Use the principle of mathematical induction to prove that?: See picture

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1 Answer
Apr 8, 2017

Please see the explanation below.

Explanation:

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#.