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Trig Identity Question

  1. Mar 30, 2012 #1
    prove
    1-(cos(x)+sin(x))(cos(x)-sin(x))=2sin^2(x)

    foil out the center
    I get
    1-cos^2(x)-cos(x)sin(x)+cos(x)sin(x)+sin^2(x)

    the -cos(x)sin(x)+cos(x)sin(x) cancels to 0 leaving

    1-cos^2(x)-sin^2(x)

    then I'm lost...

    I know I can switch 1-cos^2(x) to sin^2(x) but that doesn't help because I get
    sin^2(x)-sin^2(x)=0

    where am i going wrong?
     
  2. jcsd
  3. Mar 30, 2012 #2
    Are you sure? Looks to me like the +cos(x)sin(x) and the -cos(x)sin(x) cancel out to leave:
    1-cos^2(x)+sin^2(x)
     
  4. Mar 30, 2012 #3
    I wrote my foil wrong
    foiling leaves me
    1-cos^2(x)-cos(x)sin(x)+cos(x)sin(x) - sin^2(x)
     
  5. Mar 30, 2012 #4
    Don't remove the parentheses until after you've foiled it out or you're going to lose a negative sign.

    Foiling 1-[(cos(x)+sin(x))(cos(x)-sin(x))], we get
    1-(cos^2(x)-cos(x)sin(x)+cos(x)sin(x)-sin^2(x)). Two of the terms cancel, yielding:
    1-(cos^2(x)-sin^2(x))
    =1-cos^2(x)+sin^2(x)

    Now try using those Pythagorean identities.
     
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