Prove:
#sin theta + sin 2 theta+cdots+sin(n theta)=(sin(((n+1)theta)/2)sin((n theta)/2))/(sin (theta/2))#
Base Case (n=1)
(LHS)#=sin theta#
(RHS)#=(sin(((1+1)theta)/2)cancel sin((1 theta)/2))/cancel(sin (theta/2))=sin theta#
Hence, (RHS)=(LHS).
Induction Hypothesis (n=k)
Assume that
#sin theta + sin 2 theta+cdots+sin k theta=(sin(((k+1)theta)/2)sin((k theta)/2))/(sin (theta/2))#
Induction Step (n=k+1)
Let us show that
#sin theta + sin 2 theta+cdots+sin k theta+sin(k+1)theta=(sin(((k+2)theta)/2)sin(((k+1) theta)/2))/(sin (theta/2))#
By I.H.,
(LHS)#=(sin(((k+1)theta)/2)sin((k theta)/2))/(sin (theta/2))+sin (k+1)theta#
By taking the common denominator,
#=(sin(((k+1)theta)/2)sin((k theta)/2)+sin(k+1)theta cdot sin(theta/2))/(sin (theta/2))#
By #sin(k+1)theta=2sin(((k+1)theta)/2)cos(((k+1)theta)/2)#,
#=(sin(((k+1)theta)/2)sin((k theta)/2)+2sin(((k+1)theta)/2)cos(((k+1)theta)/2)sin(theta/2))/(sin (theta/2))#
By factoring out #sin(((k+1)theta)/2)#,
#=([sin((k theta)/2)+2cos(((k+1)theta)/2)sin(theta/2)]sin(((k+1)theta)/2))/(sin (theta/2))#
By #sin((k theta)/2)=sin(((k+1)theta)/2)cos(theta/2)-cos(((k+1)theta)/2)sin(theta/2),#
#=([sin(((k+1)theta)/2)cos(theta/2)+cos(((k+1)theta)/2)sin(theta/2)]sin(((k+1)theta)/2))/(sin (theta/2))#
By #sin(((k+2)theta)/2)=sin(((k+1)theta)/2)cos(theta/2)+cos(((k+1)theta)/2)sin(theta/2),#
#=(sin(((k+2)theta)/2)sin(((k+1) theta)/2))/(sin (theta/2))=#(RHS)
Hence, the equation is true for all natural numer #n#.
