Solving Sin theta Questions: An Example

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Homework Help Overview

The discussion revolves around solving sine equations using complex exponential forms, particularly focusing on the equation involving the imaginary unit and Euler's formula. The original poster attempts to understand how to manipulate the equation to find the angle corresponding to a given sine value.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's method of using the complex exponential form of sine and question the presence of plus/minus signs in the equation. There is an exploration of how to solve for the angle x using this form, with references to simplifying the equation and using Euler's formula.

Discussion Status

Some participants provide guidance on using Euler's formula correctly and suggest a more straightforward approach to solving for x. The discussion is ongoing, with various interpretations and methods being explored without a clear consensus on the best approach.

Contextual Notes

There is mention of confusion regarding the use of the complex exponential form and the implications of the plus/minus signs in the equation. The original poster expresses uncertainty about the results obtained from their calculations.

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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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Solving for x is as simple as taking the arcsine...are you trying to solve for x by exclusively using the complex exponential form of sine?

See this article: http://en.wikipedia.org/wiki/Euler's_formula

As far as I can tell, there should be no +/- in the equation, it should all be taken care of.
 
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
 
It is a quadratic equation in disguise.
 

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