Angle of rotation in a matrix

**smellatron** · February 4th 2009, 05:41 PM

The matrix [-0.8 -0.6
0.6 -0.8] (a 2x2 matrix)

represents a rotation. Find the angle of rotation (in radians).

This one has me stumped, any help would be appreciated. The answer is 2.5 radians but the approach is where I am lost.

**badgerigar** · February 4th 2009, 07:40 PM

You are probably expected to know that a rotation $\theta$ can be represented by the matrix
$\left[\begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right]$
This does not, however, give 2.5 radians so I suspect you have either copied this out wrong or the source is incorrect.

**smellatron** · February 5th 2009, 05:15 AM

I understand that is the rotation matrix, what I don't understand is how you come up with the final angle of rotation. Do you plug in the numbers into the corresponding thetas? Thanks again for the help. I would be more specific but I have to head to work. Any help would be appreciated.

**Opalg** · February 5th 2009, 07:09 AM

Originally Posted by badgerigar

You are probably expected to know that a rotation $\theta$ can be represented by the matrix
$\left[\begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right]$
This does not, however, give 2.5 radians so I suspect you have either copied this out wrong or the source is incorrect.

I think it does give 2.5 radians (to 2 significant figures). If $\cos\theta$ is negative and $\sin\theta$ is positive then $\pi/2<\theta<\pi$ . The angle in that interval with $\cos\theta = -0.8$ is $\pi - \arccos0.8 \approx 3.14 - 0.64 = 2.5$ .

**badgerigar** · February 6th 2009, 01:23 AM

Opalg is right, sorry to have confused you

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