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Math Help - Tan A

  1. #1
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    Tan A

    If sin A/2 = -12/13

    Find Tan A.

    I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.
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  2. #2
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    Quote Originally Posted by reiward View Post
    If sin A/2 = -12/13

    Find Tan A.

    I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.
    HI


     <br />
\tan A=\frac{2\tan \frac{A}{2}}{1-\tan^2 \frac{A}{2}}<br />

     <br />
 =\frac{2(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}})}{1-(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}})^2}<br />
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  3. #3
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    Quote Originally Posted by reiward View Post
    If sin A/2 = -12/13

    Find Tan A.

    I'm really clueless here. The A/2 made it complicated for me that I couldn't get sin A. Help please.
    Since sin is negative, so it can be in quadrant 3 and 4 . Are you sure that the question did not mention bout cos A/2 or tan A/2 ? Or else you would get 2 different answers for tan A .
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    Since sin is negative, so it can be in quadrant 3 and 4 . Are you sure that the question did not mention bout cos A/2 or tan A/2 ? Or else you would get 2 different answers for tan A .
    Oh, it mentioned that A/2 is a negative obtuse angle. So third quadrant?

    Uhm, why did you use Tan 2A identities
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  5. #5
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    Quote Originally Posted by reiward View Post
    Oh, it mentioned that A/2 is a negative obtuse angle. So third quadrant?

    Uhm, why did you use Tan 2A identities
    Because it wouldn't lead me to the answer . I used the half angles identities instead .

    Yes , A/2 would be in the 3rd quadrant .

    so sin A/2 = .... and cos A/2 = .. , then substitute into the identity i gave you just now .
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