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Use the preceding identities to evaluate the cosine-square sum.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM02

The square creates a constant 2 in each conjugate product.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Use the preceding identities to evaluate the cosine-square sum.

cos⁡22π7+cos⁡24π7+cos⁡26π7\cos^2\frac{2\pi}7+\cos^2\frac{4\pi}7+\cos^2\frac{6\pi}7

Official paper · jm02-2022 · 4(b)(iii) · PDF 6

Official original and suggested answers ↗ · Suggested answer PDF page 11

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Square (ωⁿ+ω⁻ⁿ)/2 for n=1,2,3.
Hint 2
Reduce negative exponents modulo seven.
Worked solution
  1. Expand each squared cosine.

    S=14∑n=13(ω2n+2+ω−2n)S=\frac14\sum_{n=1}^3(\omega^{2n}+2+\omega^{-2n})
  2. The six nonconstant powers are exactly ω,…,ω⁶ in a different order.

    S=14(6+ω+ω2+⋯+ω6)=14(6−1)=54S=\frac14(6+\omega+\omega^2+\cdots+\omega^6)=\frac14(6-1)=\frac54

5/4.

Checks and common pitfalls: The square creates a constant 2 in each conjugate product.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Expand each squared cosine.
    S=14∑n=13(ω2n+2+ω−2n)S=\frac14\sum_{n=1}^3(\omega^{2n}+2+\omega^{-2n})
  • The six nonconstant powers are exactly ω,…,ω⁶ in a different order.
    S=14(6+ω+ω2+⋯+ω6)=14(6−1)=54S=\frac14(6+\omega+\omega^2+\cdots+\omega^6)=\frac14(6-1)=\frac54

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Curriculum and source notes ↗