Patrick Honner on Nostr: I've always found it cool that if you double the smaller acute angle in a 3-4-5 ...
I've always found it cool that if you double the smaller acute angle in a 3-4-5 triangle you get the larger acute angle in a 7-24-25 right triangle. You can see this as a consequence of the double angle for sine. If \( \alpha \) is the smaller acute angle in a 3-4-5 triangle, then
\[ \sin (2\alpha) = 2\sin\alpha\cos\alpha=2\frac{3}{5}\frac{4}{5}=\frac{24}{25}\]
In fact, if the sine and cosine of an angle are both rational, then so will be the sine and cosine of twice that angle. This gives a way to turn Pythagorean triples into new Pythaogorean triples!
For example, suppose \( \alpha \) is an acute angle in a right triangle with \(a^2 + b^2 = c^2 \). Then
\[ \sin 2\alpha = \frac{2ab}{c^2} \]
\[ \cos 2 \alpha = \frac{a^2-b^2}{c^2} \]
By the Pythagorean identity
\[ \left(\frac{2ab}{c^2} \right)^2 + \left(\frac{a^2-b^2}{c^2} \right)^2 = 1\]
And so
\[ \left(2ab \right)^2 + \left(a^2-b^2 \right)^2 = \left(c^2\right)^2\]
which of course also follows directly from algebra.
For example, using this process
\[ (3,4,5) \mapsto (7,24,25) \mapsto(336,527,625)\]
#Math #MathEd #Mathematics #NumberTheory #Algebra #Geometry
Published at
2026-01-08 22:20:44 UTCEvent JSON
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"content": "I've always found it cool that if you double the smaller acute angle in a 3-4-5 triangle you get the larger acute angle in a 7-24-25 right triangle. You can see this as a consequence of the double angle for sine. If \\( \\alpha \\) is the smaller acute angle in a 3-4-5 triangle, then\n\n\\[ \\sin (2\\alpha) = 2\\sin\\alpha\\cos\\alpha=2\\frac{3}{5}\\frac{4}{5}=\\frac{24}{25}\\]\n\nIn fact, if the sine and cosine of an angle are both rational, then so will be the sine and cosine of twice that angle. This gives a way to turn Pythagorean triples into new Pythaogorean triples!\n\nFor example, suppose \\( \\alpha \\) is an acute angle in a right triangle with \\(a^2 + b^2 = c^2 \\). Then\n\n\\[ \\sin 2\\alpha = \\frac{2ab}{c^2} \\]\n\\[ \\cos 2 \\alpha = \\frac{a^2-b^2}{c^2} \\]\n\nBy the Pythagorean identity\n\n\\[ \\left(\\frac{2ab}{c^2} \\right)^2 + \\left(\\frac{a^2-b^2}{c^2} \\right)^2 = 1\\]\n\nAnd so\n\n\\[ \\left(2ab \\right)^2 + \\left(a^2-b^2 \\right)^2 = \\left(c^2\\right)^2\\] \nwhich of course also follows directly from algebra.\n\nFor example, using this process\n\n\\[ (3,4,5) \\mapsto (7,24,25) \\mapsto(336,527,625)\\]\n\n#Math #MathEd #Mathematics #NumberTheory #Algebra #Geometry",
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