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)`

Answer 1

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.