How do you evaluate # e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)# using trigonometric functions?

1 Answer
A. S. Adikesavan
Apr 9, 2016

#(sqrt3/2+sqrt((sqrt2+1)/(2sqrt2)))-i(1/2-sqrt((sqrt2-1)/(2sqrt2)))#

Explanation:

#e^(ix)=cos x + i sinx#.
#cos(2pi-x)=cos x, sin (2pi-x)=-sin x, cos(pi+x)=-cos x and sin(pi+x)=-sin x#.

So, #e^(11pi/6i)=cos (11pi/6)+i sin(11pi/6)=cos(2pi-pi/6)+i sin(2pi-pi/6)=cos (pi/6)-i sin (pi/6)=sqrt3/2-i(1/2)#.
#e^(9pi/8)=cos (9pi/8)+i sin(9pi/8)=cos(pi+pi/8)+i sin(pi+pi/8)=-cos (pi/8)-i sin( pi/8)#.
#cos A = sqrt((1+cos 2A)/2), sin A=sqrt((1-cos 2A)/2)#

So, #cos(pi/8)=sqrt((1+cos (pi/4))/2) = sqrt((sqrt2+1)/(2sqrt2))#.
#sin(pi/8)=sqrt((1-cos (pi/4))/2) = sqrt((sqrt2-1)/(2sqrt2))#.

Npw, #e^(9pi/8)=sqrt((sqrt2+1)/(2sqrt2))-i sqrt((sqrt2-1)/(2sqrt2))#.

So, the given expression =#(sqrt3/2+sqrt((sqrt2+1)/(2sqrt2)))-i(1/2-sqrt((sqrt2-1)/(2sqrt2)))#

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