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Question
Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
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Solution
Let P(n): cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`
Step 1: P(1): cos α = `((cos alpha)(sin beta/2))/(sin beta/2)` = cos α
Step 2: P(k): cos α + cos(α + β) + cos(α + 2β) + ... + cos[α + (k – 1)β]
= `(cos[alpha + ((k - 1)/2)beta]sin((kbeta)/2))/(sin beta/2)`. Let it be true.
Step 3: P(k + 1): cos α + cos(α + β) + cos(α + 2β) + ... + cos[α + (k – 1)β] + cos[α + (k + 1 – 1)β]
= `(cos[alpha + ((k - 1)/2)beta]sin((kbeta)/2))/(sin beta/2) + cos(alpha + kbeta)` ......(From Step 2)
= `(2cos[alpha + ((k - 1)/2)beta]sin((kbeta)/2) + 2cos(alpha + kbeta).sin beta/2)/(2 sin beta/2)`
= `(sin[alpha + kbeta - beta/2] - sin[alpha - beta/2] + sin[alpha + kbeta + beta/2] - sin[alpha + kbeta - beta/2])/(2sin beta/2)` ......[∵ 2 cosA sinB = sin(A + B) – sin(A – B)]
= `(sin[alpha + kbeta + beta/2] - sin(alpha - beta/2))/(2sin beta/2)`
= `(2cos(alpha + (kbeta)/2) sin(k + 1) beta/2)/(2sin beta/2)` ......`[because sin"A" - sin"B" = 2cos ("A" + "B")/2 . sin ("A" - "B")/2]`
= `(cos(alpha + (kbeta)/2).sin(k + 1) beta/2)/(sin beta/2)`
= `(cos[alpha + ((k + 1 - 1)/2)beta] sin((k + 1)/2)beta)/(sin beta/2)` which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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