Question

If $2 \cos \theta = x + \frac{1}{x}$, then $2 \cos 3\theta =$

• A. $x^3 - \frac{1}{x^3}$
• B. $\left(x + \frac{1}{x}\right)^3$
• C. $x + \frac{1}{x}$
• D. $x^3 + \frac{1}{x^3}$

Hint:

Remember the general complex number rule as a shortcut: if $2 \cos \theta = x + \frac{1}{x}$, then it generalizes directly to $2 \cos n\theta = x^n + \frac{1}{x^n}$ for any integer $n$. For $n=3$, you can write down $x^3 + \frac{1}{x^3}$ instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an algebraic-trigonometric identity relation $2 \cos \theta = x + \frac{1}{x}$. We need to find an equivalent expression for $2 \cos 3\theta$ in terms of powers of $x$.

Step 2: Key Formula or Approach:
We can solve this problem using De Moivre's Theorem from complex numbers, which states that if $x = \cos \theta + i \sin \theta$, then $\frac{1}{x} = \cos \theta - i \sin \theta$. Adding these expressions gives $x + \frac{1}{x} = 2 \cos \theta$. By extension, $x^n + \frac{1}{x^n} = 2 \cos n\theta$. Alternatively, we can use the standard trigonometric triple-angle identity: $$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$$ Let's solve it using the algebraic triple-angle expansion to show the step-by-step identity derivation.

Step 3: Detailed Explanation:
From the problem, we have: $$\cos \theta = \frac{1}{2}\left(x + \frac{1}{x}\right)$$ Substitute this into the triple-angle identity formula for $\cos 3\theta$: $$\cos 3\theta = 4\left[\frac{1}{2}\left(x + \frac{1}{x}\right)\right]^3 - 3\left[\frac{1}{2}\left(x + \frac{1}{x}\right)\right]$$ $$\cos 3\theta = 4 \cdot \frac{1}{8}\left(x + \frac{1}{x}\right)^3 - \frac{3}{2}\left(x + \frac{1}{x}\right)$$ $$\cos 3\theta = \frac{1}{2}\left(x + \frac{1}{x}\right)^3 - \frac{3}{2}\left(x + \frac{1}{x}\right)$$ Multiply the entire equation by 2 to isolate $2 \cos 3\theta$: $$2 \cos 3\theta = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$$ Now, expand the cubic term using the algebraic identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$: $$\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)$$ Substitute this expansion back into our equation for $2 \cos 3\theta$: $$2 \cos 3\theta = \left[x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)\right] - 3\left(x + \frac{1}{x}\right)$$ The linear terms $3\left(x + \frac{1}{x}\right)$ cancel out perfectly: $$2 \cos 3\theta = x^3 + \frac{1}{x^3}$$

Step 4: Final Answer:
The value of $2 \cos 3\theta$ is equal to $x^3 + \frac{1}{x^3}$, which corresponds to option (D).