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Welcome to NRICH.

sin((2n+1)x/4)

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By Anonymous on Wednesday, May 30, 2001 - 06:32 pm:

Find this sum:
sin(x/4) + sin(3x/4) + sin(5x/4) +...+ sin((2n+1)x/4)
Thank you.
Antonio.

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By Oliver Samson (P3202) on Wednesday, May 30, 2001 - 07:14 pm:

Do you mean p instead of x? If you do, then the answer will cycle between (1/2)^1/2, 2^1/2, (1/2)^1/2 and 0, depending on the remainder left when n is divided by 4.

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By Kerwin Hui (Kwkh2) on Thursday, May 31, 2001 - 03:24 pm:

We can use the identity

sin y = Im e^iy

and get

Sⁿ _r=0sin((2r+1)x/4) = Im Sⁿ _r=0e^i(2r+1)x/4
=Im [(e^ix/4-e^i(2r+3)x/4)/(1-e^ix/2)]
=Im (1-e^i(n+1)x/2)/(e^-ix/4-e^ix/4)
=Im (1-e^i(n+1)x/2)/[2isin(-x/4)]
=Re (1-e^i(n+1)x/2)/[2sin(x/4)]
=[1-cos((n+1)x/2)]/[2sin(x/4)]

[An alternative method is to multiply the series by sin(x/4) and use the product-to-sum formula. Details omitted]

Kerwin

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