Question

Series question. NEEED HELP PLEASE!

Answer 1

\[\sum_{1}^{\infty} (\sin(8)^n\]

Answer 2

it should be a(r)^n and since sin(8) < 1 we can use a/(1-r) but for some reason it says sin(8) is both a and r... Why is that?

Answer 3

It should be sin(8)/(1-sin(8)) but apparently that's not right....

Answer 4

0.1617 :(

Answer 5

\[\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}\]

\[\sum_{n=1}^{\infty}(\sin(8))^n=\sum_{n=1}^{\infty}\sin(8)\times(\sin(8))^{n-1}\]

Answer 6

so sin(8) x 1?

Answer 7

is the 8 in radians or degrees?

Answer 8

it is \[\frac{\sin(8)}{1-\sin(8)}\]

Answer 9

I would assume degrees... But Idk doesn't say...

Answer 10

if it was radians it would have a pi right?


It asks for the 4 decimal places, which would be 0.1617

Answer 11

if it is in degrees then you get .16167...

Answer 12

or what you have if you want 4 decimal places

Answer 13

if radians you get 92.9695

Answer 14

But apparently that answer is wrong... In the practice version it says sin(3) and does it the same way, but that $ works... 0.1643 I think...

Answer 15

so what exactly are you showing in the bottom one where you say sin(8) x (sin(8)^n-1? A and R?

Answer 16

then it is in the correct form

\[a=\sin(8)\] and \[r=\sin(8)\]

Answer 17

how do we find that? With what I said above?

Answer 18

look at my first post...match up the letters

Answer 19

yeah so sin(8) x sin(8)^n-1?

Answer 20

and yeah the radian answer worked... WHY IDK...

Answer 21

http://www.wolframalpha.com/input/?i=series+of+%28sin%288%29%29%5En+from+1+to+infinity