Use induction to prove the identity for every positive integer n.

01 / Standard#Your turn

2\sin x[\cos x+\cos3x+\cdots+\cos(2n-1)x]=\sin2nx

Official paper · jm02-2021 · 4(c)(i) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Check n=1 using the double-angle sine identity.

Hint 2

In the induction step add 2sin x cos((2k+1)x).

Worked solution

The base case holds for every real x.
$n=1:\quad2\sin x\cos x=\sin2x$
Assume the statement for a fixed positive integer k.
$2\sin x\sum_{j=1}^k\cos(2j-1)x=\sin2kx$
The addition formulas give a telescoping increment.
$2\sin x\cos(2k+1)x=\sin(2k+2)x-\sin2kx$
Add the new term to the induction hypothesis; the old right side cancels.
$2\sin x\sum_{j=1}^{k+1}\cos(2j-1)x=\sin2(k+1)x$
The base and implication establish the identity for all positive integers n.

The identity holds for all n≥1 and every real x.

Checks and common pitfalls: No division by sin x is needed in the induction proof.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The base case holds for every real x.
$n=1:\quad2\sin x\cos x=\sin2x$
Assume the statement for a fixed positive integer k.
$2\sin x\sum_{j=1}^k\cos(2j-1)x=\sin2kx$
The addition formulas give a telescoping increment.
$2\sin x\cos(2k+1)x=\sin(2k+2)x-\sin2kx$
Add the new term to the induction hypothesis; the old right side cancels.
$2\sin x\sum_{j=1}^{k+1}\cos(2j-1)x=\sin2(k+1)x$
The base and implication establish the identity for all positive integers n.

Think first. Reveal a hint when the class is ready.

2021 JM02