Find the equation of the tangent line to the curve \( y=2 \sin x \) at the point \( \left(\frac{\pi}{6}, 1\right) \). The equation of this tangent line can be written in the form \( y=m x+b \) where \( m=\square \) and \( b=\square \)

To find the equation of the tangent line to the curve \( y = 2 \sin x \) at the point \( \left(\frac{\pi}{6}, 1\right) \), we need to follow these steps: 1. **Find the derivative of the function** to determine the slope of the tangent line. 2. **Evaluate the derivative at the given point** to find the slope \( m \). 3. **Use the point-slope form of the line** to find the equation of the tangent line. ### Step 1: Find the derivative The function is given by: \[ y = 2 \sin x \] To find the derivative \( y' \): \[ y' = 2 \cos x \] ### Step 2: Evaluate the derivative at \( x = \frac{\pi}{6} \) Now we evaluate the derivative at the point \( x = \frac{\pi}{6} \): \[ y'\left(\frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{6}\right) \] We know that \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), so: \[ y'\left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \] Thus, the slope \( m \) of the tangent line at the point \( \left(\frac{\pi}{6}, 1\right) \) is: \[ m = \sqrt{3} \] ### Step 3: Use the point-slope form to find the equation of the tangent line The point-slope form of the line is given by: \[ y - y_1 = m(x - x_1) \] Substituting \( m = \sqrt{3} \), \( x_1 = \frac{\pi}{6} \), and \( y_1 = 1 \): \[ y - 1 = \sqrt{3}\left(x - \frac{\pi}{6}\right) \] Now, we can rearrange this into the slope-intercept form \( y = mx + b \): \[ y - 1 = \sqrt{3}x - \frac{\sqrt{3}\pi}{6} \] Adding 1 to both sides: \[ y = \sqrt{3}x - \frac{\sqrt{3}\pi}{6} + 1 \] ### Step 4: Identify \( b \) The equation is now in the form \( y = mx + b \), where: \[ b = 1 - \frac{\sqrt{3}\pi}{6} \] ### Final Result Thus, the values of \( m \) and \( b \) are: \[ m = \sqrt{3}, \quad b = 1 - \frac{\sqrt{3}\pi}{6} \] So, the final answer is: \[ m = \sqrt{3}, \quad b = 1 - \frac{\sqrt{3}\pi}{6} \]