Solving Sin theta Questions: An Example

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Homework Statement



sorry if the title is not that descriptive

Homework Equations


i is the imaginary number
e is 2.7...
x is any anlgle in radians that is... um see the example I explain it

sin x = +/- i/2 (1/e^(ix) +/- e^(ix))

I will use a example in atempt at a solution to show you how to use this equation

The Attempt at a Solution



the first +/- depends on were the angle is located on the unit circle

the second +/- depends on if the angle is smaller or larger than pi/4
- if it is smaller than pi/4 (by the way pi as in 3.14...) just simple enter your angle in the equation above and make the +/- a "-" in its radian measure
- if it is larger than pi/4 take your angle and subtract pi/4 from it and use that as your x and use a "+" instead of minus in the equation

- if anlge is greater than pi/2 use coresponding angle in the first quadrant if you want and just work out the "+/-" in the very beginning of the equation

EXAMPLE

I want to know what the sin of 60 degrees is... I know it is SQRT(3)/2 but i'll use the formula above for a demonstration ok...

so 60 degrees in radians is pi/3 which is greater than pi/4 so I have to add pi/4 from pi/3... I'll do this in degrees sense it is easier... 60-45 = 15 degrees

and now what I want to do is take that angle and do 45 minus that angle
45-15 = 30 degrees = pi/6

now just plug this into the equation above
sin x = +/- i/2 (1/e^(ix) +/- e^(ix))
sin x = + i/2 (1/e^((i pi)/6) + e^((i pi)/6) and the calculator gives .8660254038 i
not really sure why it gives the i? can someone answer that?
if you remove the i this is the exact value for sin pi/3 or simple SQRT(3)/2

try using the equation for other angles for example
sin 1 degree = i/2(1/e^((i pi)/180) - e^((i pi)/180)) = .0174524064 which is the exact value of sin 1 degree

MY QUESTION is how do I go backwards
for example

sin x = +/- i/2 (1/e^(ix) +/- e^(ix)) = SQRT(3)

how do I solve for x?

THANK YOU SO MUCH
 
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yes i am trying to solve for x in that formula how do i do this?
 
You've made using Euler's formula way more complicated than it needs to be. Euler's formula is

[tex]\sin x = \frac{e^{ix}-e^{-ix}}{2i}[/tex]

for any value of x. You don't have plus or minuses depending on what the value of x is.

To go backwards, let [itex]z=e^{ix}[/tex]. Then [itex]1/z = e^{-ix}[/itex], so the formula becomes:<br /> <br /> [tex]\sin x = \frac{z-\frac{1}{z}}{2i}[/tex]<br /> <br /> Solve for z. Once you have z, you can solve for x.[/itex]
 
I don't know were to go from here

2i sin x = z - 1/z