trigonometric series.

[samuel minato]

how to find the Sn of this series?

sin a + sin 2a + sin 4a + .... + sin 2^(n-1)a

[pemdas]

you've got to find r (ratio) which is sin 2a/sin a = 2 (from formula of trigon's value with Pi, Pi/2, Pi/3, Pi/4, Pi/6, etc.)

Sum of n terms can be figured using Sn = sin a(r�-1)/(r-1) OR Sn = sin a(2^(n-1))/(2-1)
ans.: sin a* 2^(n-1)

how to find the Sn of this series?

sin a + sin 2a + sin 4a + .... + sin 2^(n-1)a