How can you use trigonometric functions to simplify 7e13π12i into a non-exponential complex number?

1 Answer
Narad T.
Jul 25, 2017

The answer is =−74((√6+√2))+i74((√2−√6))

Explanation:

We apply Euler's relation

eiθ=cosθ+isinθ

Therefore,

7ei1312π=7(cos(1312π)+isin(1312π))

cos(1312π)=cos(π+112π)=−cos(112π)

=−cos(π3−π4)=−cos(π3)cos(π4)−sin(π3)sin(π4)

=−12⋅√22−√32⋅√22

=−√6+√24

sin(1312π)=sin(π+112π)=−sin(112π)

=−sin(π3−π4)=−sin(π3)cos(π4)+cos(π3)sin(π4)

=−√32⋅√22+12⋅√22

=√2−√64

Therefore,

7ei1312π=−7⋅(√6+√2)4+i⋅7(√2−√6)4

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