Topic: Engineering Mathematics
$\sqrt[3]{\cos x-j~\sin x}$ is equal to
- A. $(\cos x)^{1/3}-j(\sin x)^{1/3}$
- B. $\sqrt[3]{\cos^{2}x-j~\sin^{2}x}$
- C. $\cos\frac{x}{3}-j~\sin\frac{x}{3}$
- D. $\sin\frac{x}{3}-j~\cos\frac{x}{3}$
Answer: C. $\cos\frac{x}{3}-j~\sin\frac{x}{3}$
Explanation: Using Euler's formula, $\cos x - j\sin x = e^{-jx}$.
Taking the cube root: $(e^{-jx})^{1/3} = e^{-jx/3}$.
Applying Euler's formula again: $e^{-jx/3} = \cos\left(\frac{-x}{3}\right) + j\sin\left(\frac{-x}{3}\right) = \cos\frac{x}{3} - j\sin\frac{x}{3}$.