maybe i am reading something wrong or you are not stating what it is you want clearly. but that question seems to have nothing to do with the graphing the functions to me. it seems to have to do with the double angle formulae, do you remember them. please tell me if i am misinterpreting

if you are given sin30 = 1/2 ........which is what i think you meant to say, we can find

sin(150) = sin(30 + 120)

............= sin30cos120 + sin120cos30

............= (1/2)(-1/2) + (sqrt(3)/2)(sqrt(3)/2)

............= -1/4 + 3/4

............= 1/2

and

sin(330) = sin(30 + 300)

............= sin30cos300 + sin300cos30

............= (1/2)(1/2) + (-sqrt(3)/2)(sqrt(3)/2)

............= 1/4 - 3/4

............= -1/2

just use what you know about reference angles to find the answer.another question tat frequently pops up is something like "find two different values of x between 0 and 180 for which sin (2x)° = sin 30°". I can work out one of them as 15° but cannot find the second solution and would like to know how it can be found.

we know that sin30 = sin(180 - 30) = sin(150)

so, we can do two equations to find two values of x

sin(2x) = sin30

and

sin(2x) = sin(150)

=> 2x = 30

so x = 15

and

=> 2x = 150

so x = 75