Saturday

August 30, 2014

August 30, 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

**Related Questions**

math - Use the exact values of the sin, cos and tan of pi/3 and pi/6, and the ...

math - Use the exact values of the sin, cos and tan of pi/3 and pi/6, and the ...

math - use exact values of the sin,cos and tan of (pi/3) and (pi/6)(which I ...

Calculus - Find the area in the first quadrant bounded by the arc of the circle ...

math - use exact values of the sin,cos and tan of (pi/3) and (pi/6)(which I have...

Trig/PreCalc - Teach me how to express 5sqrt3 - 5i in polar form please. I don't...

Math - 1. On the interval [0, 2pi] what are the solutions to the equation ...

Math - complex numbers - These questions are related to de moivre's theorem: z^n...

Math(Please help) - 1)tan Q = -3/4 Find cosQ -3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5...

TRIG! - Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6...