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this picture:

http://38.media.tumblr.com/81375410b68b99c0d096e8be2b5078ff/tumblr_ng4n764sg21sszkooo1_500.gif

shows a way to construct the inverse of a number $a\ge1$. but how can we construct for a number that is less than 1?

My try::

enter image description here

  1. Q1: is my try correct?
  2. Q2: how to prove them both?
user153330
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    The original diagram actually shows the process for $a \leq 1$ as well; simply relabel the points $a \leftrightarrow 1/a$. (Your version effectively duplicates this logic.) For proof, merely consider "long leg over short leg" proportion for the red right triangle in its two extreme positions: the big right triangle has "long-over-short" = $\frac{a}{1}$; the small right triangle has "long-over-short" = $\frac{1}{1/a}$. – Blue Jan 04 '15 at 13:13
  • By the way: Where did you find the original diagram? It's pretty neat, so the creator deserves some credit. – Blue Jan 04 '15 at 13:17
  • @Blue okay write an answer then so that i acccept it (also why are those two triangles similar), i found it in tumblr – user153330 Jan 04 '15 at 14:26
  • I believe the image's creator may be Tumblr user curiosamathematica, aka Jens Bossaert. – Blue Jan 04 '15 at 15:21

1 Answers1

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Given the diagram as labeled ...

enter image description here

... we consider that construction occurred as follows:

Starting with point $P$ (or point $Q$) on $\overrightarrow{OR}$, construct $\overleftrightarrow{AP}$ (or $\overleftrightarrow{BQ}$) and let $C$ be the point where this line meets the unit circle. Then $\overleftrightarrow{BC}$ (or $\overleftrightarrow{AC}$) determines the point $Q$ (or $P$) on $\overrightarrow{OR}$.

Now, because $\angle ACB$ is inscribed in a semi-circle, it is a right angle by Thales' Theorem. Consequently, $\angle P \cong \angle B$ (as each is the complement of $\angle A$), so that $\triangle POA \sim \triangle BOQ$ and we can write $$\frac{|\overline{OP}|}{|\overline{OA}|} = \frac{|\overline{OB}|}{|\overline{OQ}|} \qquad\to\qquad \frac{|\overline{OP}|}{1} = \frac{1}{|\overline{OQ}|}$$

This proves the reciprocal relation. $\square$

Note: Even when the circle doesn't have unit radius, the relationship involves the geometric mean $$|\overline{OR}|^2 \;=\; |\overline{OP}|\;|\overline{OQ}|$$ which is important for the study of inversive geometry and such. The construction given is a nice companion to the more-common (to me) one involving the chord between points of tangency.

Blue
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    your answer is truly a gem, as always : ) !! (btw thanks for your trig addition formulas images, they helped me a lot) – user153330 Jan 04 '15 at 15:26
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    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. :) – Blue Jan 04 '15 at 15:34
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    you should most probably write a book where you collection all of your gems it will be awesome !! ; ) – user153330 Jan 04 '15 at 15:36
  • @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? – Red Banana May 02 '21 at 11:13
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    @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.) – Blue May 02 '21 at 14:48
  • @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. – Red Banana May 02 '21 at 14:55
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    @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) – Blue May 02 '21 at 15:46
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    (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? – Blue May 02 '21 at 15:46
  • @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? – Red Banana May 03 '21 at 10:14
  • @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.) – Blue May 03 '21 at 11:00