Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis [duplicate] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)$sum cos$ when angles are in arithmetic progressionFinding the integral $int_0^pidfracdtheta(2+costheta)^2$ by complex analysisCalculating the following integral using complex analysis: $int_0^pie^acos(theta)cos(asin(theta)), dtheta$complex analysis - differentiabiliityHow to use complex analysis to find the integral $int^pi_−pi frac 1 1+sin^2(theta) dtheta$?Trigonometric Expression for $1 + cos alpha + cos 2alpha + cdots + cos n alpha$ using complex numbersComplex Analysis: why does $cos(3theta)$ = $cos^3theta - 3costheta sin^2theta$.$int^pi/2_0fractheta cos(theta)1+sin^2(theta)$ through complex analysisUse the Maclaurin series to prove that $e^itheta = cos(theta) + isin(theta)$Show (via Complex Numbers): $fraccosalphacosbetacos^2theta+fracsinalphasinbetasin^2theta+1=0$ under given conditionsProving complex series $1 + costheta + cos2theta +… + cos ntheta $
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Finding $cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+…+cos(theta+nalpha)$ with complex variable analysis [duplicate]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)$sum cos$ when angles are in arithmetic progressionFinding the integral $int_0^pidfracdtheta(2+costheta)^2$ by complex analysisCalculating the following integral using complex analysis: $int_0^pie^acos(theta)cos(asin(theta)), dtheta$complex analysis - differentiabiliityHow to use complex analysis to find the integral $int^pi_−pi frac 1 1+sin^2(theta) dtheta$?Trigonometric Expression for $1 + cos alpha + cos 2alpha + cdots + cos n alpha$ using complex numbersComplex Analysis: why does $cos(3theta)$ = $cos^3theta - 3costheta sin^2theta$.$int^pi/2_0fractheta cos(theta)1+sin^2(theta)$ through complex analysisUse the Maclaurin series to prove that $e^itheta = cos(theta) + isin(theta)$Show (via Complex Numbers): $fraccosalphacosbetacos^2theta+fracsinalphasinbetasin^2theta+1=0$ under given conditionsProving complex series $1 + costheta + cos2theta +… + cos ntheta $
$begingroup$
This question already has an answer here:
$sum cos$ when angles are in arithmetic progression [duplicate]
1 answer
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 2 days ago
YuiTo Cheng
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
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marked as duplicate by Eevee Trainer, lab bhattacharjee sequences-and-series
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$begingroup$
This question already has an answer here:
$sum cos$ when angles are in arithmetic progression [duplicate]
1 answer
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 2 days ago
YuiTo Cheng
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
$endgroup$
marked as duplicate by Eevee Trainer, lab bhattacharjee sequences-and-series
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$begingroup$
This question already has an answer here:
$sum cos$ when angles are in arithmetic progression [duplicate]
1 answer
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
sequences-and-series complex-analysis complex-numbers
edited 2 days ago
YuiTo Cheng
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
$endgroup$
This question already has an answer here:
$sum cos$ when angles are in arithmetic progression [duplicate]
1 answer
We have a series as
$cos(theta)+cos(theta+alpha)+cos(theta+2alpha)+...+cos(theta+nalpha)=U$
How can we make use of complex variable analysis to arrive at the term below which is equivalent to the above series?
$U=fracsin(fracn+12alpha)sin(frac12alpha)cos(theta+frac12nalpha)$
This question already has an answer here:
$sum cos$ when angles are in arithmetic progression [duplicate]
1 answer
sequences-and-series complex-analysis complex-numbers
sequences-and-series complex-analysis complex-numbers
edited 2 days ago
YuiTo Cheng
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
edited 2 days ago
YuiTo Cheng
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
edited 2 days ago
YuiTo Cheng
2,77141138
edited 2 days ago
YuiTo Cheng
2,77141138
edited 2 days ago
YuiTo Cheng
2,77141138
2,77141138
asked 2 days ago
UnbelievableUnbelievable
1183
asked 2 days ago
UnbelievableUnbelievable
1183
asked 2 days ago
UnbelievableUnbelievable
1183
1183
marked as duplicate by Eevee Trainer, lab bhattacharjee sequences-and-series
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1 Answer
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Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 2 days ago
dialogdialog
1447
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
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$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 2 days ago
dialogdialog
1447
$endgroup$
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$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 2 days ago
dialogdialog
1447
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add a comment |
$begingroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 2 days ago
dialogdialog
1447
$endgroup$
Use the fact that this is almost a geometric series. $$beginalignU&=mathfrak R[e^itheta +e^itheta+ialpha+cdots+e^itheta+inalpha]\&=mathfrak Rleft[e^ithetasum_j=0^n e^ijalpharight]\&=mathfrak Rleft[e^ithetafrac1-e^i(n+1)alpha1-e^ialpharight]\&=mathfrak Rleft[e^ithetafrace^-i(n+1)alpha/2-e^i(n+1)alpha/2e^-ialpha/2-e^ialpha/2e^inalpha/2right]\&=mathfrak Rleft[e^i(nalpha/2+theta)fracsin[(n+1)alpha/2]sin[alpha/2]right]\&=cos(theta+tfracnalpha2)fracsinleft(frac12(n+1)alpharight)sinleft(frac12alpharight)endalign$$
answered 2 days ago
dialogdialog
1447
answered 2 days ago
dialogdialog
1447
answered 2 days ago
dialogdialog
1447
answered 2 days ago
dialogdialog
1447
1447
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