How do you evaluate e23π12i−e7π12i using trigonometric functions?

1 Answer
IDKwhatName
Jul 22, 2017

e23π12i−e7π12i=√62−i√62

Explanation:

ex⋅i=cos(x)+isin(x)

e23π12i−e7π12i=(cos(23π12)+isin(23π12))−(cos(7π12)+isin(7π12))

cos(23π12)=√6+√24
sin(23π12)=−√6+√24
cos(7π12)=−√6+√24
sin(7π12)=√6+√24

e23π12i−e7π12i=(√6+√24+i−√6+√24)−(−√6+√24+i√6+√24)

√6+√24−−√6+√24=√6+√6+√2−√24=2√64=√62

−√6+√24−√6+√24=−√6−√6+√2−√24=−2√64=−√62

e23π12i−e7π12i=√62−i√62

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