L.H.S.
f(x)=sinθ+sin2θ+sin3θ+−−−−−+sinnθ=sin(n+1)θ2sinnθ2sinθ2
Step (1):-
For n=1 L.H.S.
sinnθ=sinθ
For n=1 R.H.S.
sin(n+12)θsinnθ2sinθ2=sin2θ2sinθ2sinθ2
=sinθ
Thus then the result is true for n=1
Step (2):-
Let it is true for n=k.
Therefore,
sinθ+sin2θ+sin3θ+−−−+sinkθ=sin(k+1)2θsinkθ2sinθ2
Now in the step (3):-
To prove that, It is true for n=(k+1)
Then,
sinθ+sin2θ+sin3θ+−−−+sinkθ+sin(k+1)2θ=sin(k+12)θsinkθ2sinθ2+sin(k+1)θ
=sin(k+12)θsinkθ2sinθ2+2sin(k+1)2θcos(k+1)2θ
=sin(k+12)θsinθ2[sinkθ2+2cos(k+1)2θsinθ2]
=sin(k+12)θsinθ2[sinkθ2+2cos[(k+12)θ+θ2]−sin[(k+12)θ−θ2]]
=sin(k+12)θsinθ2[sinkθ2+sin[(k+22)θ−sink2]]
=sin(k+12)θsinθ2.sin(k+1)+12x
Then it is true for n=k+1 therefore n∈N
Hence proved.