How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{15 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{15 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2017-03-30T19:13:43

The general equation is:

$r_{1} e_{1}^{θ_{1} i} - r_{2} e_{1}^{θ_{1} i} = r_{1} cos (θ_{1}) - r_{2} cos (θ_{2}) + i (r_{1} sin (θ_{1}) - r_{2} sin (θ_{2}))$ $r_1e^(theta_1i) - r_2e^(theta_1i) = r_1cos(theta_1)-r_2cos(theta_2) + i(r_1sin(theta_1)-r_2sin(theta_2))$

Explanation:

You can derive the general equation, using Euler's Formula

$e^{θ i} = cos (θ) + i sin (θ)$ $e^(thetai) = cos(theta) + isin(theta)$

$r e^{θ i} = r cos (θ) + i (r sin (θ))$ $re^(thetai) = rcos(theta) + i(rsin(theta))$

For the given expression:

$e^{\frac{11 π}{6} i} - e^{\frac{15 π}{8} i} = cos (\frac{11 π}{6}) - cos (\frac{15 π}{8}) + i (sin (\frac{11 π}{6}) - sin (\frac{15 π}{8}))$ $e^((11pi)/6i) - e^((15pi)/8i) = cos((11pi)/6) - cos((15pi)/8) + i(sin((11pi)/6) - sin((15pi)/8))$

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{15 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i)$ using trigonometric functions?

1 Answer

Explanation:

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How do you evaluate e11π6i−e15π8i e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i) using trigonometric functions?

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{15 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i)$ using trigonometric functions?