TS EAMCET · Maths · Trigonometric Ratios & Identities
If \(P=\sin \frac{2 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{8 \pi}{7}\) and \(Q=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\), then the point \((\mathrm{P}, \mathrm{Q})\) lies on the circle of radius
- A 1
- B 0
- C 2
- D 4
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\(Q+iP = e^{i\frac{2\pi}{7}} + e^{i\frac{4\pi}{7}} + e^{i\frac{8\pi}{7}}\) Let \( \omega_k = e^{i\frac{2k\pi}{7}} \). Then \( Q+iP = \omega_1 + \omega_2 + \omega_4 \). The sum of all 7th roots of unity is \( \sum_{k=0}^{6} \omega_k = 0 \).…
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