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Math Help - trigonometry (3)

  1. #1
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    trigonometry (3)

    If 0<A<\frac{\pi}{2} , and 0<B<\frac{\pi}{2} , prove that

    (1)  <br />
\frac{1}{\cos^2 A}+\frac{1}{\sin^2 A\sin^2 B\cos^2 B}\geq 9 <br />

    (2) Prove that sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) We have

    \frac{1}{\sin^2A\sin^2B\cos^2B}=\frac{\sin^2B+\cos  ^2B}{\sin^2A\sin^2B\cos^2B}=\frac{1}{\sin^2A\cos^2  B}+\frac{1}{\sin^2A\sin^2B}

    Then we have to prove that

    \frac{1}{\cos^2A}+\frac{1}{\sin^2A\cos^2B}+\frac{1  }{\sin^2A\sin^2B}\geq 9

    Use the inequality (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\r  ight)\geq 9

    for a=\cos^2A, \ b=sin^2A\cos^2B, \ c=\sin^2A\sin^2B

    using that

    a+b+c=\cos^2A+\sin^2A\cos^2B+\sin^2A\sin^2B=

    =\cos^2A+\sin^2A(\sin^2B+\cos^2B)=\cos^2A+\sin^2A=  1
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  3. #3
    MHF Contributor
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    Quote Originally Posted by thereddevils View Post
    If 0<A<\frac{\pi}{2} , and 0<B<\frac{\pi}{2} , prove that

    (1)  <br />
\frac{1}{\cos^2 A}+\frac{1}{\sin^2 A \sin^2 B \cos^2 B} \geq 9 <br />

    (2) Prove that sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1

    For (1) \sec^2 A + \text{cosec}^2 A \, \text{cosec}^2 B \, \sec^2 B

    =(1+\tan^2 A)+(1+\cot^2 A)(1+\cot^2 B)(1+tan^2 B)

    =(1+\tan^2 A)+(1+\cot^2 A)(2+\tan^2 B+\cot^2 B)

    Consider (\tan B-\cot B)^2\geq 0

    \tan^2 B+\cot^2 B \geq 2

    Then (\tan A-2\cot A)^2\geq 0

    \tan^2 A+4\cot^2 A\geq 4

    continuing

    =(1+\tan^2 A)+(1+\cot^2 A)(4)

    =5+\tan^2 A+4\cot^2 A

    =9

    this inequality is at least 9 . Hence , proved .
    Last edited by mr fantastic; October 8th 2009 at 04:04 AM. Reason: Fixed first line of latex: \cosec is not a valid command. Use \text{cosec}.
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