Pythagorean Identities other than Trigonometric

Question

Are there any identities which follow the Pythagorean pattern, $$a^2+b^2=c^2$$ besides the standard trigonometric and hyperbolic trigonometric Pythagorean identities (e.g. $\sin^2(\theta)+\cos^2(\theta)=1$) and those derived from them? Preferably, these functions shouldn't be reducible to the trig identities, but those are acceptable as generalizations.

Alternatively, what other non-trivial functions parameterize the circle $x^2+y^2=1$? (Just like there are also functions that parameterize the Fermat cubic $x^3+y^3=1$?)

The Cosine Law for triangles: $a^2=b^2+c^2-2bc \cos A$ is an immediate consequence of Pythagoras. Your Q is a bit vague. — DanielWainfleet, Jan 07 '17 at 03:25
Yes, this was a little too vague. I'll edit it to make it more clear what I do and don't want. — Logan Clark, Jan 07 '17 at 04:16
Can we guys re-open this question? It seems clear enough now. — Tito Piezas III, Jan 07 '17 at 12:28
@Goldname I guess he/she wants identities in the form $$\text{something}^{n}+\text{something different}^{n}=\text{another thing}^{\text{which can have}\ \text{or not have} \ \text{exponents here}}.$$. — Red Banana, Jan 07 '17 at 12:29
@TitoPiezasIII Yes. My "equation" encompasses that. But I guess he/she wouldn't get hurt if it lacks an exponent after the equality. — Red Banana, Jan 07 '17 at 12:32
@LoganClark: Click on the re-open button that is below your question. — Tito Piezas III, Jan 08 '17 at 03:09

score 4 · Accepted Answer · edited Apr 13 '17 at 12:58

The Jacobi theta functions, $$\vartheta_2(q)^4+\vartheta_4(q)^4 = \vartheta_3(q)^4\tag1 $$ the Weber modular functions, $$\mathfrak{f}_1(\tau)^8+\mathfrak{f}_2(\tau)^8 = \mathfrak{f}(\tau)^8\tag2$$ the elliptic modulus, $$k^2+k'\,^2=1\tag3$$ and the Dedekind eta function, $$\left(\frac{\sqrt2\,\eta(\tfrac{\tau}2)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8+\left(\frac{\eta^2(\tfrac{\tau}2)\,\eta(2\tau)}{\eta^3(\tau)}\right)^8 = 1\tag4$$ which, with the nome $q=e^{\pi i \tau}$, has a beautiful continued fraction studied by Ramanujan, $$\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$ These four can be derived from each other.

P.S. You might also be interested in the Borwein's version, $$b(q)^3+c(q)^3=a(q)^3$$ with cubic theta functions. There's even a $26$th deg analogue, $$x_1^{26}+x_2^{26}+\dots = 1$$ using the Dedekind eta function again, but you'd need more than two addends.

score 2 · Answer 2 · answered Jan 07 '17 at 02:23

The Jacobi elliptic functions $\mathrm{sn}(z;k)$, $\mathrm{cn}(z;k)$ and $\mathrm{dn}(z;k)$ of an additional parameter $k$, which are defined by the differential equations $$\mathrm{sn}'(z;k) = \mathrm{cn}(z;k) \cdot \mathrm{dn}(z;k) \; \; \mathrm{and} \; \mathrm{sn}(0;k) = 0,$$ $$\mathrm{cn}'(z;k) = -\mathrm{sn}(z;k) \cdot \mathrm{dn}(z;k), \; \; \mathrm{cn}(0;k) = 1,$$ and $$\mathrm{dn}'(z;k) = -k^2 \mathrm{sn}(z;k) \cdot \mathrm{cn}(z;k), \; \; \mathrm{dn}(0;k) = 1$$ satisfy the "Pythagorean" formulas $$\mathrm{sn}(z;k)^2 + \mathrm{cn}(z;k)^2 = 1$$ and $$k^2 \mathrm{sn}(z;k)^2 + \mathrm{dn}(z;k)^2 = 1.$$ As $k$ approaches $0$, $\mathrm{sn}(z;k)$ approaches $\mathrm{sin}(z)$, $\mathrm{cn}(z;k)$ approaches $\mathrm{cos}(z)$, and $\mathrm{dn}(z;k)$ approaches $1$.

On the other hand, since $\operatorname{sn}(u\mid m)=\sin(\operatorname{am}(u\mid m))$ and $\operatorname{cn}(u\mid m)=\cos(\operatorname{am}(u\mid m))$, where $\operatorname{am}(u\mid m)$ is the Jacobian amplitude, this in a way violates the OP's stipulation of "these functions shouldn't be reducible to the trig identities". — J. M. ain't a mathematician, Jan 07 '17 at 08:43

score 1 · Answer 3 · answered Jan 07 '17 at 01:55

1

The hyperbolic trig functions:

$$\cosh^2(x)-\sinh^2(x)=1$$

$$\operatorname{sech}^2(x)+\tanh^2(x)=1$$

$$\coth ^{2}x-\operatorname {csch}^2x=1$$

Notice the signs are a little wacky...

answered Jan 07 '17 at 01:55

Simply Beautiful Art

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score 0 · Answer 4 · answered Jan 07 '17 at 03:59

$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2.$$ A geometric interpretation of this: In the Cartesian plane, with $X=(a,b), Y=(c,d), O=(0,0),$ we have $$a^2+b^2=XO^2\; \text { and }\; c^2+d^2=YO^2$$ $$ \text {while }\; |ac+bd|=XO\cdot YO\cdot |\cos XOY|$$ $$\text {and }\; |ad-bc|=XO\cdot YO\cdot |\sin XOY|.$$

Pythagorean Identities other than Trigonometric

4 Answers4