If `2 cos theta=x+(1)/(x)`, find the value of `2cos 3 theta`

Options

A. `x^(3)+(1)/(x^(3))`
B. `x^(2)+(1)/(x^(2))`
C. `x^(3)-(1)/(x^(3))`
D. `sqrt(x+(1)/(x))`

Answer

To solve the problem, we start with the given equation:

\[ 2 \cos \theta = x + \frac{1}{x} \]

We need to find the value of \(2 \cos 3\theta\). We can use the triple angle formula for cosine, which states:

\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]

Thus, we can express \(2 \cos 3\theta\) as:

\[ 2 \cos 3\theta = 2(4 \cos^3 \theta - 3 \cos \theta) \]

This simplifies to:

\[ 2 \cos 3\theta = 8 \cos^3 \theta - 6 \cos \theta \]

Next, we need to express \(\cos \theta\) in terms of \(x\). From the first equation, we can solve for \(\cos \theta\):

\[ \cos \theta = \frac{x + \frac{1}{x}}{2} \]

Now, we will substitute this expression into the equation for \(2 \cos 3\theta\):

1. First, calculate \(\cos^3 \theta\):

\[ \cos^3 \theta = \left(\frac{x + \frac{1}{x}}{2}\right)^3 = \frac{(x + \frac{1}{x})^3}{8} \]

2. Expanding \((x + \frac{1}{x})^3\) using the binomial theorem:

\[ (x + \frac{1}{x})^3 = x^3 + 3x^2 \cdot \frac{1}{x} + 3x \cdot \frac{1}{x^2} + \frac{1}{x^3} = x^3 + 3x + 3\frac{1}{x} + \frac{1}{x^3} \]

3. Thus, we have:

\[ \cos^3 \theta = \frac{x^3 + 3x + 3\frac{1}{x} + \frac{1}{x^3}}{8} \]

4. Now substituting this into the equation for \(2 \cos 3\theta\):

\[ 2 \cos 3\theta = 8 \left(\frac{x^3 + 3x + 3\frac{1}{x} + \frac{1}{x^3}}{8}\right) - 6 \left(\frac{x + \frac{1}{x}}{2}\right) \]

5. Simplifying this gives:

\[ 2 \cos 3\theta = x^3 + 3x + 3\frac{1}{x} + \frac{1}{x^3} - 3(x + \frac{1}{x}) \]

6. Further simplifying:

\[ 2 \cos 3\theta = x^3 + 3x + 3\frac{1}{x} + \frac{1}{x^3} - 3x - 3\frac{1}{x} \]

7. This simplifies to:

\[ 2 \cos 3\theta = x^3 - 3x + \frac{1}{x^3} - 3\frac{1}{x} \]

Thus, we have found the value of \(2 \cos 3\theta\) in terms of \(x\):

\[ 2 \cos 3\theta = x^3 - 3x + \frac{1}{x^3} - 3\frac{1}{x} \]