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How can you use trigonometric functions to simplify # 2 e^( ( 3 pi)/8 i ) # into a non-exponential complex number?

1 Answer

Shwetank Mauria

Mar 10, 2018

#2e^((3pi)/8i)=sqrt(2-sqrt2)+isqrt(2+sqrt2)#

Explanation:

Using trigonometric functions #re^(itheta)# can be written as

#r(costheta+isintheta)# or #rcostheta+irsintheta#

Hence #2e^((3pi)/8i)=2(cos((3pi)/8)+isin((3pi)/8))#

= #2(sqrt(2-sqrt2)/2+(sqrt(2+sqrt2)/2)i)#

= #sqrt(2-sqrt2)+isqrt(2+sqrt2)#

See details here for #cos67.5^@# and #sin67.5^@#

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