1. Sep 16, 2012 #1

   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

    

   cambo86, Sep 16, 2012
2. 
   Phys.org - latest science and technology news stories on Phys.org

   • • Game over? Computer beats human champ in ancient Chinese game
   • • Simplifying solar cells with a new mix of materials
   • • Imaged 'jets' reveal cerium's post-shock inner strength
3. Sep 16, 2012 #2

   Curious3141

   

   Homework Helper

   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.

    

   Curious3141, Sep 16, 2012

• Log in with Facebook
• Log in with Twitter

Your name or email address:

Do you already have an account?

• No, create an account now.
• Yes, my password is:
• Forgot your password?

Stay logged in