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anonymous
 one year ago
Trig/ Pre Cal/ identities 2 owlbucks Is this correct?
The problem
\[ \cos^4 x \sin^4 x = 2\cos^2 x  1 \]
\[ (\cos^2 x +\sin^2 x)(\cos^2 x \sin^2 x) = 2\cos^2 x  1 \]
\[ (1)(\cos^2 x \sin^2 x) = 2\cos^2 x  1 \]
\[ (1)(\cos^2 x (1cos^2 x)) = 2\cos^2 x  1 \]
\[ (1)(\cos^2 x (1+cos^2 x)) = 2\cos^2 x  1\]
\[ (1)(\cos^2 x 1+cos^2 x) = 2\cos^2 x  1 \]
\[ 2\cos^2 x  1 = 2\cos^2 x  1 \]
anonymous
 one year ago
Trig/ Pre Cal/ identities 2 owlbucks Is this correct? The problem \[ \cos^4 x \sin^4 x = 2\cos^2 x  1 \] \[ (\cos^2 x +\sin^2 x)(\cos^2 x \sin^2 x) = 2\cos^2 x  1 \] \[ (1)(\cos^2 x \sin^2 x) = 2\cos^2 x  1 \] \[ (1)(\cos^2 x (1cos^2 x)) = 2\cos^2 x  1 \] \[ (1)(\cos^2 x (1+cos^2 x)) = 2\cos^2 x  1\] \[ (1)(\cos^2 x 1+cos^2 x) = 2\cos^2 x  1 \] \[ 2\cos^2 x  1 = 2\cos^2 x  1 \]

This Question is Closed

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4You made an error when you got to \[(1)(\cos^2 x (1+cos^2 x)) = 2\cos^2 x  1\] but you have the next step of \[(1)(\cos^2 x 1+cos^2 x) = 2\cos^2 x  1\] correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I took the middle  operator and distributed it across (1  cos^2 x), which gave me \[ (1)(\cos^2 x (1cos^2 x)) = 2\cos^2 x  1 \] \[(1)(\cos^2 x (1+cos^2 x)) = 2\cos^2 x  1 \] < so I should leave this part out?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4yeah you can go from \[(1)(\cos^2 x (1cos^2 x)) = 2\cos^2 x  1\] to \[(1)(\cos^2 x 1+cos^2 x) = 2\cos^2 x  1\] so leave out that step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! Thank you. owbucks on the way :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4You can say \[(1)(\cos^2 x (1cos^2 x)) = 2\cos^2 x  1\] \[(1)(\cos^2 x (1)(cos^2 x)) = 2\cos^2 x  1\] but it's not 100% necessary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for you time :) @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4You're welcome. I'm glad to be of help.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I saw what happened... the distribution part was too much... everything was right until \[(1)(\cos^2 x (1\cos^2 x)) = 2\cos^2 x  1\] distribute the negative \[(1)\cos^2 x 1+\cos^2 x = 2\cos^2 x  1\] \[(1)\ 2cos^2 x 1= 2\cos^2 x  1\]
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