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Math Help - trig (obviously)

  1. #1
    Senior Member furor celtica's Avatar
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    Angry trig (obviously)

    find the solution for the equation tanA=3sinA for values of A in the range -180 <= A <= 180 (deg evidently)
    so i used tanA= sinA/cosA which gives cosA = sinA/3sinA = 1/3. anything wrong up until here?
    my calculator says cos^-1 1/3 = 70.5 deg, so my answer is +- 70.5 deg.
    however the answer includes +-180 as well. is this even possible? who is wrong?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by furor celtica View Post
    find the solution for the equation tanA=3sinA for values of A in the range -180 <= A <= 180 (deg evidently)
    so i used tanA= sinA/cosA which gives cosA = sinA/3sinA = 1/3. anything wrong up until here?
    my calculator says cos^-1 1/3 = 70.5 deg, so my answer is +- 70.5 deg.
    however the answer includes +-180 as well. is this even possible? who is wrong?
    The book is right. Note that \tan A=3\sin A\implies 3\sin A-\tan A=0\implies 3\sin A-\frac{\sin A}{\cos A}=0 \implies \sin A\left(3-\frac{1}{\cos A}\right)=0

    So \sin A=0 or \cos A=\frac{1}{3}. That's why \pm180^{\circ} is included in the solution set. (It seems like one more value is missing from the book's answer...can you see what it is?)

    Does this clarify things?
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  3. #3
    Senior Member furor celtica's Avatar
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    but how could i know this just by using my method?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by furor celtica View Post
    but how could i know this just by using my method?

    Chris has shown it to you . However ,

    \ tanA=3 \ sinA

    \frac{\ sinA}{\ cosA}=3\ sinA

    \ sinA=3\ sinAcosA

    3\ sinAcosA-\ sinA=0

    \ sinA(3\ cosA-1)=0

     \ sinA=0 or \ cosA=\frac{1}{3}
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