Question

Hi everyone! Can anyone help be go from [36sin(theta) plus or minus 24]/[8(1-2sin^2(theta))-2sin^(theta)] to make it look like 6/(2-3sin(theta) ? Please help!

Answer 1

you're proving identities? haaa

Answer 2

No i'm not proving identities...it is just a step in a long conversion from rectangular equation to polar equation...the answer is 6/2-3sin(theta)

Answer 3

Oh okay the way you asked it sounded like you wanted to prove it.

Answer 4

I need to factor out a sum of squares but can't quite seem to figure out how

Answer 5

I have been working on this problem for HOURS! I need a savior!

Answer 6

If you want to see all the work I have done so far, just click this link>>>

Answer 7

The work I just posted take you back a couple steps is all...I just factored the bottom

Answer 8

you can write the bottom as
8cos^2−8sin^2−2sin^2
8(cos^2−sin^2)−2sin^2
8(1−2sin^2)−2sin^2


using the identity cos^2 - sin^2 = 1-2sin^2

simplify, factor a sum of squares, and cancel terms
you get
6/(2-3sin)

Answer 9

talked about this yesterday; didn't I?

Answer 10

hi phi...yeah i got that far...then I tried to factor the sum of squares like you said but I don't know how

Answer 11

Hi prime...yeah...you gave me the answer as 6/2-3sintheta...but i need to get there on my own and cant do it

Answer 12

8 cos^2 - 10 sin^2
can be written as
8(1−2sin^2)−2sin^2 (see above)

distribute the 8:
8 -16sin^2 - 2sin^2
8 - 18 sin^2
factor out 2
2(4-9 sin^2)
2(2-3 sin)(2+3 sin)

Answer 13

Im trying it right now!!!

Answer 14

if the top is 6(3sin-2) or 6(3sin+2) depending on which sign you choose
note: 6(3sin -2)= -6(2 - 3 sin)

Answer 15

ok...i think i see it now...just trying to finish up...brb

Answer 16

GOT IT! Phis...you have no idea how thankful I am to you right now! Thank you sooo much! There were so many algebra techniques in this problem that I just couldn't see it! You are awesome!

Answer 17

I am so excited that I get to close this question finally! lol