Zoagliaj

2022-07-20

I want to find out the angle for the expression ${a}^{3}+{b}^{3}={c}^{3}$.

kamphundg4

Expert

Step 1
There is no single angle corresponding to the relationship ${a}^{3}+{b}^{3}={c}^{3}$.
Suppose that a triangle has sides of lengths a,b, and c such that ${a}^{3}+{b}^{3}={c}^{3}$. We know from the law of cosines that if θ is the angle opposite the side of length c, then ${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta$, so $\mathrm{cos}\theta =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\phantom{\rule{thickmathspace}{0ex}}.$.
Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]{2}$, and $\mathrm{cos}\theta =\frac{2-{2}^{2/3}}{2}\approx 0.20630\phantom{\rule{thickmathspace}{0ex}}.$.
Step 2
If $a=1$ and $b=2$, then $c=\sqrt[3]{9}$, and $\mathrm{cos}\theta =\frac{5-{9}^{2/3}}{4}\approx 0.16831\phantom{\rule{thickmathspace}{0ex}}.$.
If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $\mathrm{cos}\theta =\frac{10-{28}^{2/3}}{6}\approx 0.12985\phantom{\rule{thickmathspace}{0ex}}.$.
As you can see, these values of $\mathrm{cos}\theta$ are all different, so the angles themselves are also different.

suchonosdy

Expert

Step 1
You do get slightly different answers for the Pythagorean Theorem on the surface of a sphere of radius 1, or the hyperbolic plane of curvature -1. On the sphere, a right triangle with geodesic lengths of legs a,b and hypotenuse c obeys $\mathrm{cos}c=\mathrm{cos}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}b,$, while in the hyperbolic plane with curvature -1 it becomes $\mathrm{cosh}c=\mathrm{cosh}a\phantom{\rule{thickmathspace}{0ex}}\mathrm{cosh}b.$.
Step 2
In both cases, if you write out the functions as power series in a, b, c, you see that the limit as a, b, c all shrink to nearly 0 is the traditional Pythagorean Theorem.

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