Prove that:cosx+cos(2x)+⋯+cos(nx)=sin(nx2)cos(n+1)x2sin(x2). (1)
My attempt:sin(x2)n∑k=1cos(kx)=n∑k=1sin(x2)cos(kx) (Applying sinxcosy=12[sin(x+y)+sin(x−y)])
=n∑k=112sin(x2−kx)+12sin(x2+kx).Multiplying (1) by sin(x2) ,gives: n∑k=1sin(x2)cos(kx)=sin(nx2)cos(n+1)x2 ⇔n∑k=1[sin(x2−kx)+sin(x2+kx)]=sin(−x2)+sin(x2+nx).Then, by induction on n and using (sinxsiny) and (sinx+siny) formulas, I end in here: n+1∑k=1sin(x2)cos(kx)=[2sin(nx2)cos(n+1)x2]+[2sin(x2)cos(n+1)x]. Any help would be appreciated, thanks!
Answer
Here is the induction step: it comes down to proving
sinnx2cos(n+1)x2sinx2+cos(n+1)x=sin(n+1)x2cos(n+2)x2sin(x2)orsinnx2cos(n+1)x2+sinx2cos(n+1)x=sin(n+1)x2cos(n+2)x2
Now use the linearisation formulae:
{sinnx2cos(n+1)x2=12(sin(2n+1)x2−sinx2),sinx2cos(n+1)x=12(sin(2n+3)x2−sin(2n+1)x2),
whence the l.h.s. is
12(sin(2n+3)x2−sinx2)=sin(n+1)x2cos(n+2)x2
by the factorisation formula: sinp−sinq=2sinp−q2cosp+q2.
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