Trig / Pre Cal / identities These things are killing me. Please help guide me. Thank you. \[ (\sin^4 x-\cos^4 x) = 2\sin^2 x -1 \] \[ (\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 \] Am I on the right path?

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Trig / Pre Cal / identities These things are killing me. Please help guide me. Thank you. \[ (\sin^4 x-\cos^4 x) = 2\sin^2 x -1 \] \[ (\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 \] Am I on the right path?

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apply special identity \[\huge\rm sin^2 x + \cos^2 =1 \] solve for sin^2
  • phi
yes that is a good start
so we have \[ (1)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 \]

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Other answers:

  • phi
I would next replace the cos^2
sin^2 x + cos ^2 = 1 solve this identity for cos^2
\[ (1)(\sin^2 x + 1-\sin^2 x )= 2\sin^2 x -1 \]
\[ \rm (\sin^2 x-(\color{reD}{{1-sin^2x}}))\] write that in the parentheses and distribute by negative one
  • phi
close, but you lost a minus sign -(1 - sin^2)
\[ (1)(\sin^2 x - (1-\sin^2 x))= 2\sin^2 x -1 \] Now what?
  • phi
-(a +b) means -1*(a+b) = -1*a + -1*b
Ah so \[ (1)(\sin^2 x - 1+\sin^2 x )= 2\sin^2 x -1 \] I see it now.
Thank you both!!!
:-)

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