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  1. #1
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    exact values practice question

    Suppose cos A = -\frac{7}{25} and tan B = \frac{15}{8}
    where A is in the second quadrant and B is in the third quadrant. Find the exact values of:

    sin(A-B):


    tan(A-B):

    any help much appreciated.

    jv
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  2. #2
    MHF Contributor red_dog's Avatar
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    \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}

    \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}

    \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}

    \tan B=\frac{15}{8}

    \sin(A-B)=\sin A\cos B-\sin B\cos A

    \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
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  3. #3
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    Quote Originally Posted by red_dog View Post
    \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}

    \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}

    \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}

    \tan B=\frac{15}{8}

    \sin(A-B)=\sin A\cos B-\sin B\cos A

    \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
    ahh yeah question how did u get \frac{24}{25} from \sqrt{1-\cos^2A} ?

    EDIT: sorry i understand now \sqrt{1-(\frac{-7}{25})^2}]
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  4. #4
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    Quote Originally Posted by red_dog View Post
    \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}

    \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}

    \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}

    \tan B=\frac{15}{8}

    \sin(A-B)=\sin A\cos B-\sin B\cos A

    \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
    sorry question how did u get sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}} 3rd quadrant is tan, yes but then u got another tan ontop where numerator is . why?
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