Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

(6) show : (sinx + acosx)(sinx + bcosx) ≤ 1 + [(a + b)/2]² .?

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  • Indica
    Lv 7
    2 years ago
    Favorite Answer

    LHS = sin²x + (a+b)sinxcosx + abcos²x = ½{ (1+ab) + (a+b)sin(2x) + (ab−1)cos(2x) }

    (a+b)sin(2x) + (ab−1)cos(2x) = √{(a+b)²+(ab−1)²}sin(2x+α) where tanα = (ab−1)/(a+b)

    ∴ LHS ≤ = ½{ (1+ab) + √{(a+b)²+(ab−1)²} } since sin≤1

    But √{(a+b)²+(ab−1)²} = √{(1+a²)(1+b²)} ≤ (2+a²+b²)/2 = 1+½(a²+b²)

    ∴ LHS ≤ ½{ 1+ab+1+½(a²+b²)) } = 1+((a+b)/2)²

    Note overall equality is only when sin(2x+α)=1 and a=b, so generally equality is not achieved.

    • Indica
      Lv 7
      2 years agoReport

      http://www.wolframalpha.com/input/?i=plot+1%2B(a%2Bb)%5E2%2F4+-+(sinx+%2B+a*cosx)(sinx+%2B+b*cosx)++for+a%3D7,+b%3D3

      http://www.wolframalpha.com/input/?i=plot+1%2B(a%2Bb)%5E2%2F4+-+(sinx+%2B+a*cosx)(sinx+%2B+b*cosx)++for+a%3D3,+b%3D3

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