Question

If sin \( \theta \) = \( \frac{1}{2}\left(x + \frac{1}{x}\right) \), then sin \( 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) \) =}

• A. 0
• B. 1
• C. \( \frac{1}{4} \)
• D. 2

Hint:

$x + 1/x$ is always $\ge 2$ or $\le -2$. This often forces $x$ to be 1 or $-1$ in trig equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question provides a relationship between \(\sin\theta\) and a variable \(x\), and asks for the value of an expression involving \(\sin(3\theta)\) and \(x^3\). The key is to find a relationship between \(\frac{1}{2}\left(x^3 + \frac{1}{x^3}\right)\) and \(\sin(3\theta)\) using the given equation.
Step 2: Key Formula or Approach:
We will use the given equation and the cubic expansion formula \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). We will also use the triple angle identity for sine: \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\).
Step 3: Detailed Explanation:
We are given:
\[ \sin\theta = \frac{1}{2}\left(x + \frac{1}{x}\right) \implies 2\sin\theta = x + \frac{1}{x} \] Let's cube both sides of this equation:
\[ (2\sin\theta)^3 = \left(x + \frac{1}{x}\right)^3 \] \[ 8\sin^3\theta = x^3 + \left(\frac{1}{x}\right)^3 + 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) \] \[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 3(1)\left(x + \frac{1}{x}\right) \] We know that \(x + \frac{1}{x} = 2\sin\theta\). Substitute this back into the equation:
\[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 3(2\sin\theta) \] \[ 8\sin^3\theta = \left(x^3 + \frac{1}{x^3}\right) + 6\sin\theta \] Now, isolate the term \(\left(x^3 + \frac{1}{x^3}\right)\):
\[ x^3 + \frac{1}{x^3} = 8\sin^3\theta - 6\sin\theta \] The expression we need to evaluate is \( \sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) \).
Let's find the value of \( \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) \):
\[ \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = \frac{1}{2}(8\sin^3\theta - 6\sin\theta) = 4\sin^3\theta - 3\sin\theta \] This expression is related to the triple angle formula for sine. Recall:
\[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \] Therefore,
\[ 4\sin^3\theta - 3\sin\theta = -(3\sin\theta - 4\sin^3\theta) = -\sin(3\theta) \] So, we have found that:
\[ \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = -\sin(3\theta) \] Now, substitute this into the original expression we need to find:
\[ \sin 3\theta + \frac{1}{2}\left(x^3 + \frac{1}{x^3}\right) = \sin 3\theta + (-\sin 3\theta) = 0 \] Step 4: Final Answer:
The value of the expression is 0.