# B Trigonometry Identity Question

1. Oct 8, 2017

Can someone please tell me how sin(180 - x) = sin x?

Here my attempt:

sin (180 - x) = sin 180 . cos x - cos 180 . sin x

Next? I have no idea...

2. Oct 8, 2017

### phinds

I suggest that you draw the XY axis, draw show zero degrees, X degrees, and 180-X degrees. Or do you mean that you need to show it algebraically?

3. Oct 8, 2017

Yes, I need it algebraically.

4. Oct 8, 2017

### Staff: Mentor

What are $\sin 180°$ and $\cos 180°$?

5. Oct 8, 2017

6. Oct 8, 2017

### Staff: Mentor

You should know the values of cosine for common angles. Specifically for the angles of 0, 30, 45,60,90,120,135, 150 and 180 degrees at least and their equivalents in radian measure otherwise you will never solve these kinds of problems.

Did you learn about the unit circle?

https://en.m.wikipedia.org/wiki/Unit_circle

Last edited: Oct 8, 2017
7. Oct 8, 2017

I only know the value of sin x and cos x in the range from 0 degree to 90 degree.

I don't know the value of sin x and cos x greater than 90 degree.

8. Oct 8, 2017

### Staff: Mentor

Then find out. Look at my unit circle reference. In truth you should know them for 360 degrees and beyond as that in itself is also important to understand the periodicity of the functions.

Last edited: Oct 8, 2017
9. Oct 8, 2017

### Staff: Mentor

Or calculate $\sin (90°+90°)$ by the formulas you already used.

10. Oct 9, 2017

### Svein

Or.. you know (or ought to know) that sin(90°-x) = cos(x) when 0≤x≤90°. Use that fact and the fact that 180°-x = 90° + (90°-x) and use your formula on that.