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Question Number 60156 by Tawa1 last updated on 18/May/19
Prove by principle of mathematical induction          sin(x) + sin(2x) + sin(3x) + ... + sin(nx)  =  ((cos((1/2)x) − cos(n + (1/2))x)/(2 sin((1/2)x)))
Provebyprincipleofmathematicalinductionsin(x)+sin(2x)+sin(3x)++sin(nx)=cos(12x)cos(n+12)x2sin(12x)
Commented by Smail last updated on 19/May/19
sinx+sin2x+...+sin(nx)=Σ_(k=0) ^n sin(kx)  =Im(Σ_(k=0) ^n e^(ikx) )  Σ_(k=0) ^n e^(ikx) =Σ_(k=0) ^n (e^(ix) )^k =((1−(e^(ix) )^(n+1) )/(1−e^(ix) ))  =((1−e^(i(n+1)x) )/(1−e^(ix) ))=((1−cos((n+1)x)−isin((n+1)x))/(1−cosx−isinx))  =((2sin^2 (((n+1)/2)x)−2isin(((n+1)/2)x)cos(((n+1)/2)x))/(2sin^2 (x/2)−2isin(x/2)cos(x/2)))  =((−i×sin(((n+1)/2)x)(isin(((n+1)/2)x)+cos(((n+1)/2)x)))/(−isin(x/2)×(isin(x/2)+cos(x/2))))  =((sin(((n+1)/2)x)e^(i((n+1)/2)x) )/(sin(x/2)×e^(i(x/2)) ))=((sin(((n+1)/2)x))/(sin(x/2)))e^(i(n/2)x)   Im(Σ_(k=0) ^n e^(ikx) )=((sin(((n+1)/2)x)sin((n/2)x))/(sin(x/2)))  sinx+sin2x+...+sinnx=((sin(((n+1)/2)x)sin((n/2)x))/(sin(x/2)))  =((cos(((n+1)/2)x−(n/2)x)−cos(((n+1)/2)x+(n/2)x))/(2sin(x/2)))  =((cos((x/2))−cos(((2n+1)/2)x))/(2sin(x/2)))
sinx+sin2x++sin(nx)=nk=0sin(kx)=Im(nk=0eikx)nk=0eikx=nk=0(eix)k=1(eix)n+11eix=1ei(n+1)x1eix=1cos((n+1)x)isin((n+1)x)1cosxisinx=2sin2(n+12x)2isin(n+12x)cos(n+12x)2sin2x22isinx2cosx2=i×sin(n+12x)(isin(n+12x)+cos(n+12x))isinx2×(isinx2+cosx2)=sin(n+12x)ein+12xsinx2×eix2=sin(n+12x)sinx2ein2xIm(nk=0eikx)=sin(n+12x)sin(n2x)sinx2sinx+sin2x++sinnx=sin(n+12x)sin(n2x)sinx2=cos(n+12xn2x)cos(n+12x+n2x)2sinx2=cos(x2)cos(2n+12x)2sinx2
Commented by Tawa1 last updated on 19/May/19
God bless you sir
Godblessyousir
Answered by Kunal12588 last updated on 18/May/19
P(n): sin(x)+sin(2x)+sin(3x)+...+sin(nx)=((cos((1/2)x)−cos(n+(1/2))x)/(2 sin((1/2)x)))  P(1) :   LHS=sin(x)  RHS=((cos((1/2)x)−cos(1 + (1/2))x)/(2 sin((1/2)x)))=((sin((((1/2)+1+(1/2))/2))x sin(((1+(1/2)−(1/2))/2))x)/(sin((1/2))x))  =((sin(x) sin((1/2))x)/(sin((1/2))x))=sin(x)  ⇒LHS=RHS  ∴P(1) is true  Let P(k)  be true ; k∈N  P(k): sin(x)+sin(2x)+sin(3x)+...+sin(kx)=((cos((1/2))x−cos(k+ (1/2))x)/(2 sin((1/2))x))  Now,  P(k+1):  sin(x)+sin(2x)+...+sin(kx)+sin((k+1)x)  =((cos((1/2))x−cos(k+ (1/2))x)/(2 sin((1/2))x))+sin((k+1)x)  =((cos((1/2))x−cos(k+ (1/2))x)/(2 sin((1/2))x))+((2sin((k+1)x)sin((1/2))x)/(2sin((1/2))x))  =p+((cos((1/2)−k−1)x−cos((1/2)+k+1)x)/(2sin((1/2))x))  =p+((cos(k+(1/2))x−cos(k+1+(1/2))x)/(2sin((1/2))x))  =((cos((1/2))x−cos(k+(1/2))x+cos(k+(1/2))x−cos(k+1+(1/2))x)/(2sin((1/2))x))  =((cos((1/2))x−cos(k+1+(1/2))x)/(2sin((1/2))x))  ∴P(n) is true from Principle of Mathematical Induction
