# Sine, cosine and tan graphs

1. Apr 29, 2007

### 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.

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. Apr 29, 2007

### Fredrik

Staff Emeritus
You know how to define cos and sin as the x and y coordinates of points on the unit circle, right? Then it should be easy. Just draw the unit circle and lines through the origin that make 30°, 150° and 330° angles with the x axis. If you know that sin is the y coordinate of the point where the line intersects the unit circle, the answer will be almost obvious once you've drawn the picture.

You solved the first part of the second problem by realizing that two 15° rotations are equal to a 30° rotation. To solve the second part, the first thing you have to do is to find a second point on the unit circle with the y coordinate equal sin 30°. What's the angle between the x axis and a line from the origin to that point? The answer you seek is half of that.

3. Apr 29, 2007

### turdferguson

Either brush up on the unit circle or do a bunch of memorizing:

4. Apr 29, 2007

### christianjb

If you can draw sin(x) and cos(x) with reasonable accuracy over 1 period, then the answers are obvious from looking at the symmetry of the functions.

Try and draw sin such that it crosses the x axis at 0,180 and 360 and it has a maximum at 90, minimum at 270 (half-way between 180 and 360).

Draw cos such that it crosses the x axis at 90 and 270 and has a maximum at 0,360 and a minimum at 180.

I find that to be much easier than memorizing the above formulae.

Fredrik's method might be easier for you. Whatever works best, but it does require some practice.

5. Apr 29, 2007

### Feldoh

You should familiarize yourself with the unit circle: http://www.humboldt.edu/~dlj1/PreCalculus/Images/UnitCircle.html

Once you do you'll start to realize things like:
$$sin(30) = 1/2$$
$$sin(150) = 1/2$$
$$sin(330) = -1/2$$

To find sin(2x) = sin(30) over the interval (0,180] becomes simple from above:
sin(2x) = 1/2 and x = 15 and 75 (half of 30 and 150)

Last edited: Apr 29, 2007
6. Apr 29, 2007

### ZioX

Man, these guys are nuts.

All you need to do is to be able to trigonometry with the 30-60-90 and the 45-45-90 triangle. Everything else follows from the unit circle.

7. Apr 30, 2007

### Feldoh

What's the difference? If you know trig with those two triangles you've memorized like 80% of the basic unit circle anyways :tongue: