How can you use trigonometric functions to simplify # 6 e^( ( pi)/8 i ) # into a non-exponential complex number?

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Answer 1 · 2016-03-25T12:41:12

#6 e^{i pi/8} = 6 (cos(pi/8) + i sin(pi/8))#

#= 3sqrt{2+sqrt2} + i(3sqrt{2-sqrt2})#

Euler showed that for a complex number #z = r e^{i theta}#,

#r e^{i theta} = r (cos(theta) + i sin(theta))#

In this case

Therefore,

#6 e^{i pi/8} = 6 (cos(pi/8) + i sin(pi/8))#

You can use half angle formulas to find #sin(pi/8)# and #cos(pi/8)#.

#6 (cos(pi/8) + i sin(pi/8)) = 6(sqrt{2+sqrt2}/2 + i(sqrt{2-sqrt2}/2))#

#= 3sqrt{2+sqrt2} + i(3sqrt{2-sqrt2})#