P(n):sin(x)+sin(2x)+sin(3x)++sin(nx)=cos(12x)cos(n+12)x2sin(12x)P(1):LHS=sin(x)RHS=cos(12x)cos(1+12)x2sin(12x)=sin(12+1+122)xsin(1+12122)xsin(12)x=sin(x)sin(12)xsin(12)x=sin(x)LHS=RHSP(1)istrueLetP(k)betrue;kNP(k):sin(x)+sin(2x)+sin(3x)++sin(kx)=cos(12)xcos(k+12)x2sin(12)xNow,P(k+1):sin(x)+sin(2x)++sin(kx)+sin((k+1)x)=cos(12)xcos(k+12)x2sin(12)x+sin((k+1)x)=cos(12)xcos(k+12)x2sin(12)x+2sin((k+1)x)sin(12)x2sin(12)x=p+cos(12k1)xcos(12+k+1)x2sin(12)x=p+cos(k+12)xcos(k+1+12)x2sin(12)x=cos(12)xcos(k+12)x+cos(k+12)xcos(k+1+12)x2sin(12)x=cos(12)xcos(k+1+12)x2sin(12)xP(n)istruefromPrincipleofMathematicalInduction
Commented by Tawa1 last updated on 18/May/19
God bless you sir
Godblessyousir
Answered by tanmay last updated on 18/May/19
by general method−1  rule is→see the values of angles→they are in   A.P and common difference is x  now multiply each term by sin((x/2)) ←main concept  now  2sinx.sin((x/2))=cos(x/2)−cos(((3x)/2))  2sin2x.sin((x/2))=cos((3x)/2)−cos((5x)/2)  ...  ...  2sinnx.sin(x/2)=cos(((2n−1)/2))x−cos(((2n+1)/2))x  now add them  2sin(x/2)×S=cos(x/2)−cos(((2n+1)/2))x  S=((cos((x/2))−cos(n+(1/2))x)/(2sin((x/2))))
bygeneralmethod1ruleisseethevaluesofanglestheyareinA.Pandcommondifferenceisxnowmultiplyeachtermbysin(x2)mainconceptnow2sinx.sin(x2)=cosx2cos(3x2)2sin2x.sin(x2)=cos3x2cos5x22sinnx.sinx2=cos(2n12)xcos(2n+12)xnowaddthem2sinx2×S=cosx2cos(2n+12)xS=cos(x2)cos(n+12)x2sin(x2)
Commented by Kunal12588 last updated on 18/May/19
that's actually great
Commented by Tawa1 last updated on 18/May/19
God bless you sir
Godblessyousir
Answered by tanmay last updated on 18/May/19
method−2  p=cosx+cos2x+cos3x+...+cosxnx  q=sinx+sin2x+sin3x+...+sinnx  p+iq=e^(ix) +e^(i2x) +...+e^(inx)   p+iq=((e^(ix) (e^(inx) −1))/(e^(ix) −1))  p+iq=((e^(i(n+1)x) −e^(ix) )/(e^(ix) −1))  =((cos(n+1)x+isin(n+1)x−cosx−isinx)/(cosx+isinx−1))  =(([cos(n+1)x−cosx]+i[sin(n+1)x−sinx])/((cosx−1)+isinx))  =((2sin(x+((nx)/2))sin(((−nx)/2))+i[2cos(x+((nx)/2))sin(((nx)/2)))/(−2sin^2 (x/2)+i2sin(x/2)cos(x/2)))  =((2sin((nx)/2))/(2sin(x/2)))×((−sin(x+((nx)/2))+icos(x+((nx)/2)))/(−sin(x/2)+icos(x/2)))  =((sin((nx)/2))/(sin(x/2)))×(([sin(x+((nx)/2))−icos(x+((nx)/2))][sin(x/2)+icos(x/2)])/((sin(x/2)−icos(x/2))(sin(x/2)+icos(x/2))))  =((sin((nx)/2))/(sin(x/2)))×(([sin(x+((nx)/2))sin(x/2)+isin(x+((nx)/2))cos(x/2)−icos(x+((nx)/2))sin(x/2)+cos(x+((nx)/2))cos(x/2)])/(sin^2 (x/2)+cos^2 (x/2)))  =((sin((nx)/2))/(sin(x/2)))×((cos(((nx)/2)+(x/2))+isin(((nx)/2)+(x/2)))/1)  =[((sin((nx)/2))/(sin(x/2)))×cos(n+1)(x/2)]+i[((sin((nx)/2))/(sin(x/2)))×sin(n+1)(x/2)]  =[((sin(((nx)/2)+((nx)/2)+(x/2))−sin((x/2)))/(2sin(x/2)))]+i[((cos(x/2)−cos(((nx)/2)+((nx)/2)+(x/2))/(2sin(x/2)))]  So  p=Σ_(r=1) ^n cosrx=[((sin(nx+(x/2)))/(2sin(x/2)))]  q=Σ_(r=1) ^n sinrx=[((cos(x/2)−cos(nx+(x/2)))/(2sin(x/2)))]  this is second method...          =
