Timeline for Constructing the inverse of a number geometrically.
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2021 at 11:00 | comment | added | Blue | @BillyRubina: ("Extended discussion" warning. This may be my last comment.) "Are you saying that [...] similar squares [...] have congruent angles but not corresponding sides proportional?" No. Any similar polygons have both congruent angles & proportional sides. I'm saying that congruent angles alone (or proportional sides alone) don't guarantee similarity of polygons w/more than 3 sides. Eg, any two rectangles have congruent angles, but they may not be similar. (Limiting attention to squares does imply similarity, but to limit attention is to assume more than just congruent angles.) | |
| May 3, 2021 at 10:14 | comment | added | Red Banana | @Blue I'm a bit confused: Are you saying that if we have two similar squares, then we have congruent angles but not corresponding sides proportional? | |
| May 2, 2021 at 15:46 | comment | added | Blue | (continuing) Trig exploits this convenience, in more-or-less the way you describe it: trig functions "depending" on angles puts the focus on (1); the definitions (eg, "tan=opp-over-adj") simply assign names to ratios that follow from (2). Above, $\triangle POA$ & $\triangle BOQ$ satisfy (1) because $P=B$, $90=90$, & $A=Q$. Thus, (2) tells us, eg, $OA/OP=OQ/OB$. All trig does here is recognize "$OA/OP$" & "$OQ/OB$" as the "opp-over-adj" ratios relative to the equal angles $P$ & $B$ in those triangles, allowing us to express the proportion by the shorthand "$\tan P=\tan B$". ... Does this help? | |
| May 2, 2021 at 15:46 | comment | added | Blue | @BillyRubina: Similarity is about having the same "shape" and not (necessarily) the same "size". Same-shape-ness amounts to (1) corresponding angles being congruent, and (2) corresponding sides being proportional. Conveniently for triangles, if you know (1), then you get (2) for free, and vice-versa. (Other polygons don't do that. For instance, a square and a (non-square) rectangle have (1), but not (2); a square and a (non-square) rhombus have (2), but not (1).) (continued) | |
| May 2, 2021 at 14:55 | comment | added | Red Banana | @Blue I remember reading somewhere that there is a proof for that the trigonometric functions depend only on the angle and not on the sizes of the sides of the triangle. Does this follows immediately from triangle similarity? It always seemed to me that something else is needed but I know next to nothing about geometry. | |
| May 2, 2021 at 14:48 | comment | added | Blue | @BillyRubina: The relation follows from the similarity of $\triangle POA$ and $\triangle BOQ$. Because the triangles happen to be right triangles, the relation matches an equality of $\tan\angle P$ and $\tan\angle Q$, so it's fine to approach the relation that way. (Note: If you use cotangent instead of tangent, then you get the relation without an extra multiplication step. But that amounts to a trade-off between a very-slightly-less-familiar trig function vs very-slightly-more algebra, so neither is really better or worse.) | |
| May 2, 2021 at 11:13 | comment | added | Red Banana | @Blue Why can we write $\frac{|\overline{OP}|}{|\overline{OA}|} = \frac{|\overline{OB}|}{|\overline{OQ}|}$? My guess is the following, as $\angle P \cong \angle B$, then $\tan(\angle P)=\tan(\angle B)$, with this we obtain $\frac{|\overline{OA}|}{|\overline{OP}|}=\frac{|\overline{OQ}|}{|\overline{OB}|}$. Now we just need to multiply in the obvious way to obtain the relation you gave. Is that correct? | |
| Jan 4, 2015 at 15:36 | comment | added | user153330 | you should most probably write a book where you collection all of your gems it will be awesome !! ; ) | |
| Jan 4, 2015 at 15:34 | comment | added | Blue | Glad to be of service. "Trigonography" (my name for diagrams that make trig relations clear) is a bit of a passion of mine. This reciprocal construction will make a nice addition to my collection. :) | |
| Jan 4, 2015 at 15:26 | comment | added | user153330 | your answer is truly a gem, as always : ) !! (btw thanks for your trig addition formulas images, they helped me a lot) | |
| Jan 4, 2015 at 15:26 | vote | accept | user153330 | ||
| Jan 4, 2015 at 15:12 | history | answered | Blue | CC BY-SA 3.0 |