11just simply put the values
Then later add them
sin(\frac{2pi}{11}+\frac{4pi}{11}.........)+icos(\frac{2pi}{11}+\frac{4pi}{11}...........)
now add the values
u will get
sin 5pi + icos5pi
hence
-i
please give me as many alternative ways as possible
the value of
\sum_{k=1}^{10}{sin(2k\pi /11)+icos(2k\pi/11)}
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15 Answers
1multiply n divide by i^3
then u get it as summation frm 1 to 10 e^(- i2kpi/11)
iske aage
gp ke funde laga
u get d answer
111yaar aise mat bolo oye gp ke funde lagane hote to kyun puchta bata de yaar
and mani yaar please explain [2]
11what may i explain????
u have to add
2+4+6+8...............20
i did that to jump onto 2nd last step!!!!!!
1arre wahi kar raha tha nahi hua aur ab tumhe lag raha hai ki main silly dbts puch raha hun [2]
1kya bolte ho tum
summation frm 1 to 10 e^(- i2kpi/11)
itz
e-2pi/11 +e-4pi/11........... +e-20pi/11
so nw gp banta hai.........
so sumation of this gp =a(r^n-1)/(r-1)
wer r is d common ratio yane second term/first term
11#3
It should be
sin(2Î /11)+sin(4Î /11)+sin(6Î /11)+...........
there is formula for this
1boss!! i don find anyother alternative than d above mentioned....
