# Thread: Sine, cosine and tan graphs

1. ## Sine, cosine and tan graphs

I'm having a lot of problems with this topic. I know what the sine, cosine and tan graphs look like.

one question i come across fequently is "given that sin 30°, what is a) sin 150° b) sin 330°. - i can work it out on a calculator but the questions on a non-calc paper. Im presuming it's something about the graph having repetitions above and below the x axis.

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.

2. Originally Posted by david18
I'm having a lot of problems with this topic. I know what the sine, cosine and tan graphs look like.

one question i come across fequently is "given that sin 30°, what is a) sin 150° b) sin 330°. - i can work it out on a calculator but the questions on a non-calc paper. Im presuming it's something about the graph having repetitions above and below the x axis.
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

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.
just use what you know about reference angles to find the answer.

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