Solve # (sinx-cosx)^2=1-sin2x #?
1 Answer
Apr 28, 2017
You have asked how to solve
# (sinx-cosx)^2=1-sin2x #
But in ffact the above is an identity not an equation, which we can show as follows:
Multiplying out the LHS we get:
# (sinx-cosx)^2 -= sin^2x-2sinxcosx+cos^2x #
# " " -= sin^2x+cos^2x-2sinxcosx #
So then using the identities:
# sin^2x+cos^2x -= 1 #
# sin2x -= 2sinxcosx #
We have:
# (sinx-cosx)^2 -= 1-sin2x \ \ \ \ # QED
