Can someone check proof for |sin x-cos x|≤√2

[cambo86]

I'm not confident with my proofs and was wondering if someone could just check this one.

Show that |sin x-cos x|≤√2 for all x.

Attempt:
LHS = |sin x-cos x|
=√(sin x-cos x)²
=√(sin² x - 2sin x cos x + cos² x)
since sin² x + cos² 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

 

[Curious3141]

I'm not confident with my proofs and was wondering if someone could just check this one.

Show that |sin x-cos x|≤√2 for all x.

Attempt:
LHS = |sin x-cos x|
=√(sin x-cos x)²
=√(sin² x - 2sin x cos x + cos² x)
since sin² x + cos² 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

The idea is sound, although I would have expressed the last part more formally starting with $-1 \leq \sin 2x \leq 1$, then manipulating that with the rules of inequalities to give the desired bounds.

BTW, another way to do the proof is to express $\sin x - \cos x$ in the form $R\sin(x - \theta)$ (where $R$ and $\theta$ are to be found). I prefer this method, as it's more elegant.