How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( 4 pi)/3 i)# using trigonometric functions?

1 Answer
bp
Feb 13, 2016

# 2[sin ((5pi)/4)sin (pi/12) + i cos ((5pi)/4) sin (pi/12)]#

Explanation:

#e^((7pi i)/6 )= cos ((7pi)/6) - i sin ((7pi)/6)#

# e^((4pi i)/3)= cos ((4pi)/3) -i sin ((4pi)/3)#

The required simplification would be

#cos ((7pi )/6) - cos((4pi )/3) +i[sin ((4pi)/3) - sin ((7pi)/6)]#

= # 2 sin( ((7pi)/6+ (4pi)/3)/2) sin( ((4pi)/3 -(7pi)/6)/2) +i[2 cos(((4pi)/3 +(7pi)/6)/2) sin (((4pi)/3 -(7pi)/6)/2)]#

= # 2[sin ((5pi)/4)sin (pi/12) + i cos ((5pi)/4) sin (pi/12)]#

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