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Math Help - Prove this inequality

  1. #1
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    Prove this inequality

    - \sqrt{a ^{2}+b ^{2}  } \leqslant  a \cdot cos \alpha + b \cdot sin \alpha  \leqslant \sqrt{a ^{2}+b ^{2}  }

    My solution:

    \begin{cases} - \sqrt{ a^{2}+b^{2}  } \leqslant a \cdot cos \alpha +b \cdot sin \alpha  \\a \cdot cos \alpha +b \cdot sin \alpha \leqslant \sqrt{ a^{2}+b^{2}  }\end{cases}

    After adding up both equalities we have:

    a \cdot cos \alpha +b \cdot sin \alpha-\sqrt{ a^{2}+b^{2}  } \leqslant \sqrt{ a^{2}+b^{2}  } +  a \cdot cos \alpha +b \cdot sin \alpha

    2\sqrt{ a^{2}+b^{2}  } \geqslant 0

    My question is as follows:
    Is this solution correct, or is it after adding up the two equalities still obvious to write two?
    I hope you understand what I mean :]
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Smile ...

    u can try this
    LHS=[acos(x)+bsin(x)]
    dividing andd multiplying with

    take [a^2+b^2]^(1/2)=k
    [acos(x)+bsin(x)]*k/k..(1)

    when we take
    a/[a^2+b^2]^(1/2)=sin(y)
    we get
    b/[a^2+b^2]^(1/2)=cos(y)
    equation (1) than becomes
    k*(sin(x+y))
    here
    -1=< sin(x+y) <=1
    so
    -k<=LHS<=k
    thus proved



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