Find the sum of $sin^2(n)$












2














I have no clue how to solve this a detailed solution would be great$$sum_{n=1}^N sin^2(n)=? $$






calculus sequences-and-series algebra-precalculus summation discrete-calculus







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    math.stackexchange.com/questions/17966/…
    – lab bhattacharjee
    Nov 27 '18 at 8:44










  • @labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:46


















2














I have no clue how to solve this a detailed solution would be great$$sum_{n=1}^N sin^2(n)=? $$






calculus sequences-and-series algebra-precalculus summation discrete-calculus







share|cite|improve this question




















  • 2




    math.stackexchange.com/questions/17966/…
    – lab bhattacharjee
    Nov 27 '18 at 8:44










  • @labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:46
















2












2








2


0





I have no clue how to solve this a detailed solution would be great$$sum_{n=1}^N sin^2(n)=? $$






calculus sequences-and-series algebra-precalculus summation discrete-calculus







share|cite|improve this question















I have no clue how to solve this a detailed solution would be great$$sum_{n=1}^N sin^2(n)=? $$






calculus sequences-and-series algebra-precalculus summation discrete-calculus




calculus sequences-and-series algebra-precalculus summation discrete-calculus






share|cite|improve this question















share|cite|improve this question













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edited Nov 27 '18 at 8:49









A.Γ.

22.6k32656




22.6k32656










asked Nov 27 '18 at 8:44









RuvikRuvik

438




438








  • 2




    math.stackexchange.com/questions/17966/…
    – lab bhattacharjee
    Nov 27 '18 at 8:44










  • @labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:46
















  • 2




    math.stackexchange.com/questions/17966/…
    – lab bhattacharjee
    Nov 27 '18 at 8:44










  • @labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:46










2




2




math.stackexchange.com/questions/17966/…
– lab bhattacharjee
Nov 27 '18 at 8:44




math.stackexchange.com/questions/17966/…
– lab bhattacharjee
Nov 27 '18 at 8:44












@labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
– Ruvik
Nov 27 '18 at 8:46






@labbhattacharjee sorry i forgot the square ahh never mind i can use trigonometric identities to reduce the power,thnx for the answer
– Ruvik
Nov 27 '18 at 8:46












1 Answer
1






active

oldest

votes


















6














NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.

You can try to consider that $sin(n) = Im(cos(n)+i*sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.



If your question is for the sum of $sin^2(n)$ (as written in the title) remember that:
$sin^2(x) = frac{1-cos(2x)}{2}$ and then you can use the same method as above



Solution below (for $sin^2$)




$sum_{n=1}^N sin^2(n) = sum_{n=1}^N frac{1-cos(2n)}{2}$

And $sum_{n=1}^N cos(2n) = ReBigl(sum_{n=1}^N cos(2n)+isin(2n)Bigl)$
$=sum_{n=1}^N e^{2in}=sum_{n=1}^N (e^{2i})^n=e^{2i}sum_{n=0}^{N-1} (e^{2i})^n$
$=e^{2i}*frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$
$=e^{iN+2}frac{sin(N+1)}{sin(1)}$

Thus $ReBigl(sum_{n=1}^N e^{2in}Bigl) = frac{cos(N+2)sin(n+1)}{sin(1)}$

Finally $sum_{n=1}^N sin^2(n) = frac N2-frac{cos(N+2)sin(n+1)}{2sin(1)}$




Hope i didn't make any mistakes :)






share|cite|improve this answer























  • no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:52











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6














NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.

You can try to consider that $sin(n) = Im(cos(n)+i*sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.



If your question is for the sum of $sin^2(n)$ (as written in the title) remember that:
$sin^2(x) = frac{1-cos(2x)}{2}$ and then you can use the same method as above



Solution below (for $sin^2$)




$sum_{n=1}^N sin^2(n) = sum_{n=1}^N frac{1-cos(2n)}{2}$

And $sum_{n=1}^N cos(2n) = ReBigl(sum_{n=1}^N cos(2n)+isin(2n)Bigl)$
$=sum_{n=1}^N e^{2in}=sum_{n=1}^N (e^{2i})^n=e^{2i}sum_{n=0}^{N-1} (e^{2i})^n$
$=e^{2i}*frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$
$=e^{iN+2}frac{sin(N+1)}{sin(1)}$

Thus $ReBigl(sum_{n=1}^N e^{2in}Bigl) = frac{cos(N+2)sin(n+1)}{sin(1)}$

Finally $sum_{n=1}^N sin^2(n) = frac N2-frac{cos(N+2)sin(n+1)}{2sin(1)}$




Hope i didn't make any mistakes :)






share|cite|improve this answer























  • no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:52
















6














NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.

