Question

how do you transform
1-2cos^2A+(1-sin^2A)^2/cos^4
to
(2-2cos^4A-2sin^2A+sin^4A)/cos4A

Answer 1

Please Loser66 be my hero :D

Answer 2

huh?
I tried
1-2cos^2A+1-2sin^2A+sin^4A/cos^4 how ever from there I don't know what to do

Answer 3

let re write yours.
\[\frac{1-2cos^2A+(1-sin^2A)^2}{cos^4A}\]right? is it yours?

Answer 4

yeah

Answer 5

\[(1-sin^2A) = cos^2A\\therefore~yours~will~become~ \frac{1-2cos^2A +cos^4A}{cos^4A}\]right?

Answer 6

yeah

Answer 7

wait...

Answer 8

yours second one is \[\frac{2-2cos^4A-2sin^2A+sin^4A}{cos^4A}\]my question is the denominator is cos^4 A or cos (4A)

Answer 9

cos^4 A

Answer 10

nope, it is (1-cos^2A)^2

Answer 11

let me re arrange them

Answer 12

but I cannot get your second one. hehehe... my turns to tan^4 A

Answer 13

it is supposed to go to tan^4 A

Answer 14

I'm just trying to make it larger

Answer 15

I got tan^4 A only. hehehe... sorry.

Answer 16

but how?

Answer 17

\[\frac{1-2cos^2A +cos^4A}{cos^4A}= \frac{(1-sin^2A)^2}{cos^4A}=\frac{sin^4A}{cos^4A}=tan^4A\]

Answer 18

\[1-2cos^2A +cos^4A\]form the form of (a -2ab+b^2 ) =(a-b)^2
your a =1 your b =cos^2

Answer 19

I have to go.

Answer 20

however how can you transform \[\frac{ 1-2 \cos ^{2}A+1-2\sin ^{2}A+\sin ^{4}A}{ \cos ^{4}A }\]

Answer 21

from numerator,\[1-2cos^2A+1-2sin^2A+sin^4A\\= 2-2(cos^2 A+sin^2A)+sin^4A\\= 2-2+sin^4A \\= sin^4A\]
but I really not know why do you have to go on that way.
Sorry for suddenly logged out, this morning I was late for class.