How can you use trigonometric functions to simplify # 16 e^( ( 23 pi)/12 i ) # into a non-exponential complex number?

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Answer 1 · 2016-01-13T01:10:06

#z = 8sqrt(2+sqrt(3)) -8sqrt(2-sqrt(3))i#

#e^((23pi)i/12) = cos(23pi/12) + isin(23pi/12)#

Since #23pi/12 = 2pi - pi/12# so we have

#e^((23pi)i/12) = cos(2pi - pi/12) + isin(2pi - pi/12)#

But since #2pi# is the period, we can safely ignore it

#e^((23pi)i/12) = cos(-pi/12) + isin(-pi/12)#

Remember that #cos(-theta) = cos(theta)# and #sin(-theta) = -sin(theta)#

#e^((23pi)i/12) = cos(pi/12) - isin(pi/12)#

From there, it's using half angle formulas to figure out these values

#sin(u/2) = sqrt((1-cos(u))/2)# and #cos(u) = sqrt((1+cos(u))/2)#

So

#sin(pi/12) = sqrt((1 - sqrt(3)/2)/2) = sqrt((2-sqrt(3))/4) = sqrt(2-sqrt(3))/2#

#cos(pi/12) = sqrt((1 + sqrt(3)/2)/2) = sqrt((2+sqrt(3))/4) = sqrt(2+sqrt(3))/2#

#e^((23pi)i/12) = sqrt(2+sqrt(3))/2 -isqrt(2-sqrt(3))/2#

And lastly, multiply by #16# so we have

#z = 8sqrt(2+sqrt(3)) -8sqrt(2-sqrt(3))i#