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Gauss Sum cos2π/7+cos4π/7+cos8π/7 Value | JEE Complex

JEE Maths question with a full step-by-step solution.

Question

Let

C = \cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{8\pi}{7}

and

S = \sin\dfrac{2\pi}{7} + \sin\dfrac{4\pi}{7} + \sin\dfrac{8\pi}{7}

, then

C = \dfrac{1}{2}

S = \dfrac{1}{2}

C = \dfrac{\sqrt{7}}{2}

S = \dfrac{\sqrt{7}}{2}

correct

Solution

Step 1: Combine

C

and

S

into one complex sum using

e^{i\theta} = \cos\theta + i\sin\theta

C + iS = e^{i2\pi/7} + e^{i4\pi/7} + e^{i8\pi/7}

Step 2: Let

\zeta = e^{i2\pi/7}

, a primitive 7th root of unity. Note that

e^{i8\pi/7} = e^{i2\pi\cdot 4/7} = \zeta^4

, so:

C + iS = \zeta + \zeta^2 + \zeta^4

Step 3: Use the fact that the seven 7th roots of unity sum to zero:

1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 = 0 \implies \zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 = -1

Split this into two triples,

y_1 = \zeta + \zeta^2 + \zeta^4

and

y_2 = \zeta^3 + \zeta^5 + \zeta^6

. Then:

y_1 + y_2 = -1

Step 4: Compute the product

y_1 y_2

. Expanding and reducing every exponent modulo

7

(using

\zeta^7 = 1

y_1 y_2 = (\zeta + \zeta^2 + \zeta^4)(\zeta^3 + \zeta^5 + \zeta^6)

= \zeta^4 + \zeta^6 + \zeta^7 + \zeta^5 + \zeta^7 + \zeta^8 + \zeta^7 + \zeta^9 + \zeta^{10}

= \zeta^4 + \zeta^6 + 1 + \zeta^5 + 1 + \zeta + 1 + \zeta^2 + \zeta^3 = 3 + (\zeta + \zeta^2 + \cdots + \zeta^6) = 3 + (-1) = 2

Step 5: So

y_1

and

y_2

are roots of the quadratic

t^2 - (y_1 + y_2)t + y_1 y_2 = 0

t^2 + t + 2 = 0 \implies t = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm i\sqrt{7}}{2}

Step 6: Since

y_1 = \zeta + \zeta^2 + \zeta^4

has positive imaginary part (the arguments

\tfrac{2\pi}{7}, \tfrac{4\pi}{7}, \tfrac{8\pi}{7}

give a net upward sum), we take:

C + iS = \frac{-1 + i\sqrt{7}}{2} \implies C = -\frac{1}{2}, \quad S = \frac{\sqrt{7}}{2}

Step 7: Compare with the options.

C = -\tfrac{1}{2}

rules out (A) and (C);

S = \tfrac{\sqrt{7}}{2}

matches (4). Correct answer: (4)

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