- #1

- 25

- 0

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