How do you evaluate #e^( ( pi)/12 i) - e^( ( pi)/8 i)# using trigonometric functions?

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Answer 1 · 2018-07-27T14:12:39

#color(chocolate)(=> 0.042 - 0.1239#, IV Quadrant

#e^(i theta) = cos theta + i sin theta#

#e^(((pi)/12) i )= cos ((pi)/12 + i sin ((pi)/12)#

#~~> 0.9659 + 0.2588 i#, I Quadrant

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

#=> 0.9239 + 0.3827 i#, I Quadrant.

#e^(((pi)/12)i) - e^(((pi)/8)i) = 0.9659 - 0.9239 + 0.2588 i - 0.3827#

#color(chocolate)(=> 0.042 - 0.1239#, IV Quadrant.