Question

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 (P,Q) lies on the circle of radius

• A. 1
• B. 0
• C. $\sqrt{2}$
• D. 4

Hint:

Sums of sines and cosines of angles like $2\pi k/n$ are often solved using the properties of the $n^{th}$ roots of unity. The fact that the sum of all $n^{th}$ roots of unity is zero is a very powerful tool.

Answer 1

The Correct Option is C

Step 1: Express as complex numbers.
\[ S = Q + iP = e^{i\frac{2\pi}{7}} + e^{i\frac{4\pi}{7}} + e^{i\frac{8\pi}{7}} \]
Step 2: Use roots of unity.
\[ \bar{S} = e^{-i\frac{2\pi}{7}} + e^{-i\frac{4\pi}{7}} + e^{-i\frac{8\pi}{7}} = e^{i3\frac{2\pi}{7}} + e^{i5\frac{2\pi}{7}} + e^{i6\frac{2\pi}{7}} \]
Step 3: Apply sum of roots of unity.
\[ 1 + S + \bar{S} = 0 \implies Q = -\frac12 \]
Step 4: Compute radius.
\[ |S|^2 = S\bar{S} = 3 + 2(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}) = 2 \] \[ \text{Radius} = |S| = \sqrt{2} \]