Finding the Value of Theta for Maximum Trigonometric Expression

The question asks for the value of $\theta$ in the interval $0 < \theta < \frac{\pi}{2}$ that maximizes the expression $f(\theta) = \sin \theta + \sin \theta \cos \theta$. To find the maximum value of a function, we can use calculus by finding the critical points. Critical points occur where the first derivative of the function is zero or undefined. The function $f(\theta)$ is differentiable for all $\theta$ in the given interval.

Steps to Find the Maximum Value

We will follow these steps:

1. Find the first derivative of the function $f(\theta)$ with respect to $\theta$.
2. Set the first derivative equal to zero and solve for $\theta$ to find the critical points.
3. Check which critical points fall within the given interval $0 < \theta < \frac{\pi}{2}$.
4. Determine if the critical point corresponds to a maximum value within the interval.

Calculating the First Derivative

The function is $f(\theta) = \sin \theta + \sin \theta \cos \theta$. We use the sum rule and the product rule for differentiation:

$\frac{d}{d\theta}(\sin \theta) = \cos \theta$

$\frac{d}{d\theta}(\sin \theta \cos \theta) = \frac{d}{d\theta}(\sin \theta) \cos \theta + \sin \theta \frac{d}{d\theta}(\cos \theta)$

= $(\cos \theta) \cos \theta + \sin \theta (-\sin \theta)$

= $\cos^2 \theta - \sin^2 \theta$

So, the first derivative $f'(\theta)$ is:

$f'(\theta) = \cos \theta + \cos^2 \theta - \sin^2 \theta$

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we can rewrite $f'(\theta)$ in terms of $\cos \theta$ only:

$f'(\theta) = \cos \theta + \cos^2 \theta - (1 - \cos^2 \theta)$

$f'(\theta) = \cos \theta + \cos^2 \theta - 1 + \cos^2 \theta$

$f'(\theta) = 2\cos^2 \theta + \cos \theta - 1$

Finding Critical Points

To find the critical points, we set the first derivative equal to zero:

$f'(\theta) = 0$

$2\cos^2 \theta + \cos \theta - 1 = 0$

Let $x = \cos \theta$. The equation becomes a quadratic equation in $x$:

$2x^2 + x - 1 = 0$

We can factor this quadratic equation:

$(2x - 1)(x + 1) = 0$

This gives two possible values for $x$:

• $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
• $x + 1 = 0 \Rightarrow x = -1$

Substituting back $x = \cos \theta$, we get:

• $\cos \theta = \frac{1}{2}$
• $\cos \theta = -1$

Evaluating Critical Points in the Given Interval

The given interval for $\theta$ is $0 < \theta < \frac{\pi}{2}$.

For $\cos \theta = -1$, the principal value is $\theta = \pi$. This value is outside the interval $0 < \theta < \frac{\pi}{2}$.

For $\cos \theta = \frac{1}{2}$, the principal value in the interval $[0, \pi]$ is $\theta = \frac{\pi}{3}$. This value $\theta = \frac{\pi}{3}$ is in the interval $0 < \theta < \frac{\pi}{2}$ (since $0 < \frac{\pi}{3} < \frac{\pi}{2}$).

Thus, $\theta = \frac{\pi}{3}$ is the only critical point within the specified interval.

Determining if it's a Maximum

We can check if $\theta = \frac{\pi}{3}$ corresponds to a maximum by analyzing the sign of the first derivative $f'(\theta) = 2\cos^2 \theta + \cos \theta - 1$ around $\theta = \frac{\pi}{3}$ within the interval $0 < \theta < \frac{\pi}{2}$.

• Consider a value of $\theta$ slightly less than $\frac{\pi}{3}$, e.g., $\theta = \frac{\pi}{4}$. $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \approx 0.707$. $f'(\frac{\pi}{4}) = 2(\frac{\sqrt{2}}{2})^2 + \frac{\sqrt{2}}{2} - 1 = 2(\frac{2}{4}) + \frac{\sqrt{2}}{2} - 1 = 1 + \frac{\sqrt{2}}{2} - 1 = \frac{\sqrt{2}}{2} > 0$. Since $f'(\theta) > 0$ for $\theta < \frac{\pi}{3}$, the function is increasing.
• Consider a value of $\theta$ slightly greater than $\frac{\pi}{3}$, e.g., $\theta = \frac{\pi}{2}$ (though the interval is open, we can consider approaching it, or pick $\theta$ very close to $\pi/3$ like $2\pi/5$). Let's pick $\theta$ between $\pi/3$ and $\pi/2$. Say $\cos \theta = 0.3$ (since $\cos(\pi/3)=0.5$ and $\cos(\pi/2)=0$). $f'(\theta) = 2(0.3)^2 + 0.3 - 1 = 0.18 + 0.3 - 1 = -0.52 < 0$. Since $f'(\theta) < 0$ for $\theta > \frac{\pi}{3}$, the function is decreasing.

Since the function $f(\theta)$ is increasing before $\theta = \frac{\pi}{3}$ and decreasing after $\theta = \frac{\pi}{3}$ in the interval $0 < \theta < \frac{\pi}{2}$, the critical point $\theta = \frac{\pi}{3}$ corresponds to a local maximum. As it is the only critical point in the open interval, it represents the absolute maximum within this interval.

Conclusion

The value of $\theta$ in the interval $0 < \theta < \frac{\pi}{2}$ for which the expression $\sin \theta + \sin \theta \cos \theta$ has a maximum value is $\theta = \frac{\pi}{3}$.

Revision Table: Key Concepts

Additional Information: Trigonometric Optimization

Trigonometric optimization problems often involve finding the maximum or minimum values of functions that include trigonometric terms. Calculus is a powerful tool for solving such problems. The process typically involves differentiation to find critical points. It's crucial to pay attention to the given interval, as the maximum or minimum might occur at a critical point within the interval or at the boundaries of a closed interval.

In this specific problem, the function is $f(\theta) = \sin \theta + \sin \theta \cos \theta$. Another approach to potentially simplify the function before differentiation could be to use trigonometric identities, but in this case, direct differentiation was straightforward and led to a solvable quadratic in $\cos \theta$. Recognizing the derivative as a quadratic in $\cos \theta$ is a key step.

The interval $0 < \theta < \frac{\pi}{2}$ is the first quadrant, where $\sin \theta$ and $\cos \theta$ are both positive. The critical point $\theta = \frac{\pi}{3}$ is a standard angle whose cosine value ($\frac{1}{2}$) is well-known, simplifying the verification steps.