Question 1125669
Consider {{{P (n): sin (alpha)+ sin (alpha+ beta) + sin (alpha + 2beta)}}} + ... + {{{sin (alpha + (n - 1) beta)= cos (alpha+((n-1)beta/2)).(sin(n*beta/2))/(sin(n*beta/2))}}} for all natural number {{{n}}}.


We observe that {{{P (1)}}} is true, since
{{{P (1) : sin (alpha)=(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 (alpha)+ sin (alpha+ beta)+ sin (alpha+ 2beta)}}} + ... +{{{ sin (alpha + (k - 1)beta)=(sin (alpha+((k-1)/2)beta)sin(k*beta/2))/ sin(beta/2)}}}


Now, to prove that {{{P(k+ 1)}}} is true, we have


{{{P (k+ 1) : sin (alpha)+ sin (alpha+ beta) + sin (alpha+ 2beta)}}}+ ... + {{{sin (alpha+ (k- 1)beta)+ sin (alpha+ k*beta)=sin (alpha+((k-1)/2)sin(k*beta/2))/sin(beta/2) +sin (alpha+ k*beta)}}}


={{{(sin (alpha+((k-1)/2)sin(k*beta/2))+sin (alpha+ k*beta)sin(beta/2))/sin(beta/2)}}}

={{{(cos(alpha -beta/2)-cos(alpha +k*beta-beta/2)+cos(alpha +k*beta-beta/2)-cos(alpha +k*beta+beta/2))/2sin(beta/2)}}}

={{{(cos(alpha -beta/2)-cos(alpha +k*beta+beta/2))/2sin(beta/2)}}}

={{{(sin(alpha+k*beta/2)*sin((k*beta+beta)/2))/sin(beta/2)}}}

={{{(sin(alpha+k*beta/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}}}.