JEE MainMathematics

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Sum of Cosines of Angles in Arithmetic Progression

Q: Why multiply specifically by 2sin(π/7)?

Because the product 2cos(kθ)sin(θ/2) = sin((k + 1/2)θ) - sin((k - 1/2)θ) forms a telescoping sum where intermediate terms cancel out.

Q: Why does α^6 + α = 2cos(2π/7)?

Since α = e^iθ, α^6 = e^(i(6θ)) = e^(i(2π - θ)) = e^(-iθ) = α bar. The sum of a complex number and its conjugate is 2Re(α) = 2cosθ.

The value of cos ((2 π)/7)+cos ((4 π)/7)+cos ((6 π)/7) is equal to:

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A-1B-1/2C-1/3D-1/4

Step-by-step solutionView

Correct answer

The sum $\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ equals $- \frac{1}{2}$ .

Option analysis

Why each option works or fails

A · -1

Believing the sum of the three cosines directly equals the sum of the non-trivial roots of unity, forgetting to divide by $2$ after pairing complex conjugates. Recall that $\sum_{k = 1}^{6} e^{i \frac{2 k π}{7}} = - 1$ . Because $\cos (\frac{2 k π}{7}) = \cos (\frac{2 (7 - k) π}{7})$ , the sum of the three cosines is half of the real part of this sum, giving $- \frac{1}{2}$ .

B · -1/2

Correct choice. Multiplying the sum $S$ by $2 \sin (\frac{π}{7})$ produces a telescoping difference $\sin (π) - \sin (\frac{π}{7}) = - \sin (\frac{π}{7})$ , which yields $S = - \frac{1}{2}$ .

C · -1/3

Assuming that averaging three cosine terms with sum $- 1$ yields a denominator of $3$ . Do not divide the sum of roots by the number of terms; use the exact trigonometric identity or conjugate pairing to find the coefficient.

D · -1/4

Introducing an extra factor of $\frac{1}{2}$ when converting the product to sum or applying the half-angle formula. Carefully track the coefficient $2 \sin (\frac{π}{7})$ used to telescope the terms: $2 S \sin (\frac{π}{7}) = - \sin (\frac{π}{7})$ , so $S = - \frac{1}{2}$ .

Reviewed route

Solution

StepWorking

01Given

The expression to evaluate is $S = \cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ .

02Goal

Compute the exact numerical value of $S$ .

03Approach

The angles form an AP with first term $a = \frac{2 π}{7}$ , common difference $d = \frac{2 π}{7}$ , and number of terms $n = 3$ . We multiply and divide by $2 \sin (\frac{d}{2}) = 2 \sin (\frac{π}{7})$ to apply the telescoping formula $\sum_{k = 1}^{n} \cos (a + (k - 1) d) = \frac{\sin (n d / 2)}{\sin (d / 2)} \cos (a + \frac{(n - 1) d}{2})$ .

04Execute

Applying the formula: $S = \frac{\sin (3 \times \frac{π}{7})}{\sin (\frac{π}{7})} \cos (\frac{\frac{2 π}{7} + \frac{6 π}{7}}{2}) = \frac{\sin (\frac{3 π}{7}) \cos (\frac{4 π}{7})}{\sin (\frac{π}{7})}$ . Multiplying numerator and denominator by 2 gives: $S = \frac{2 \sin (\frac{3 π}{7}) \cos (\frac{4 π}{7})}{2 \sin (\frac{π}{7})} = \frac{\sin (π) + \sin (- \frac{π}{7})}{2 \sin (\frac{π}{7})} = \frac{0 - \sin (\frac{π}{7})}{2 \sin (\frac{π}{7})} = - \frac{1}{2}$ .

✓Verify

Alternatively, consider the 7th roots of unity: $1 + \sum_{k = 1}^{6} e^{i \frac{2 k π}{7}} = 0$ . Taking the real part, $1 + 2 \sum_{k = 1}^{3} \cos (\frac{2 k π}{7}) = 0$ , which yields $S = - \frac{1}{2}$ . This confirms the result.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

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Students also ask

Why multiply specifically by

2 \sin (π / 7)

Because the product $2 \cos (k θ) \sin (θ / 2) = \sin ((k + 1 / 2) θ) - \sin ((k - 1 / 2) θ)$ forms a telescoping sum where intermediate terms cancel out.

