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To find the equation of the tangent line to the curve \( y = 6 \sin(x) \) at the point \( \left( \frac{\pi}{6}, 3 \right) \), we follow these steps:

### Step 1: Differentiate the function to find the slope of the tangent line.
The function is \( y = 6 \sin(x) \). The derivative \( y' \) gives the slope of the tangent line at any point \( x \).
\[
\frac{dy}{dx} = \frac{d}{dx} \left( 6 \sin(x) \right) = 6 \cos(x)
\]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{6} \).
To find the slope at \( \left( \frac{\pi}{6}, 3 \right) \), substitute \( x = \frac{\pi}{6} \) into the derivative:
\[
\text{slope} = 6 \cos\left( \frac{\pi}{6} \right)
\]
Since \( \cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \), the slope becomes:
\[
\text{slope} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}
\]
Thus, the slope \( m = 3\sqrt{3} \).

### Step 3: Use the point-slope form to find the equation of the tangent line.
The point-slope form of a line is:
\[
y - y_1 = m(x - x_1)
\]
We know \( m = 3\sqrt{3} \) and the point \( \left( x_1, y_1 \right) = \left( \frac{\pi}{6}, 3 \right) \). Substituting these values:
\[
y - 3 = 3\sqrt{3}\left(x - \frac{\pi}{6}\right)
\]

### Step 4: Solve for \( y \) in the slope-intercept form \( y = mx + b \).
Expand the equation:
\[
y - 3 = 3\sqrt{3}x - 3\sqrt{3}\cdot\frac{\pi}{6}
\]
\[
y - 3 = 3\sqrt{3}x - \frac{\pi\sqrt{3}}{2}
\]
Now, add 3 to both sides to isolate \( y \):
\[
y = 3\sqrt{3}x - \frac{\pi\sqrt{3}}{2} + 3
\]

Thus, the equation of the tangent line is:
\[
y = 3\sqrt{3}x + \left( 3 - \frac{\pi\sqrt{3}}{2} \right)
\]

So, \( m = 3\sqrt{3} \) and \( b = 3 - \frac{\pi\sqrt{3}}{2} \).

Would you like more details on any of the steps?

Here are five related questions to further your understanding:
1. What is the geometric meaning of the slope of the tangent line?
2. How would the equation change if the function were \( y = 6 \cos(x) \) instead?
3. What happens to the tangent line equation if we change the point of tangency?
4. Can we generalize this process to any sinusoidal function, such as \( y = A \sin(x) \)?
5. How does the value of \( b \) change if the point of tangency is altered?

**Tip:** In calculus, the derivative of a function at a point gives the slope of the tangent line at that specific point, which helps in constructing the tangent equation.

Find the equation of the tangent line to the curve y = 6 sin(x) at the point (π/6, 3). The equation of this tangent line can be written in the form y = mx + b where m = __ and b = __.

Learn how to find the equation of the tangent line to the curve y = 6 sin(x) at the point (π/6, 3). This problem involves key calculus concepts like differentiation and the derivative as the slope of the tangent line. Suitable for students in Grades 11-12.

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