• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter cambo86
  • Start date
  • #1
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
 

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
86
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 [itex]-1 \leq \sin 2x \leq 1[/itex], then manipulating that with the rules of inequalities to give the desired bounds.

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

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

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
9
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
960
  • Last Post
Replies
5
Views
718
Top