anonymous
  • anonymous
how do you transform 1-2cos^2A+(1-sin^2A)^2/cos^4 to (2-2cos^4A-2sin^2A+sin^4A)/cos4A
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Please Loser66 be my hero :D
anonymous
  • anonymous
huh? I tried 1-2cos^2A+1-2sin^2A+sin^4A/cos^4 how ever from there I don't know what to do
Loser66
  • Loser66
let re write yours. \[\frac{1-2cos^2A+(1-sin^2A)^2}{cos^4A}\]right? is it yours?

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More answers

anonymous
  • anonymous
yeah
Loser66
  • Loser66
\[(1-sin^2A) = cos^2A\\therefore~yours~will~become~ \frac{1-2cos^2A +cos^4A}{cos^4A}\]right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
wait...
Loser66
  • Loser66
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)
anonymous
  • anonymous
cos^4 A
Loser66
  • Loser66
nope, it is (1-cos^2A)^2
anonymous
  • anonymous
let me re arrange them
Loser66
  • Loser66
but I cannot get your second one. hehehe... my turns to tan^4 A
anonymous
  • anonymous
it is supposed to go to tan^4 A
anonymous
  • anonymous
I'm just trying to make it larger
Loser66
  • Loser66
I got tan^4 A only. hehehe... sorry.
anonymous
  • anonymous
but how?
Loser66
  • Loser66
\[\frac{1-2cos^2A +cos^4A}{cos^4A}= \frac{(1-sin^2A)^2}{cos^4A}=\frac{sin^4A}{cos^4A}=tan^4A\]
Loser66
  • Loser66
\[1-2cos^2A +cos^4A\]form the form of (a -2ab+b^2 ) =(a-b)^2 your a =1 your b =cos^2
Loser66
  • Loser66
I have to go.
anonymous
  • anonymous
however how can you transform \[\frac{ 1-2 \cos ^{2}A+1-2\sin ^{2}A+\sin ^{4}A}{ \cos ^{4}A }\]
Loser66
  • Loser66
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.

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