method2p=cosx+cos2x+cos3x++cosxnxq=sinx+sin2x+sin3x++sinnxp+iq=eix+ei2x++einxp+iq=eix(einx1)eix1p+iq=ei(n+1)xeixeix1=cos(n+1)x+isin(n+1)xcosxisinxcosx+isinx1=[cos(n+1)xcosx]+i[sin(n+1)xsinx](cosx1)+isinx=2sin(x+nx2)sin(nx2)+i[2cos(x+nx2)sin(nx2)2sin2x2+i2sinx2cosx2=2sinnx22sinx2×sin(x+nx2)+icos(x+nx2)sinx2+icosx2=sinnx2sinx2×[sin(x+nx2)icos(x+nx2)][sinx2+icosx2](sinx2icosx2)(sinx2+icosx2)=sinnx2sinx2×[sin(x+nx2)sinx2+isin(x+nx2)cosx2icos(x+nx2)sinx2+cos(x+nx2)cosx2]sin2x2+cos2x2=sinnx2sinx2×cos(nx2+x2)+isin(nx2+x2)1=[sinnx2sinx2×cos(n+1)x2]+i[sinnx2sinx2×sin(n+1)x2]=[sin(nx2+nx2+x2)sin(x2)2sinx2]+i[cosx2cos(nx2+nx2+x22sinx2]Sop=nr=1cosrx=[sin(nx+x2)2sinx2]q=nr=1sinrx=[cosx2cos(nx+x2)2sinx2]thisissecondmethod=
Commented by Tawa1 last updated on 18/May/19
God bless you sir
Godblessyousir
Commented by Tawa1 last updated on 18/May/19
Sir, how to prove the second question too, without induction.    (1/2) + cos(x) + cos(2x) + cos(3x) + ... + cos(nx)  =  ((sin(n + (1/2))x)/(2 sin((1/2)x)))
Sir,howtoprovethesecondquestiontoo,withoutinduction.12+cos(x)+cos(2x)+cos(3x)++cos(nx)=sin(n+12)x2sin(12x)
Commented by Tawa1 last updated on 18/May/19
Thanks for your time.
Thanksforyourtime.
Commented by tanmay last updated on 18/May/19
S=(1/2)+cosx+cos2x+..+cosnx  S−(1/2)=cosx+cosx+...+cosnx  same trick...angles are in A.P  common difference=x  multiply each term by sin((x/2))←always sin(((c.d)/2))  2cosxsin(x/2)=sin((3x)/2)−sin(x/2)  2cos2x.sin(x/2)=sin((5x)/2)−sin((3x)/2)  2cos3x.sin(x/2)=sin((7x)/2)−sin((5x)/2)  ...  ...  2cosnx.sin(x/2)=sin(((2n+1)/2))x−sin(((2n−1)/2))x  now add them  2sin(x/2)(S−(1/2))=sin(((2n+1)/2))x−sin(x/2)  S−(1/2)=((sin(((2n+1)/2))x−sin(x/2))/(2sin(x/2)))  S=(1/(2 ))+((sin(((2n+1)/2))x−sin(x/2))/(2sin(x/2)))  S=((sin(x/2)+sin(((2n+1)/2))x−sin(x/2))/(2sin(x/2)))  S=((sin(((2n+1)/(2 )))x)/(2sin(x/2)))
S=12+cosx+cos2x+..+cosnxS12=cosx+cosx++cosnxsametrickanglesareinA.Pcommondifference=xmultiplyeachtermbysin(x2)alwayssin(c.d2)2cosxsinx2=sin3x2sinx22cos2x.sinx2=sin5x2sin3x22cos3x.sinx2=sin7x2sin5x22cosnx.sinx2=sin(2n+12)xsin(2n12)xnowaddthem2sinx2(S12)=sin(2n+12)xsinx2S12=sin(2n+12)xsinx22sinx2S=12+sin(2n+12)xsinx22sinx2S=sinx2+sin(2n+12)xsinx22sinx2S=sin(2n+12)x2sinx2
Commented by Tawa1 last updated on 18/May/19
I really appreciate sir.  God bless you.
Ireallyappreciatesir.Godblessyou.Ireallyappreciatesir.Godblessyou.
Commented by malwaan last updated on 19/May/19
FANTASTIC sir
FANTASTICsirFANTASTICsir

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