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Math Help - Trig

  1. #1
    Member srirahulan's Avatar
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    Arrow Trig

    Use only cos rule to prove this cotA:cotB:cotC= (b^2+c^2-a^2): (a^2+c^2-b^2): (a^2+b^2-c^2)

    In this case i use cos rule but, after few step i cannot simplify, but i have finish this sum with sine rule, If anyone have a idea to do simplify this sum through cos rule>>>>>>
    Last edited by srirahulan; October 21st 2013 at 05:58 PM.
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  2. #2
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    Re: Trig

    Hey srirahulan.

    Since cot A = cos(A)/sin(A), what are you using to calculate the sines if you can only use the cosine rule?
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  3. #3
    Member srirahulan's Avatar
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    Re: Trig

    i cannot understand, what you say?
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  4. #4
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    Re: Trig

    Quote Originally Posted by srirahulan View Post
    Use only cos rule to prove this cotA:cotB:cotC= (b^2+c^2-a^2): (a^2+c^2-b^2): (a^2+b^2-c^2)

    In this case i use cos rule but, after few step i cannot simplify, but i have finish this sum with sine rule, If anyone have a idea to do simplify this sum through cos rule>>>>>>
    According to the Law of Cosines,
    \begin{align*}a^2 & = b^2+c^2-2bc\cos A \\ \Rightarrow \cos A & = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot A & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{b^2+c^2-a^2}{\sqrt{4b^2c^2-(b^2+c^2-a^2)^2}} \\ b^2 & = a^2+c^2-2ac\cos B \\ \Rightarrow \cos B & = \dfrac{a^2+c^2-b^2}{2ac} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot B & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{a^2+c^2-b^2}{\sqrt{4a^2c^2-(a^2+c^2-b^2)^2}} \\ c^2 & = a^2+b^2-2ab\cos C \\ \Rightarrow \cos C & = \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot C & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{a^2+b^2-c^2}{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}} \end{align*}

    Show that the denominators for \cot A, \cot B, \cot C are all equal.
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  5. #5
    Member srirahulan's Avatar
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    Re: Trig

    Great Thanks For you>>>>
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