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I have this question regarding a rotation about the $y$ axis given as the following:

\begin{bmatrix} 0.6&0&-0.8\\ 0&1&0\\ 0.8&0&0.6\\ \end{bmatrix}

I realized this takes the same form as the rotation about the $y$ axis and proceeded to calculate the value of theta that it represented. So I calculated the inverse cosine of $0.6$ and got $53.13$ degrees however when I calculated my value of theta for $-0.8$ I got $-53.13$ degrees. I am taught that if the values of theta are different, we must find the next possible value that fits both Cosine and Sine and use this as the angle of rotation.

My question is, does it matter if we obtain a value of theta that is just the negative of the other? I then calculated that a suitable value of theta for which it fits both Cosine and Sine which was $-53.13$ degrees. The answer is apparently $53.13$ degrees and not negative or can this negative be ignored?

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  • $\begingroup$ $\cos(\theta) = \cos(-\theta)$ $\endgroup$ – Henry May 20 '17 at 9:58
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In three dimensions, the angle of rotation is only determined up to sign. Viewed from "above" a rotation may appear to be through an angle $\theta$ clockwise, but from "below" it will appear to be through an angle $\theta$ anticlockwise.

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  • $\begingroup$ Ok, so the fact we have different signs of the same angle for this rotation doesn't make a difference, the rotation is still generically $53.13$ degrees dependent on the way it is "viewed"? $\endgroup$ – JayVB May 20 '17 at 9:59
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    $\begingroup$ That's right. You can't consistently attach a sign to the rotation. $\endgroup$ – Lord Shark the Unknown May 20 '17 at 10:00
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Angles extend on four quadrants.

Knowing one of sine, cosine or tangent alone leaves an undeterminacy between two quadrants. To lift the ambiguity, you need to check the sign of another trigonometric function.

$$\begin{matrix}Quadrant&I&II&III&IV\\Sine&+&+&-&-\\Cosine&+&-&-&+\\Tangent&+&-&+&-\end{matrix}$$

But as said by @LordShark, you need to know how the positive direction of rotation is defined.

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