How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/12 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2016-06-10T05:22:32

$e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i} = - (1 + cos (\frac{π}{12})) + i sin (\frac{π}{12})$ $e^((3pi)/2i)-e^(pi/12i)=-(1+cos(pi/12))+isin(pi/12)$

Explanation:

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

Hence $e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i}$ $e^((3pi)/2i)-e^(pi/12i)$

= $[cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2})] - [cos (\frac{π}{12}) + i sin (\frac{π}{12})]$ $[cos((3pi)/2)+isin((3pi)/2)]-[cos(pi/12)+isin(pi/12)]$

But $cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2)=0$ and $sin (\frac{3 π}{2}) = - 1$ $sin((3pi)/2)=-1$

Hence $e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i}$ $e^((3pi)/2i)-e^(pi/12i)$

= $[0 - 1] - [cos (\frac{π}{12}) + i sin (\frac{π}{12})]$ $[0-1]-[cos(pi/12)+isin(pi/12)]$

= $- (1 + cos (\frac{π}{12})) + i sin (\frac{π}{12})$ $-(1+cos(pi/12))+isin(pi/12)$

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How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{12} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/12 i)$ using trigonometric functions?

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