Posted by Jeremy on Monday, August 17, 2009 at 3:58pm.
Using the identity
cos(x)+i sin(x) = e^{ix}
the expression simplifies considerably:
3(cos (5pi/12) + i sin (5pi/12) / 6(cos (pi/12) + i sin (pi/12))
=3(e^{5iπ/12})/6(e^{iπ/12})
=(1/2)e^{(5iπ-iπ)/12}
=(1/2)e^{iπ/3}
=(1/2)(cos(π/3)+i sin(π/3))
=(1/2)(1/2+(√3/3)i)
=1/4+(√3/6)i
Note: the identity can be derived by the expansion of e^{ix}
=1 + ix + (ix)²/2! + (ix)³/3! + ...
=1 +ix -x²/2! - x³/3! + ...
=1 -x²/2! + x⁴/4! - x⁶/6! + ...
+ i( x - x³/3! + x⁵/5! - ...)
= cos(x) + i sin(x)
Alternatively, multiply both numerator and denominator by the conjugate of the denominator, namely (cos (pi/12)- i sin (pi/12))
to reduce the denominator to:
6(cos (pi/12)+i sin (pi/12))(cos (pi/12)-i sin (pi/12))
=6(cos²(π/12)+sin²(π/12))
=6
The numerator becomes
3(cos (5pi/12) + i sin (5pi/12)(cos (pi/12)-i sin (pi/12))
=3(cos(5&pi/12)cos(&pi/12)+sin(5&pi/12)sin(&pi/12))
+ 3i(sin(5&pi/12)cos(&pi/12)-cos(5π/12)sin(&pi/12))
=3(cos(&pi/3)+isin(&pi/3))
So the result is also
3(cos(&pi/3)+isin(&pi/3))/6
=(1/2)(1/2+(√3/3)i)
=1/4+(√3/6)i