes.

1. **For the first curve**:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{6}} = 2\sin\left(2 \cdot \frac{\pi}{6}\right) = 2\sin\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

2. **For the second curve**:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{6}} = -2\sin\left(2 \cdot \frac{\pi}{6}\right) = -2\sin\left(\frac{\pi}{3}\right) = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} \]

### Step 3: Use the slopes to find the angle of intersection
Let \( m_1 = \sqrt{3} \) and \( m_2 = -\sqrt{3} \). The formula for the angle \( \theta \) between two curves is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values:
\[ \tan \theta = \left| \frac{\sqrt{3} - (-\sqrt{3})}{1 + \sqrt{3} \cdot (-\sqrt{3})} \right| = \left| \frac{\sqrt{3} + \sqrt{3}}{1 - 3} \right| = \left| \frac{2\sqrt{3}}{-2} \right| = \left| -\sqrt{3} \right| = \sqrt{3} \]

### Step 4: Find the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}(\sqrt{3}) \]

Since \( \tan \frac{\pi}{3} = \sqrt{3} \), we have:
\[ \theta = \frac{\pi}{3} \text{ radians} = 60^\circ \]

Thus, the angle of intersection of the curves at \( x = \frac{\pi}{6} \) is \( 60^\circ \).