Question

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

Answer 1

You 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

Answer 2

I took the middle - operator and distributed it across (1 - cos^2 x), which gave me
\[ (1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1 \]
\[(1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1 \] <--- so I should leave this part out?

Answer 3

yeah you can go from
\[(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1\]
to
\[(1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1\]
so leave out that step

Answer 4

Awesome! Thank you. owbucks on the way :-)

Answer 5

You can say
\[(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1\]
\[(1)(\cos^2 x -(1)-(-cos^2 x)) = 2\cos^2 x - 1\]
but it's not 100% necessary

Answer 6

Thanks for you time :-) @jim_thompson5910

Answer 7

You're welcome. I'm glad to be of help.

Answer 8

Jim is correct

Answer 9

I 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

\[(1)\cos^2 x -1+\cos^2 x = 2\cos^2 x - 1\]
\[(1)\ 2cos^2 x -1= 2\cos^2 x - 1\]