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Math Help - [SOLVED] Prove the trigonometric inequality?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Prove the trigonometric inequality?

    Prove that cot x/2 - cot x ≥ 1 for x between 0 and pi.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Prove that cot x/2 - cot x ≥ 1 for x between 0 and pi.
    Note that \frac{\cos x}{\sin x} = \frac{\cos^2 \left( \frac{x}{2} \right) - \sin^2 \left( \frac{x}{2}\right) }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}.

    Therefore the left hand side becomes

    \frac{\cos \left( \frac{x}{2} \right) }{\sin \left( \frac{x}{2} \right) } - \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{2\cos^2 \left( \frac{x}{2} \right) }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} - \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}


     = \frac{1}{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{1}{\sin x}.


    Now note that 0 \leq \sin x \leq 1 over 0 \leq x \leq \pi.
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  3. #3
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    Hello, fardeen_gen!

    Another approach . . .

    Identity: . \cot\frac{x}{2} \;=\;\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;=\;\frac{\sqrt{\frac{1+\cos x}{2}}}{\sqrt{\frac{1-\cos x}{2}}} \;=\;\sqrt{\frac{1 + \cos x}{1-\cos x}}


    Prove that: . \cot\frac{x}{2} - \cot x \:\geq \:1\quad\text{ for }0 \leq x \leq \pi

    We have: . \cot\frac{x}{2} - \cot x \;\;=\;\;<br />
\sqrt{\frac{1+ \cos x}{1-\cos x}} - \frac{\cos x}{\sin x}


    Rationalize: . \sqrt{\frac{1+\cos x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}}} -\frac{\cos x}{\sin x}\;\;=\;\;\sqrt{\frac{(1+\cos x)^2}{1-\cos^2x}} - \frac{\cos x}{\sin x}

    . . = \;\sqrt{\frac{(1+\cos x)^2}{\sin^2\!x}} - \frac{\cos x}{\sin x} \;\;=\;\;\frac{1+\cos x}{\sin x} - \frac{\cos x}{\sin x} \;\;\frac{1}{\sin x} \;\;=\;\;\csc x


    And: . \csc x \:\geq \:1\;\text{ for }0 \:\leq x\;\leq \pi\quad\hdots\quad\text{Q.E.D.}

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