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Question
Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`
Solution
Consider P(n): sinα + sin(α + β) + sin(α + 2β) + ... + sin(α + (n – 1)β)
= `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`, for all natural number n.
We observe that P(1) is true.
Since P(1): sin α = `(sin(alpha + 0) sin beta/2)/(sin beta/2)`
Assume that P(n) is true for some natural numbers k.
i.e., P(k): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β)
= `(sin (alpha + (k - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2))`
Now, to prove that P(k + 1) is true.
We have P(k + 1): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β) + sin(α + kβ)
= `(sin (alpha + ("k" - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2)) + sin(alpha + kbeta)`
= `(sin(alpha + (k - 1)/2 beta) sin (kbeta)/2 + sin(alpha + kbeta) sin beta/2)/(sin beta/2)`
= `(cos(alpha - beta/2) - cos(alpha + kbeta - beta/2) + cos(alpha + kbeta - beta/2) - cos(alpha + kbeta + beta/2))/(2sin beta/2)`
= `(cos(alpha - beta/2) - cos(alpha + kbeta + beta/2))/(2sin beta/2)`
= `(sin (alpha + (kbeta)/2)sin ((kbeta + beta)/2))/(sin beta/2)`
= `(sin(alpha + (kbeta)/2) sin(k + 1)(beta/2))/(sin beta/2)`
Thus P(k + 1) is true whenever P(k) is true.
Hence, by the Principle of Mathematical Induction P(n) is true for all natural number n.
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