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Math Help - complex trigo

  1. #1
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    complex trigo

    the value of 169e^(pi+sin(inverse)(12/13)+cos(inverse)(5/13))

    i know de moivres theorem cosa+isina

    but how to proceed with inverses
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  2. #2
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    Is there i=\sqrt{-1} somewhere in this formula?
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  3. #3
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    Quote Originally Posted by prasum View Post
    the value of 169e^(pi+sin(inverse)(12/13)+cos(inverse)(5/13))

    i know de moivres theorem cosa+isina

    but how to proceed with inverses
    With no "i" Demoivre's theorem does not apply. I see nothing to do but evaluate that using a calculator.

    If that was supposed to be 169e^{i(\pi+ sin^{-1}(5/13)+ cos^{-1}(12/13))}
    then use the fact that e^{a+ b}= e^{a}e^b

    Of course, e^{i sin^{-1}(5/13)}= cos(sin^{-1}(5/13))+ i sin(sin^{-1}(5/13)). You should recognize that sin^{-1}(5/13) refers to an angle in a right triangle with "opposite side" 5 and "hypotenuse" 13. What is the length of the other leg? So what are the sine and cosine?
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  4. #4
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    Hello, prasum!

    \text}Find the value of: }\:169e^{\left(\pi+\sin^{\text{-}1}\!\frac{12}{13}+\cos^{\text{-}1}\!\frac{5}{13}\right)}

    \text{Examine that exponent: }\:\pi + \underbrace{\sin^{\text{-}1}\!\!\left(\tfrac{12}{13}\right)}_{\alpha} + \underbrace{\cos^{\text{-}1}\!\!\left(\tfrac{5}{13}\right)}_{\beta}

    \text{We have: }\:\begin{Bmatrix}\alpha \:=\:\sin^{-1}\frac{12}{13} & \Rightarrow & \sin\alpha \:=\:\frac{12}{13} \\ \\[-3mm]<br />
\beta \:=\:\cos^{-1}\frac{5}{13} & \Rightarrow & \cos\beta \:=\:\frac{5}{12} \end{Bmatrix}


    Note that \alpha and \beta are the same angle!

    Code:
    
                *
               **
              * *
          13 *  * 12
            *   *
           *θ   *
          *******
             5 

    Hence, the exponent is: . \pi + \alpha + \alpha \:=\:\pi + 2\sin^{\text{-}1}\frac{12}{13}


    But what is the point of all this?

    We could have cranked out the answer on our calculator.

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