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Math Help - Proving identities

  1. #1
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    Proving identities

    a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

    b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

    thanks.
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  2. #2
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    Let B = A/2

    <br />
cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B}<br />

    <br />
= \frac { cos^2 B - sin^2B}{sin B cos B}<br />

    <br />
= \frac { cos2 B}{(sin 2B)/2}<br />

    <br />
 = 2 cot 2B<br />
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  3. #3
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA
    Quote Originally Posted by Oasis1993 View Post
    thanks.



    Hi all

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  4. #4
    Junior Member mrmohamed's Avatar
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    Quote Originally Posted by qmech View Post
    Let B = A/2

    <br />
cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B}<br />

    <br />
= \frac { cos^2 B - sin^2B}{sin B cos B}<br />

    <br />
= \frac { cos2 B}{(sin 2B)/2}<br />

    <br />
= 2 cot 2B<br />
    sorry sir
    i didnt see your solution
    mrmohamed
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  5. #5
    MHF Contributor
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    Quote Originally Posted by Oasis1993 View Post
    a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

    b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

    thanks.
    HI

    \cot \frac{1}{2}A-\tan \frac{1}{2}A=\frac{\cos \frac{1}{2}A}{\sin \frac{1}{2}A}-\frac{\sin \frac{1}{2}A}{\cos \frac{1}{2}A}

    \frac{\cos^2 \frac{1}{2}A-\sin^2 \frac{1}{2}A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}

    \frac{\cos A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}\cdot \frac{2}{2}

    \frac{2\cos A}{\sin A}

    2\cos A (proved)

    Now , from this identity ..

    \tan \frac{1}{2}A=\cot \frac{1}{2}A-2\cot A

    \tan \frac{1}{2}(30)=\cot \frac{1}{2}(30)-2\cot 30

    \tan 15=\frac{1}{\tan 15}-\frac{2}{\tan 30}

    \tan 15=\frac{1}{\tan 15}-2\sqrt{3}

    \tan^2 15+2\sqrt{3}\tan 15-1=0

    let \tan 15 be t

    t^2+2\sqrt{3}t-1=0

    Then by using the quadratic formula

    t=\frac{-2\sqrt{3}\pm \sqrt{16}}{2}

    t=-\sqrt{3}\pm 2
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