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

I get this.

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)#