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Question
Prove that using the Mathematical induction
`sin(alpha) + sin (alpha + pi/6) + sin(alpha + (2pi)/6) + ... + sin(alpha + (("n" - 1)pi)/6) = (sin(alpha + (("n" - 1)pi)/12) xx sin(("n"pi)/12))/(sin (pi/12)`
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Solution
P(n) is the statement
`sin(alpha) + sin (alpha + pi/6) + sin(alpha + (2pi)/6) + ... + sin(alpha + (("n" - 1)pi)/6)`
= `(sin(alpha + (("n" - 1)pi)/12) xx sin(("n"pi)/12))/(sin (pi/12)`
Put n = 1
⇒ P(1) = sin α = L.H.S
R.H.S = `(sin(alpha + ((1 - 1)pi)/12) sin pi/12)/(sin pi/12)`
= sin α
L.H.S = R.H.S
⇒ P(1) is true
Assume that the statement is true for n = k
(i.e.) P(k) = `sin(alpha) + sin(alpha + pi/6) + sin(alpha + (2pi)/6) + ... + sin(alpha + (("k" - 1)pi)/6)`
= `(sin(alpha + (("k" - 1)pi)/12) xx sin(("k"pi)/12))/(sin(pi/12)` is true
To prove P(k + 1) is true
Now P(k + 1) = P(k) + t(k + 1)
Now P(k + 1) = `"P"("k") + sin(alpha + ("k"pi)/6)`
= `(sin[alpha + ("k" - 1) pi/12] sin ("k"pi)/12)/(sin pi/12) + sin(alpha + ("k"pi)/6)`
= `(sin[alpha + ("k" - 1) pi/12]sin ("k"pi)/12 + sin(alpha + ("k"pi)/6)sin pi/12)/(sin pi/12)`
Nr. = `1/2[cos(alpha + ("k" - 1) pi/12 - ("k"pi)/12) - cos(alpha + (("k" - 1)pi)/12 + ("k"pi)/12) + cos(alpha + ("k"pi)/6 - pi/12) - cos(alpha + ("k"pi)/6 + pi/12)]`
= `1/2[cos(alpha + ("k"pi)/12 - pi/12 - ("k"pi)/12) - cos(alpha + ("k"pi)/12 - pi/12 + ("k"pi)/12) + cos(alpha + (2"k"pi)/12 - pi/12) - cos(alpha + (2"k"pi)/12 + pi/12)]`
= `1/2[cos(alpha - pi/12) - cos(alpha + (2"k"pi)/12 - pi/12) + cos(alpha + (2"k"pi)/12 - pi/12) - cos(alpha + (2"k"pi)/12 + pi/12)]`
= `1/2[cos(alpha - pi/12) - cos(alpha + (2"k"pi)/12 + pi/12)] + sin pi/12`
= `- sin 1/2 (2alpha + (2"k"pi)/12) sin 1/2(-(2"k"pi - 2pi)/12)`
= `sin(alpha + pi/12) sin("k" + 1) pi/12`
Dr. = `sin pi/12`
P(k + 1) = `(sin(alpha + pi/12) sin("k" + 1) pi/12)/(sin pi/12)`
⇒ P(k + 1) is true
Whenever P(k) is true.
So by the principle of mathematical induction
P(n) is true.
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