You can try to consider that $sin(n) = Im(cos(n)+i*sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.



If your question is for the sum of $sin^2(n)$ (as written in the title) remember that:
$sin^2(x) = frac{1-cos(2x)}{2}$ and then you can use the same method as above



Solution below (for $sin^2$)




$sum_{n=1}^N sin^2(n) = sum_{n=1}^N frac{1-cos(2n)}{2}$

And $sum_{n=1}^N cos(2n) = ReBigl(sum_{n=1}^N cos(2n)+isin(2n)Bigl)$
$=sum_{n=1}^N e^{2in}=sum_{n=1}^N (e^{2i})^n=e^{2i}sum_{n=0}^{N-1} (e^{2i})^n$
$=e^{2i}*frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$
$=e^{iN+2}frac{sin(N+1)}{sin(1)}$

Thus $ReBigl(sum_{n=1}^N e^{2in}Bigl) = frac{cos(N+2)sin(n+1)}{sin(1)}$

Finally $sum_{n=1}^N sin^2(n) = frac N2-frac{cos(N+2)sin(n+1)}{2sin(1)}$




Hope i didn't make any mistakes :)






share|cite|improve this answer























  • no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:52














6












6








6






NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.

You can try to consider that $sin(n) = Im(cos(n)+i*sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.



If your question is for the sum of $sin^2(n)$ (as written in the title) remember that:
$sin^2(x) = frac{1-cos(2x)}{2}$ and then you can use the same method as above



Solution below (for $sin^2$)




$sum_{n=1}^N sin^2(n) = sum_{n=1}^N frac{1-cos(2n)}{2}$

And $sum_{n=1}^N cos(2n) = ReBigl(sum_{n=1}^N cos(2n)+isin(2n)Bigl)$
$=sum_{n=1}^N e^{2in}=sum_{n=1}^N (e^{2i})^n=e^{2i}sum_{n=0}^{N-1} (e^{2i})^n$
$=e^{2i}*frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$
$=e^{iN+2}frac{sin(N+1)}{sin(1)}$

Thus $ReBigl(sum_{n=1}^N e^{2in}Bigl) = frac{cos(N+2)sin(n+1)}{sin(1)}$

Finally $sum_{n=1}^N sin^2(n) = frac N2-frac{cos(N+2)sin(n+1)}{2sin(1)}$




Hope i didn't make any mistakes :)






share|cite|improve this answer














NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.

You can try to consider that $sin(n) = Im(cos(n)+i*sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.



If your question is for the sum of $sin^2(n)$ (as written in the title) remember that:
$sin^2(x) = frac{1-cos(2x)}{2}$ and then you can use the same method as above



Solution below (for $sin^2$)




$sum_{n=1}^N sin^2(n) = sum_{n=1}^N frac{1-cos(2n)}{2}$

And $sum_{n=1}^N cos(2n) = ReBigl(sum_{n=1}^N cos(2n)+isin(2n)Bigl)$
$=sum_{n=1}^N e^{2in}=sum_{n=1}^N (e^{2i})^n=e^{2i}sum_{n=0}^{N-1} (e^{2i})^n$
$=e^{2i}*frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$
$=e^{iN+2}frac{sin(N+1)}{sin(1)}$

Thus $ReBigl(sum_{n=1}^N e^{2in}Bigl) = frac{cos(N+2)sin(n+1)}{sin(1)}$

Finally $sum_{n=1}^N sin^2(n) = frac N2-frac{cos(N+2)sin(n+1)}{2sin(1)}$




Hope i didn't make any mistakes :)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 12:10

























answered Nov 27 '18 at 8:50









TheD0ubleTTheD0ubleT

39218




39218












  • no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:52


















  • no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
    – Ruvik
    Nov 27 '18 at 8:52
















no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
– Ruvik
Nov 27 '18 at 8:52




no it is not i need this sum for a physics problem later there is discrete sum but thank i already know what to do thnx for the answer
– Ruvik
Nov 27 '18 at 8:52


















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