How can you use trigonometric functions to simplify $16 e^{\frac{5 π}{6} i}$ $16 e^( ( 5 pi)/6 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $16 e^{\frac{5 π}{6} i}$ $16 e^( ( 5 pi)/6 i )$ into a non-exponential complex number?

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Answer 1 · 2016-01-29T12:01:26

$16 [cos (\frac{5 π}{6}) + i sin (\frac{5 π}{6})]$ $16 [ cos((5pi)/6) + isin((5pi)/6 ) ]$

Explanation:

Euler's theorem states : $e^{i θ} = cos θ + i sin θ$ $e^(itheta) = costheta + isintheta$

Here $θ = \frac{5 π}{6}$ $theta =( 5pi)/6$

hence $16 e^{\frac{5 π}{6} i = 16 [cos (\frac{5 π}{6}) + i sin (\frac{5 π}{6})]}$ $16e^((5pi)/6 i = 16 [ cos((5pi)/6) + isin((5pi)/6) ]$

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How can you use trigonometric functions to simplify $16 e^{\frac{5 π}{6} i}$ $16 e^( ( 5 pi)/6 i )$ into a non-exponential complex number?

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Explanation:

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Science

Math

Humanities

... and beyond

How can you use trigonometric functions to simplify 16e5π6i 16 e^( ( 5 pi)/6 i ) into a non-exponential complex number?

1 Answer

Explanation:

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How can you use trigonometric functions to simplify $16 e^{\frac{5 π}{6} i}$ $16 e^( ( 5 pi)/6 i )$ into a non-exponential complex number?