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Math Help - simple trig identities

  1. #1
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    simple trig identities

    tan(90+X) = -cotx

    i don't understand the 90+x part..
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Using the identity tan (A+B)=\frac{tan A+tan B}{1-tan Atan B}, we get:

    tan (90^0+X)=\frac{tan 90^0+tan X}{1-tan 90^0 tan X}

    = \frac{1+\frac{tan x}{tan 90^o}}{\frac{1}{tan 90^0}-tan X} (Dividing throughout by tan 90^o)

    = \frac{1+0}{0-tan X}

    = \frac{1}{-tan x}

    = -cot X
    Last edited by alexmahone; July 5th 2009 at 06:52 AM.
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  3. #3
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    sorry
    then??
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by z1llch View Post
    sorry
    then??
    I re-edited my post. Now do you understand?
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  5. #5
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    why do them become zero?
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by z1llch View Post
    why do them become zero?
    \frac{1}{tan 90^0}=\frac{cos 90^0}{sin 90^0}=\frac{0}{1}=0
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  7. #7
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    Quote Originally Posted by z1llch View Post
    tan(90+X) = -cotx

    i don't understand the 90+x part..
    \tan(90+x) =

    \frac{\sin(90+x)}{\cos(90+x)} =

    \frac{\sin(90)\cos{x} + \cos(90)\sin{x}}{\cos(90)\cos{x} - \sin(90)\sin{x}} =

    \frac{1 \cdot \cos{x} + 0 \cdot \sin{x}}{0 \cdot \cos{x} - 1 \cdot \sin{x}} =

    \frac{\cos{x}}{-\sin{x}} = -\cot{x}
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  8. #8
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    Quote Originally Posted by alexmahone View Post
    Using the identity tan (A+B)=\frac{tan A+tan B}{1-tan Atan B}, we get:

    tan (90^0+X)=\frac{tan 90^0+tan X}{1-tan 90^0 tan X}

    = \frac{1+\frac{tan x}{tan 90^o}}{\frac{1}{tan 90^0}-tan X} (Dividing throughout by tan 90^o)
    Actually, you can't do that, because tan 90^{\circ} is undefined. You have to change \tan (90^0+X) into \frac{\sin (90^0+X)}{\cos (90^0+X)} as skeeter showed above.


    01
    Last edited by yeongil; July 5th 2009 at 07:47 AM. Reason: Big oops!
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