I guess this is a really dumb question, but i've been trying quite a lot and I can't figure out how to determine the solutinons.

I need to determine the values of two angles and they don't seem to be ok. But I do know for sure that the values are right. The equation is as it follows:

$$\begin{align*} &\sin(\alpha)=1/3\;;\\ &\cos(\alpha)\cos(\beta)=0\;;\text{ and}\\ &\cos(\alpha)\sin(\beta)=0.94280899908173\;. \end{align*}$$

Because if $\sin(\alpha)$ is not $1$, neither will $\cos(\alpha)$ be $0$. Also means that $\cos(\beta)$ is $0$, resulting that $\sin(\beta)$ is $1$. Determined by calculus that $\cos(\alpha)$ is not $0.94280899908173$, I seem to be in a real trouble..

  • $\begingroup$ Maybe whatever process produced the value $0.94280899908173$ had some rounding errors of about $10^{-7}$? Or your $\frac13$ is really rather $0.333333453\ldots$? $\endgroup$ – Hagen von Eitzen Dec 1 '12 at 10:43
  • $\begingroup$ Might be so, I took from wiki the values representing coordinates of Hydrogens in methane molecule and I'm making an app to represent this in 3d. Also all values are calculated with computer so rounding errors may occur when much calculus is done (such as many sin, cos, arcsin and arccos). My $1/3$ is really $1/3$. ($0.363/1.089$). The $0.94280899908173$ is produced from $1.026719/1.089$ $\endgroup$ – Mihai Bujanca Dec 1 '12 at 10:46
  • $\begingroup$ You are aware that a number given with three to six decimals incurs an error that is easily compatible with the discrepancy you observed? And why don't you simply take the vertices and center of a regular tetrahedron? $\endgroup$ – Hagen von Eitzen Dec 1 '12 at 12:39
  • $\begingroup$ It is much more complicated, because I will recursively make larger molecule and i need all the coordinates and angles so I will be able to have the very same algorithm when dealing with other types of bounds (double or triple which have different angles). And there are also other details that are part of my program mechanism in favor of this type of approaching the problem $\endgroup$ – Mihai Bujanca Dec 1 '12 at 13:39

Your reasoning to the effect that $\cos\beta=0$ is correct, and it follows that $\sin\beta=\pm1$. Since $\sin\alpha=\frac13$, you know that $\cos^2\alpha=1-\sin^2\alpha=\frac89$, and therefore $$\cos\alpha=\pm\frac{2\sqrt2}3\approx\pm0.9428090415821$$ and $$\cos\alpha\sin\beta\approx0.9428090415821$$ as well. Your figure of $0.94280899908173$ simply isn’t quite correct, though it’s close: the three values that you give aren’t quite consistent.

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  • $\begingroup$ Thank you, seemed like I had some calculus mistake $\endgroup$ – Mihai Bujanca Dec 1 '12 at 10:50
  • $\begingroup$ @Bujanca: You’re welcome. $\endgroup$ – Brian M. Scott Dec 1 '12 at 10:50

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