y=5sin^2 (x) find the equation of the tangent at x= pi/6

We are tasked with finding the equation of the tangent line to the curve \( y = 5 \sin^2(x) \) at \( x = \frac{\pi}{6} \). To do this, we need to:

1. Differentiate \( y \) with respect to \( x \) to find the slope of the tangent line at \( x = \frac{\pi}{6} \).
2. Use the point-slope form of the tangent line equation to find the equation of the tangent.

### Step 1: Differentiate \( y = 5 \sin^2(x) \)

First, rewrite the equation using a chain rule-friendly form:
\[
y = 5 (\sin(x))^2
\]
Now apply the chain rule to differentiate:
\[
\frac{dy}{dx} = 5 \cdot 2 \sin(x) \cdot \cos(x) = 10 \sin(x) \cos(x)
\]
Using the double-angle identity \( \sin(2x) = 2 \sin(x) \cos(x) \), we simplify the derivative:
\[
\frac{dy}{dx} = 5 \sin(2x)
\]

### Step 2: Find the slope at \( x = \frac{\pi}{6} \)

Substitute \( x = \frac{\pi}{6} \) into the derivative to find the slope of the tangent line:
\[
\frac{dy}{dx}\bigg|_{x = \frac{\pi}{6}} = 5 \sin\left(2 \cdot \frac{\pi}{6}\right) = 5 \sin\left(\frac{\pi}{3}\right) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}
\]
Thus, the slope of the tangent line is \( \frac{5\sqrt{3}}{2} \).

### Step 3: Find the point on the curve at \( x = \frac{\pi}{6} \)

We now find the \( y \)-coordinate of the point where \( x = \frac{\pi}{6} \):
\[
y = 5 \sin^2\left(\frac{\pi}{6}\right) = 5 \left(\frac{1}{2}\right)^2 = 5 \cdot \frac{1}{4} = \frac{5}{4}
\]
So the point on the curve is \( \left(\frac{\pi}{6}, \frac{5}{4}\right) \).

### Step 4: Use the point-slope form to find the equation of the tangent

The point-slope form of a line is:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the curve.

Substitute the slope \( \frac{5\sqrt{3}}{2} \) and the point \( \left(\frac{\pi}{6}, \frac{5}{4}\right) \) into the formula:
\[
y - \frac{5}{4} = \frac{5\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right)
\]
This is the equation of the tangent line.

Would you like me to simplify this further or provide additional details?

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Here are some related questions:
1. How do we use the product rule for differentiating trigonometric functions?
2. What is the geometric meaning of a tangent line to a curve?
3. How would the tangent equation change if \( x = \frac{\pi}{3} \) instead?
4. What is the significance of the double-angle identity in simplifying derivatives?
5. How do we find the normal line to a curve at a given point?

**Tip:** When differentiating trigonometric functions, identities like the double-angle identity can often simplify your work.

y = 5sin^2(x) find the equation of the tangent at x = pi/6

Learn how to find the tangent line to the curve y = 5sin^2(x) at x = pi/6 using differentiation and the point-slope form. This solution covers key trigonometric identities and calculus concepts. Suitable for students in Grades 10-12.

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