- #1
cambo86
- 25
- 0
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
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