Question

If \(\cos\theta + i \sin\theta, \, \theta \in \mathbb{R}\), is a root of the equation
\[ a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1}x + a_n = 0, \, a_0, a_1, \ldots, a_n \in \mathbb{R}, \, a_0 \neq 0 \]
then the value of \(a_1 \sin\theta + a_2 \sin 2\theta + \cdots + a_n \sin n\theta\) is:

• A. \(2n\)
• B. \(n\)
• C. \(0\)
• D. \(n + 1\)

Hint:

When dealing with roots of complex numbers in polynomial equations, use Euler’s formula to convert the trigonometric terms into exponential form and simplify the equation.

Answer 1

The Correct Option is C

The given polynomial equation is:

\[ a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1}x + a_n = 0. \]

The root \( \cos \theta + i \sin \theta \) can be represented in exponential form using Euler’s formula:

\[ x = e^{i\theta}. \]

Substitute \( x = e^{i\theta} \) into the polynomial:

\[ a_0 (e^{i\theta})^n + a_1 (e^{i\theta})^{n-1} + \cdots + a_{n-1}(e^{i\theta}) + a_n = 0. \]

Simplify the powers of \( e^{i\theta} \):

\[ a_0 e^{in\theta} + a_1 e^{i(n-1)\theta} + \cdots + a_{n-1} e^{i\theta} + a_n = 0. \]

Separate the real and imaginary parts of the equation:

\[ \text{Real part: } a_0 \cos(n\theta) + a_1 \cos((n-1)\theta) + \cdots + a_{n-1} \cos(\theta) + a_n = 0, \] \[ \text{Imaginary part: } a_0 \sin(n\theta) + a_1 \sin((n-1)\theta) + \cdots + a_{n-1} \sin(\theta) = 0. \]

From the imaginary part:

\[ a_0 \sin(n\theta) + a_1 \sin((n-1)\theta) + \cdots + a_n \sin(0) = 0. \]

Since \( \sin(0) = 0 \), the term involving \( a_n \) vanishes. Therefore:

\[ a_1 \sin(\theta) + a_2 \sin(2\theta) + \cdots + a_n \sin(n\theta) = 0. \]

Conclusion: The value of \( a_1 \sin \theta + a_2 \sin 2\theta + \cdots + a_n \sin n\theta \) is:

\[ 0. \]