To find the equation of the tangent line to the graph of $f(\theta) = \tan \theta \sin \theta$ at $\theta = \frac{\pi}{4}$, we need to follow these steps:

1. Differentiate $f(\theta) = \tan \theta \sin \theta$ using the product rule (or quotient rule, if rewritten as a quotient).
2. Evaluate $f'(\theta)$ at $\theta = \frac{\pi}{4}$.
3. Find the point on the curve at $\theta = \frac{\pi}{4}$.
4. Construct the tangent line using the point-slope form of a line.

Step 1: Differentiate $f(\theta) = \tan \theta \sin \theta$

We can use the product rule for differentiation: $f'(\theta) = \frac{d}{d\theta} (\tan \theta) \cdot \sin \theta + \tan \theta \cdot \frac{d}{d\theta} (\sin \theta)$

The derivatives are: $\frac{d}{d\theta} (\tan \theta) = \sec^2 \theta \quad \text{and} \quad \frac{d}{d\theta} (\sin \theta) = \cos \theta$

Thus, $f'(\theta) = (\sec^2 \theta) \sin \theta + (\tan \theta) \cos \theta$

Step 2: Evaluate $f'(\theta)$ at $\theta = \frac{\pi}{4}$

At $\theta = \frac{\pi}{4}$:

• $\tan \frac{\pi}{4} = 1$
• $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$
• $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$
• $\sec^2 \frac{\pi}{4} = 2$ (since $\sec \frac{\pi}{4} = \sqrt{2}$)

Plugging these values into $f'(\theta)$: $f'\left( \frac{\pi}{4} \right) = (2) \cdot \frac{\sqrt{2}}{2} + (1) \cdot \frac{\sqrt{2}}{2} = \sqrt{2} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$

Step 3: Find the point on the curve at $\theta = \frac{\pi}{4}$

To find $f\left( \frac{\pi}{4} \right)$: $f\left( \frac{\pi}{4} \right) = \tan \frac{\pi}{4} \cdot \sin \frac{\pi}{4} = 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$

Thus, the point on the curve is $\left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \right)$.

Step 4: Write the equation of the tangent line

Using the point-slope form $y - y_1 = m(x - x_1)$ where $m = f'(\frac{\pi}{4})$ and $(x_1, y_1) = \left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \right)$: $y - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \left( \theta - \frac{\pi}{4} \right)$

This is the equation of the tangent line at $\theta = \frac{\pi}{4}$.

---

Would you like more details or have any questions about the solution?

Related Questions

1. How would the process change if $f(\theta) = \tan \theta + \sin \theta$ instead?
2. What is the slope of the tangent to $f(\theta) = \cos \theta \tan \theta$ at $\theta = \frac{\pi}{3}$?
3. Can we find a similar tangent line if $f(\theta) = \sin \theta \cdot \cos \theta$?
4. How would we differentiate $f(\theta) = \frac{\tan \theta}{\sin \theta}$ using the quotient rule?
5. What is the general method for finding tangent lines of trigonometric functions?

Tip:

When differentiating products like $\tan \theta \sin \theta$, consider using product or quotient rules as appropriate to simplify