Question

How do you evaluate `e^( ( pi)/2 i) - e^( ( 19 pi)/12 i)` using trigonometric functions?

Answer 1

I get this.

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

`\mathbf(e^(ix) = isinx+cosx)`

Thus, you can substitute `x = pi/2` or `(19pi)/12` to give:

`e^(i*pi"/"2) - e^(i*19pi"/"12)`

`= [isin(pi/2) + cos(pi/2)] - [isin((19pi)/12) + cos((19pi)/12)]`

We know that `(19pi)/12 = [(19*180)/12]^@ = 285^@`, so...

`= color(green)([isin(90^@) + cos(90^@)] - [isin(285^@) + cos(285^@)])`

Okay, so what is `sin` or `cos(285^@)`? We could use some trig relationships to get:

`sin(285^@) = sin(-75^@) = -sin(75^@) = -sin(30^@+45^@)`

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

`=> -[sin30^@cos45^@ + cos30^@sin45^@]`

`= -(1/2*sqrt2/2 + sqrt3/2*sqrt2/2)`

`= (-sqrt2-sqrt6)/4 ~~ -0.9659`

Then, consider `cos(285^@)` to be:

`cos(285^@) = cos(-285^@) = cos(75^@) = cos(30^@+45^@)`

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

`=> cos30^@cos45^@ - sin30^@sin45^@`

`= sqrt3/2*sqrt2/2 - 1/2*sqrt2/2`

`= (sqrt6 - sqrt2)/4 ~~ 0.2588`

So overall, we have:

`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)`