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

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.

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.

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

We can use the identity

siny=Áe^{i y}

and get

n
å
r=0
sin((2r+1)x/4)=Á n
å
r=0
e^i(2r+1)x/4

=Á[(e^{i x/4}-e^i(2r+3)x/4)/(1-e^{i x/2})]

=Á(1-e^i(n+1)x/2)/(e^{-i x/4}-e^{i x/4})

=Á(1-e^i(n+1)x/2)/[2isin(-x/4)]

=Â(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