Tuesday, June 19, 2018

trigonometry - Induction proof of the identity cosx+cos(2x)+cdots+cos(nx)=fracsin(fracnx2)cosfrac(n+1)x2sin(fracx2)





Prove that:cosx+cos(2x)++cos(nx)=sin(nx2)cos(n+1)x2sin(x2). (1)




My attempt:sin(x2)nk=1cos(kx)=nk=1sin(x2)cos(kx) (Applying sinxcosy=12[sin(x+y)+sin(xy)])
=nk=112sin(x2kx)+12sin(x2+kx).Multiplying (1) by sin(x2) ,gives: nk=1sin(x2)cos(kx)=sin(nx2)cos(n+1)x2 nk=1[sin(x2kx)+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+1k=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)x2sinx2),sinx2cos(n+1)x=12(sin(2n+3)x2sin(2n+1)x2),
whence the l.h.s. is
12(sin(2n+3)x2sinx2)=sin(n+1)x2cos(n+2)x2
by the factorisation formula: sinpsinq=2sinpq2cosp+q2.



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