# Vector components

1. Oct 18, 2005

### th77

I have a problem that includes an acceleration vector 'a' located in the 3 quadrant and it makes an angle (theta) 30 degrees to the Negative Y axis.
The solution manual shows the the vector resolved like
a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j
Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.

2. Oct 18, 2005

### vanesch

Staff Emeritus

Because the angle is here defined differently than usual...

3. Oct 18, 2005

### whozum

The reason you flip them is because the reference angle is made with the y axis instead of the x. Usually we measure the angle from the x axis.. The reason it works when you flip then is due to the two properties of sin and cos:

$\sin(x-90) = \cos(x)$

$\cos(x-90) = \sin(x)$

4. Oct 18, 2005

### th77

Thanks!! That leands me to another question...
In this problem, the angle is 30 degrees with the negative axis so shouldn't cos 240 be equal to sin 30? They come to -0.5 and 0.5 respectively.

5. Oct 18, 2005

### vanesch

Staff Emeritus
You're turning in the wrong direction: 30 degrees starting from the negative Y axis (=270 degrees) gives you 300 degrees, not 240...

6. Oct 18, 2005

### th77

the vector is in the 3rd quadrant so isn't it 240?

7. Oct 18, 2005

### whozum

$\cos(240)$ measures the (60 deg) reference angle from the negative x axis in the third quadrant, where cosine is always negative. However if you are going to refer to the angle from hte positive x axis, then you are also taking $\sin(240)$ and not $\sin(30)$. Those relations are technically supposed to be used in the first quadrant, or just as a relative measure.