How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{17 π}{12} i}$ $e^( ( 13 pi)/8 i) - e^( (17 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2016-03-26T05:01:48

$e^{\frac{13 π}{8} i} - e^{\frac{17 π}{12} i} = 0.0003 + 0.01142 i$ $e^((13pi)/8i)-e^((17pi)/12i)=0.0003+0.01142i$ #

As $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$

$e^{\frac{13 π}{8} i} = cos (\frac{13 π}{8}) + i sin (\frac{13 π}{8})$ $e^((13pi)/8i)=cos((13pi)/8)+isin((13pi)/8)$

= $cos (\frac{13 π}{8} - 2 π) + i sin (\frac{13 π}{8} - 2 π)$ $cos((13pi)/8-2pi)+isin((13pi)/8-2pi)$

= $cos (\frac{- 3 π}{8}) + i sin (\frac{- 3 π}{8})$ $cos((-3pi)/8)+isin((-3pi)/8)$

= $0.99979 - 0.02056 i$ $0.99979-0.02056i$ (using scientific calculator)

Similarly $e^{\frac{17 π}{12} i} = cos (\frac{17 π}{12}) + i sin (\frac{17 π}{12})$ $e^((17pi)/12i)=cos((17pi)/12)+isin((17pi)/12)$

= $cos (\frac{17 π}{12} - 2 π) + i sin (\frac{17 π}{12} - 2 π)$ $cos((17pi)/12-2pi)+isin((17pi)/12-2pi)$

= $cos (\frac{- 7 π}{12}) + i sin (\frac{- 7 π}{12})$ $cos((-7pi)/12)+isin((-7pi)/12)$

= $0.99949 - 0.03198 i$ $0.99949-0.03198i$ (using scientific calculator)

Hence $e^{\frac{13 π}{8} i} - e^{\frac{17 π}{12} i}$ $e^((13pi)/8i)-e^((17pi)/12i)$

= $(0.99979 - 0.02056 i) - (0.99949 - 0.03198 i)$ $(0.99979-0.02056i)-(0.99949-0.03198i)$

= $(0.99979 - 0.99949) - (0.02056 - 0.03198) i$ $(0.99979-0.99949)-(0.02056-0.03198)i$

= $0.0003 - (- 0.01142) i$ $0.0003-(-0.01142)i$

= $0.0003 + 0.01142 i$ $0.0003+0.01142i$

How do you evaluate e13π8i−e17π12i e^( ( 13 pi)/8 i) - e^( (17 pi)/12 i) using trigonometric functions?