http://i146.photobucket.com/albums/r...6/problem1.jpg

I've never done any like this before (Worried) help!

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- November 30th 2008, 02:22 PMNeverendingweird trig problem
http://i146.photobucket.com/albums/r...6/problem1.jpg

I've never done any like this before (Worried) help! - November 30th 2008, 02:46 PMShivian
Not the best at explaining things but here goes:

First use the unit circle to find all angles in which sine equals 1/2 and write them down in radians.

Sine works with the interval equation angle + 2*pi*k

Since it is double angle (sine2theta) you must use the equation for both of the radians you wrote down before and divide by 2 to get your formulas.

You then plug in numbers to K to find all occurrences where theta equals anything in the interval 0 to 2pi

sin2theta = 1/2

2theta = pi/6 , 5pi/6

first equation

2theta = pi/6 + 2*pi*k

theta = pi/12 + pi*k

second equation

2theta = 5pi/6 + 2*pi*k

theta = 5pi/12 + pi*k

Try plugging in 0,1,2,3 and so on to both equations and your answer will be all the numbers that are between your interval of 0 and 2pi.

Hope that helps somewhat, but I'm sure someone else can explain it much better. - November 30th 2008, 03:59 PMNeverending
surprisingly.. I understand your explanation =D I came out with:

pi/12, 13pi/12, 5pi/12, and 17pi/12

I think that's it. Thanks so much for the help!