Question Number 60156 by Tawa1 last updated on 18/May/19

Commented by Smail last updated on 19/May/19

Commented by Tawa1 last updated on 19/May/19

Answered by Kunal12588 last updated on 18/May/19

Commented by Tawa1 last updated on 18/May/19

Answered by tanmay last updated on 18/May/19

Commented by Kunal12588 last updated on 18/May/19
that's actually great
Commented by Tawa1 last updated on 18/May/19

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... =](https://www.tinkutara.com/question/Q60177.png)
Commented by Tawa1 last updated on 18/May/19

Commented by Tawa1 last updated on 18/May/19

Commented by Tawa1 last updated on 18/May/19

Commented by tanmay last updated on 18/May/19

Commented by Tawa1 last updated on 18/May/19

Commented by malwaan last updated on 19/May/19
