How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{3 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 3 pi)/4 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{3 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 3 pi)/4 i)$ using trigonometric functions?

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Answer 1 · 2017-02-07T09:33:25

$= \frac{1}{\sqrt{2}} (1 - \sqrt{1 - \frac{1}{\sqrt{2}}} + i (\sqrt{1 + \frac{1}{\sqrt{2}}} - 1))$ $=1/sqrt2(1-sqrt(1-1/sqrt2)+i(sqrt(1+1/sqrt2) -1))$
= $0.3242 + i 0.2168$ $0.3242+i0.2168$ , nearly. After some time, I would edit my answer myself, for bugs, if any.

Explanation:

Use

$e^{i 2 π} = cos 2 π + i sin 2 π = 1$ $e^(i2pi)=cos2pi+isin2pi=1$ and

$e^{i π} = cos π + i sin π = - 1$ $e^(ipi)=cospi+isinpi=-1$

The given expression is

$e^{i π} e^{i \frac{5}{8} π} - e^{i π} e^{- i \frac{1}{4} π}$ $e^(ipi)e^(i5/8pi)-e^(ipi)e^(-i1/4pi)$

$= (cos ({112.5}^{o}) + i sin ({112.5}^{o})) + ({cos 45}^{0} - i {sin 45}^{o}))$ $=(cos(112.5^o)+isin(112.5^o))+(cos45^0-isin45^o))$

$= - {sin 22.5}^{o} + {cos 45}^{\oplus} i ({cos 22.5}^{o} - {sin 45}^{o})$ $=-sin22.5^o +cos45^o+i(cos22.5^o -sin45^o)$

$= \frac{1}{\sqrt{2}} - \sqrt{\frac{1 - {cos 45}^{0}}{2}} + i ⎛ ⎝ - \frac{1}{\sqrt{2}} + \sqrt{\frac{1 + {cos 45}^{0}}{2}} ⎞ ⎠$ $=1/sqrt2-sqrt((1-cos45^0)/2)+i(-1/sqrt2+sqrt((1+cos45^0)/2))$

$= \frac{1}{\sqrt{2}} (1 - \sqrt{1 - \frac{1}{\sqrt{2}}} + i (\sqrt{1 + \frac{1}{\sqrt{2}}} - 1))$ $=1/sqrt2(1-sqrt(1-1/sqrt2)+i(sqrt(1+1/sqrt2) -1))$

$= 0.3244 + i 0.2168$ $=0.3244+i0.2168$ , nearly.

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How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{3 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 3 pi)/4 i)$ using trigonometric functions?

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