I'm not confident with my proofs and was wondering if someone could just check this one.(adsbygoogle = window.adsbygoogle || []).push({});

Show that |sin x-cos x|≤√2 for all x.

Attempt:

LHS = |sin x-cos x|

=√(sin x-cos x)^{2}

=√(sin^{2}x - 2sin x cos x + cos^{2}x)

since sin^{2}x + cos^{2}x=1, and 2sin x cos x = sin(2x)

LHS = √(1 - sin(2x))

since sin(2x) oscilates between -1 and 1, LHS oscilates between 0 and √2

∴LHS ≤ √2

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# Homework Help: Can someone check proof for |sin x-cos x|≤√2

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