Solve ${(sin x - cos x)}^{2} = 1 - sin 2 x$ $(sinx-cosx)^2=1-sin2x$ ?

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Solve ${(sin x - cos x)}^{2} = 1 - sin 2 x$ $(sinx-cosx)^2=1-sin2x$ ?

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Answer 1 · 2017-04-28T01:28:52

You have asked how to solve

${(sin x - cos x)}^{2} = 1 - sin 2 x$ $(sinx-cosx)^2=1-sin2x$

But in ffact the above is an identity not an equation, which we can show as follows:

Multiplying out the LHS we get:

${(sin x - cos x)}^{2} \equiv {sin}^{2} x - 2 sin x cos x + {cos}^{2} x$ $(sinx-cosx)^2 -= sin^2x-2sinxcosx+cos^2x$
$\equiv {sin}^{2} x + {cos}^{2} x - 2 sin x cos x$ $" " -= sin^2x+cos^2x-2sinxcosx$

So then using the identities:

${sin}^{2} x + {cos}^{2} x \equiv 1$ $sin^2x+cos^2x -= 1$
$sin 2 x \equiv 2 sin x cos x$ $sin2x -= 2sinxcosx$

We have:

$(sinx-cosx)^2 -= 1-sin2x \ \ \ \$ QED

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Solve ${(sin x - cos x)}^{2} = 1 - sin 2 x$ $(sinx-cosx)^2=1-sin2x$ ?

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Humanities

... and beyond

Solve (sinx-cosx)^2=1-sin2x (sinx−cosx)2=1−sin2x (sinx-cosx)^2=1-sin2x ?

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Solve ${(sin x - cos x)}^{2} = 1 - sin 2 x$ $(sinx-cosx)^2=1-sin2x$ ?