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December 19, 2014

December 19, 2014

Posted by **Jeremy** on Monday, August 17, 2009 at 3:58pm.

I have no idea what I did but I got (1/4) - 433i/1000.

- Math -
**MathMate**, Monday, August 17, 2009 at 5:06pmUsing 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

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