1. ## complex

express in trig form

$\displaystyle (-2 Ix)/(e^-Ix - e^(Ix)$ $\displaystyle - (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix)$ $\displaystyle - (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3$ $\displaystyle = 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix)$ $\displaystyle - e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix)$ $\displaystyle - e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3$

sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking larger than its supposed to be.

nvm figured it out

express in trig trig form

$\displaystyle ((-2 I) x)/(E^((-I) x) - E^(I x)) - ((8 I) (E^((-I) x) + E^(I x)) ((I/2) (E^((-I) x) - E^(I x)) - ((E^((-I) x) + E^(I x)) x)/2))/(E^((-I) x) - E^(I x))^3 == 4/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) - (4 E^((2 I) x))/(E^((-I) x) - E^(I x))^3 + ((12 I) x)/(E^((-I) x) - E^(I x))^3 + ((2 I) x)/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) + ((2 I) E^((2 I) x) x)/(E^((-I) x) - E^(I x))^3$
Recall that $\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}$.

You should be able to convert them all to sines and cosines, and then use trigonometric identities to simplify.

express in trig form

$\displaystyle (-2 Ix)/(e^-Ix - e^(Ix)$ $\displaystyle - (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix)$ $\displaystyle - (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3$ $\displaystyle = 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix)$ $\displaystyle - e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix)$ $\displaystyle - e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3$

sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking large

Also an exponent should be written e^{ix} in LaTeX.

Also^2 try asking WolframAlpha to simplify this.

CB