Why does

α^{6} + α = 2 \cos (2 π / 7)

Since $α = e^{i θ}$ , $α^{6} = e^{i (6 θ)} = e^{i (2 π - θ)} = e^{- i θ} = \overline{α}$ . The sum of a complex number and its conjugate is $2 Re (α) = 2 \cos θ$ .

Answer

The sum $\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ equals $- \frac{1}{2}$ .

Why each option works or fails

A: -1 - Believing the sum of the three cosines directly equals the sum of the non-trivial roots of unity, forgetting to divide by $2$ after pairing complex conjugates. Recall that $\sum_{k = 1}^{6} e^{i \frac{2 k π}{7}} = - 1$ . Because $\cos (\frac{2 k π}{7}) = \cos (\frac{2 (7 - k) π}{7})$ , the sum of the three cosines is half of the real part of this sum, giving $- \frac{1}{2}$ .
B · correct: -1/2 - Correct choice. Multiplying the sum $S$ by $2 \sin (\frac{π}{7})$ produces a telescoping difference $\sin (π) - \sin (\frac{π}{7}) = - \sin (\frac{π}{7})$ , which yields $S = - \frac{1}{2}$ .
C: -1/3 - Assuming that averaging three cosine terms with sum $- 1$ yields a denominator of $3$ . Do not divide the sum of roots by the number of terms; use the exact trigonometric identity or conjugate pairing to find the coefficient.
D: -1/4 - Introducing an extra factor of $\frac{1}{2}$ when converting the product to sum or applying the half-angle formula. Carefully track the coefficient $2 \sin (\frac{π}{7})$ used to telescope the terms: $2 S \sin (\frac{π}{7}) = - \sin (\frac{π}{7})$ , so $S = - \frac{1}{2}$ .

Step-by-step solution

Given: The expression to evaluate is $S = \cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ .
Goal: Compute the exact numerical value of $S$ .
Approach: The angles form an AP with first term $a = \frac{2 π}{7}$ , common difference $d = \frac{2 π}{7}$ , and number of terms $n = 3$ . We multiply and divide by $2 \sin (\frac{d}{2}) = 2 \sin (\frac{π}{7})$ to apply the telescoping formula $a = \frac{2 π}{7}$ .
Execute: Applying the formula: $S = \frac{\sin (3 \times \frac{π}{7})}{\sin (\frac{π}{7})} \cos (\frac{\frac{2 π}{7} + \frac{6 π}{7}}{2}) = \frac{\sin (\frac{3 π}{7}) \cos (\frac{4 π}{7})}{\sin (\frac{π}{7})}$ . Multiplying numerator and denominator by 2 gives: $S = \frac{2 \sin (\frac{3 π}{7}) \cos (\frac{4 π}{7})}{2 \sin (\frac{π}{7})} = \frac{\sin (π) + \sin (- \frac{π}{7})}{2 \sin (\frac{π}{7})} = \frac{0 - \sin (\frac{π}{7})}{2 \sin (\frac{π}{7})} = - \frac{1}{2}$ .
Verify: Alternatively, consider the 7th roots of unity: $1 + \sum_{k = 1}^{6} e^{i \frac{2 k π}{7}} = 0$ . Taking the real part, $1 + 2 \sum_{k = 1}^{3} \cos (\frac{2 k π}{7}) = 0$ , which yields $S = - \frac{1}{2}$ . This confirms the result.

Shortcut:

When to use it: When angles are multiples of $2 π / n$ , using complex roots of unity gives the answer in two lines.

Given: The expression $S = \cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ .

Goal: Evaluate $S$ using properties of roots of unity.

Approach: Let $α = e^{i \frac{2 π}{7}}$ , which is a primitive 7th root of unity. The sum of all 7th roots of unity is $1 + α + α^{2} + α^{3} + α^{4} + α^{5} + α^{6} = 0$ .

Execute: Notice that $α^{6} = \overline{α}$ , $α^{5} = \overline{α^{2}}$ , and $α^{4} = \overline{α^{3}}$ . Taking the real part of $1 + \sum_{k = 1}^{6} α^{k} = 0$ : $1 + 2 [\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})] = 0 ⟹ 1 + 2 S = 0 ⟹ S = - \frac{1}{2}$ .

Verify: The polynomial equation $x^{7} - 1 = (x - 1) (x^{6} + x^{5} + \dots + 1) = 0$ confirms the sum of roots is zero, exactly verifying $1 + 2 S = 0$ .

Sum of Cosines of Angles in Arithmetic Progression

What feels right?

The sum $\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$ equals $- \frac{1}{2}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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