Question

\(\frac{(cos2θ+isin2θ)^7}{(cos4θ+isin4θ)^3} =\)

• A. \(cos2θ+isin2θ\)
• B. \((cos2θ+isin2θ)^4\)
• C. \(cos4θ+isin4θ\)
• D. \((cos4θ+isin4θ)^{-2}\)

Hint:

Reduce the powers of omega using omega cubed = 1 and 1 + omega + omega squared = 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
If \(\omega\) is a complex cube root of unity, then \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

Step 2: Reduce the powers:
\(\omega^{10} = \omega^{9}\cdot\omega = \omega\) and \(\omega^{23} = \omega^{21}\cdot\omega^2 = \omega^2\).
So \(\omega^{10} + \omega^{23} = \omega + \omega^2 = -1\).

Step 3: Evaluate the sine:
\[ \sin\left[\pi(-1) - \frac{\pi}{4}\right] = -\sin\left(\pi + \frac{\pi}{4}\right) = -\left(-\sin\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Here \(\sin(\pi + \theta) = -\sin\theta\) was used. A negative answer would result if the sign of the inner angle were mishandled.

Final Answer:
The value is \(\frac{1}{\sqrt{2}}\), option (C). \[ \boxed{\frac{1}{\sqrt{2}}} \]