If `f (theta)=4/3 (1- cos ^(6) theta - sin ^(6)theta),` then
`lim _(ntooo) 1/n [sqrt(f ((1)/(n)))+sqrt(f ((2)/(n)))+sqrt(f((n)/(n)))]=`

Options

A. `(1- cos 1)/(2)`
B. `1-cos 2`
C. `(sin 2)/(2)`
D. `(1- cos 2)/(2)`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} \frac{1}{n} \left[ \sqrt{f\left(\frac{1}{n}\right)} + \sqrt{f\left(\frac{2}{n}\right)} + \sqrt{f\left(\frac{n}{n}\right)} \right] \]

where \( f(\theta) = \frac{4}{3} \left( 1 - \cos^6 \theta - \sin^6 \theta \right) \).

### Step 1: Simplify \( f(\theta) \)

We start with the function:

\[ f(\theta) = \frac{4}{3} \left( 1 - \cos^6 \theta - \sin^6 \theta \right) \]

Using the identity \( \cos^6 \theta + \sin^6 \theta = (\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \) and knowing that \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ \cos^6 \theta + \sin^6 \theta = 1 \cdot \left( (\cos^2 \theta + \sin^2 \theta)^2 - 3 \cos^2 \theta \sin^2 \theta \right) = 1 - 3 \cos^2 \theta \sin^2 \theta \]

Thus, we can rewrite \( f(\theta) \):

\[ f(\theta) = \frac{4}{3} \left( 1 - (1 - 3 \cos^2 \theta \sin^2 \theta) \right) = \frac{4}{3} \cdot 3 \cos^2 \theta \sin^2 \theta = 4 \cos^2 \theta \sin^2 \theta \]

Using the double angle identity \( 2 \cos \theta \sin \theta = \sin(2\theta) \):

\[ f(\theta) = 4 \left( \frac{1}{2} \sin(2\theta) \right)^2 = \sin^2(2\theta) \]

### Step 2: Substitute into the limit

Now, we substitute \( f\left(\frac{k}{n}\right) \):

\[ f\left(\frac{k}{n}\right) = \sin^2\left(\frac{2k}{n}\right) \]

Thus, we have:

\[ \sqrt{f\left(\frac{k}{n}\right)} = \sin\left(\frac{2k}{n}\right) \]

The limit becomes:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sin\left(\frac{2k}{n}\right) \]

### Step 3: Recognize the Riemann Sum

This limit can be interpreted as a Riemann sum for the integral of \( \sin(2x) \) from 0 to 1:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sin\left(\frac{2k}{n}\right) \to \int_0^1 \sin(2x) \, dx \]

### Step 4: Evaluate the integral

Now we compute the integral:

\[ \int_0^1 \sin(2x) \, dx \]

Using the antiderivative of \( \sin(2x) \):

\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]

Evaluating from 0 to 1:

\[ \left[-\frac{1}{2} \cos(2x)\right]_0^1 = -\frac{1}{2} \cos(2) - \left(-\frac{1}{2} \cos(0)\right) = -\frac{1}{2} \cos(2) + \frac{1}{2} \]

This simplifies to:

\[ \frac{1}{2} - \frac{1}{2} \cos(2) = \frac{1 - \cos(2)}{2} \]

### Final Answer

Thus, the final answer is:

\[ \lim_{n \to \infty} \frac{1}{n} \left[ \sqrt{f\left(\frac{1}{n}\right)} + \sqrt{f\left(\frac{2}{n}\right)} + \sqrt{f\left(\frac{n}{n}\right)} \right] = \frac{1 - \cos(2)}{2} \]