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how to find the minimum and maximum value of a angle on the following equation:

$$\theta = \displaystyle\arccos\bigg[\dfrac{\frac x{255}}{\sqrt{(\frac{x}{255})^2+(\frac{y}{255})^2+(\frac{z}{255})^2}}\bigg]$$

Assuming $x,y ,z$ are values from ranges from $1\; \text{to}\; 255$ each.

$x_{Min} = 1, y_{Min} = 1, z_{Min} = 1$

$x_{Max} = 255, y_{Max} = 255, z_{Max} = 255$

I wanted to know what are the maximum and minimum values for $\theta$ (In radians)

Many thanks

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  • $\begingroup$ At least you can remove the $255$ giving $$\theta=\arccos\left(\frac x{\sqrt{x^2+y^2+z^2}}\right)$$ $\endgroup$ – TheSimpliFire Jun 2 '18 at 9:18
  • $\begingroup$ Try plugging the max and min values for $x,y,z$ $\endgroup$ – John Glenn Jun 2 '18 at 12:51
  • $\begingroup$ Thanks, i tried with a combination of the maximum and minimum values for x,y,x and reached this: MIn:0.005545878680246213278454190712838542737925083234725246248. Max = 1.566874808423639808746573123890695629216279098460007008786. Which seems to be a formula such as: ThethaMIn = arccos(255/(sqrt(255^2 + 2))) and thethaMax = arccos(1/(sqrt(255^2 + 2))) But, i want to make sure these are the correct values and formula for min/max $\endgroup$ – Gustavo Trigueiros Jun 2 '18 at 17:13
  • $\begingroup$ @Gustavo Trigueiros without the boundary constraints, the max value is $\pi$ with ($x\to-1,y\to0,z\to0$) but since they are outside the boundary, you should check the values at the boundary $\endgroup$ – John Glenn Jun 4 '18 at 7:33

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