How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{19 π}{12} i}$ $e^( ( pi)/2 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{19 π}{12} i}$ $e^( ( pi)/2 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2016-04-27T08:41:01

I get this.

Interestingly enough, you can use Euler's formula, which states:

$e^{i x} = i sin x + cos x$ $\mathbf(e^(ix) = isinx+cosx)$

Thus, you can substitute $x = \frac{π}{2}$ $x = pi/2$ or $\frac{19 π}{12}$ $(19pi)/12$ to give:

$e^{i \cdot π / 2} - e^{i \cdot 19 π / 12}$ $e^(i*pi"/"2) - e^(i*19pi"/"12)$

$= [i sin (\frac{π}{2}) + cos (\frac{π}{2})] - [i sin (\frac{19 π}{12}) + cos (\frac{19 π}{12})]$ $= [isin(pi/2) + cos(pi/2)] - [isin((19pi)/12) + cos((19pi)/12)]$

We know that $\frac{19 π}{12} = {[\frac{19 \cdot 180}{12}]}^{\circ} = 285^{\circ}$ $(19pi)/12 = [(19*180)/12]^@ = 285^@$ , so...

$= [i sin (90^{\circ}) + cos (90^{\circ})] - [i sin (285^{\circ}) + cos (285^{\circ})]$ $= color(green)([isin(90^@) + cos(90^@)] - [isin(285^@) + cos(285^@)])$

Okay, so what is $sin$ $sin$ or $cos (285^{\circ})$ $cos(285^@)$ ? We could use some trig relationships to get:

$sin (285^{\circ}) = sin (- 75^{\circ}) = - sin (75^{\circ}) = - sin (30^{\circ} + 45^{\circ})$ $sin(285^@) = sin(-75^@) = -sin(75^@) = -sin(30^@+45^@)$

Using the addition identity $sin (u + v) = sin u cos v + cos u sin v$ $sin(u + v) = sinucosv + cosusinv$ ,

$\Rightarrow - [{sin 30}^{\circ} {cos 45}^{\circ} + {cos 30}^{\circ} {sin 45}^{\circ}]$ $=> -[sin30^@cos45^@ + cos30^@sin45^@]$

$= - (\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2})$ $= -(1/2*sqrt2/2 + sqrt3/2*sqrt2/2)$

$= \frac{- \sqrt{2} - \sqrt{6}}{4} \approx - 0.9659$ $= (-sqrt2-sqrt6)/4 ~~ -0.9659$

Then, consider $cos (285^{\circ})$ $cos(285^@)$ to be:

$cos (285^{\circ}) = cos (- 285^{\circ}) = cos (75^{\circ}) = cos (30^{\circ} + 45^{\circ})$ $cos(285^@) = cos(-285^@) = cos(75^@) = cos(30^@+45^@)$

Using the addition identity $cos (u + v) = cos u cos v - sin u sin v$ $cos(u + v) = cosucosv - sinusinv$ ,

$\Rightarrow {cos 30}^{\circ} {cos 45}^{\circ} - {sin 30}^{\circ} {sin 45}^{\circ}$ $=> cos30^@cos45^@ - sin30^@sin45^@$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ $= sqrt3/2*sqrt2/2 - 1/2*sqrt2/2$

$= \frac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588$ $= (sqrt6 - sqrt2)/4 ~~ 0.2588$

So overall, we have:

$e^{i \cdot π / 2} - e^{i \cdot 19 π / 12}$ $color(blue)(e^(i*pi"/"2) - e^(i*19pi"/"12))$

$= [icancel(sin(90^@)) + cancel(cos(90^@))^0] - [isin(285^@) + cos(285^@)]$

$= i - i[(-sqrt2-sqrt6)/4] - [(sqrt6 - sqrt2)/4]$

$= -(sqrt6 - sqrt2)/4 + (1 - (-sqrt2-sqrt6)/4)i$

$= color(blue)(-(sqrt6 - sqrt2)/4 + (1 + (sqrt2+sqrt6)/4)i)$

$~~ color(blue)(-0.2588 + 1.9659i)$

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{19 π}{12} i}$ $e^( ( pi)/2 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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