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Context

It's pretty easy to prove said identity for angles smaller than 90 degrees, because we can use a right-angled triangle, and the result falls out of the definitions of $\sin$ and $\cos$ inside the triangle.

What I'd like to do, is prove it more generally, but I'm unsure of how I can visualize $\frac\pi2 - \theta$ if $\theta > \frac\pi2$.

As an example, if we draw an angle in the 4th quadrant, we get the angles $\frac\pi2, \quad 2\pi - \theta, \quad x - \frac{3\pi}2$.

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Question

What's a nice visual proof of $\sin(\frac\pi2 - \theta) = \cos(\theta)$ for angles larger than $\frac\pi2$?

Caveat

I'm aware of proofs that involve algebraically deducing it using other identities like angle sums etc. I'm specifically seeking direct, visual proofs for this one.

Alec
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  • $\sin(\frac\pi2-\theta)=\pm\cos\theta$ depending on the quadrant your angle is in. – Andrew Chin May 04 '20 at 12:27
  • Related (duplicate?): "How to remember a particular class of trig identities.". In particular, see my answer. – Blue May 04 '20 at 13:36
  • @Blue - First off, that's a fantastic answer, and the windmill-style illustrations are something I'll take with me for sure. But I guess the follow-up here is; can the idea of $\pi/2 - \theta$ be illustrated in such a nice way when $\theta \in [\frac\pi2, \pi]$? That's sort of the big difference between that answer, and the one I'm looking for. – Alec May 06 '20 at 23:25
  • @Alec: "can the idea of $\pi/2−\theta$ be illustrated [...] when $\theta\in[\frac{\pi}{2},\pi]$?" Sure. Generally, adding/subtracting to/from a right angle merely swaps horizontals and verticals; the only fiddly part is handling the signs. For your obtuse $\theta$, consider: Let the axes meet the unit circle at $X$ & $Y$. If going CCW from $X$ by angle $\theta$ gets you to $A$, and going CW from $Y$ by $\theta$ gets you to $B$, then $A$ goes just as far "beyond" $Y$ as $B$ goes "beyond" $X$; thus, the sine/cosine triangles match, as in the windmill, except they face opposite directions. – Blue May 06 '20 at 23:50
  • @Blue - Here's as far as I've gotten: https://i.imgur.com/wrAMixb.png - I'm on some shaky ground here because for one, I seem to conclude that within the green right-angled triangle, the angle $\beta = \alpha < \pi/2$ despite starting off with defining $\alpha > \pi/2$. However, if I accept that $\beta = \alpha < \pi/2$, then the identity $\sin(\frac\pi2 - \alpha) = \cos\alpha$ shows up regardless. Am I missing something here? – Alec May 07 '20 at 00:04
  • @Alec: You seem to be running-up against sign issues. The angle measure "$\pi/2-\alpha$ is a "signed" measure; when $\alpha>\pi/2$, that measure is negative, which corresponds to the clockwise direction you followed in making it. The "absolute" measure of that angle —as in, the unsigned size— is, in this case, $\alpha-\pi/2$ (a positive number), and that is what you should use to calculate your $\beta$, although $\beta$ isn't really important for this story. (continued) – Blue May 07 '20 at 19:19
  • (continuing) What is important is that what we might call the "reference triangle" (my answer's windmill triangles) corresponding to the (negative) angle $\pi/2-\alpha$ is congruent to the ref triangle for $\alpha$ itself. The first (as you've drawn) is the right-hand triangle in my 2nd windmill; the latter is the top triangle in my 1st windmill. The vertical & horizontal legs of these triangles have swapped, so that sine of one angle is the cosine of the other, & vice versa ... w/ potential sign issues, although luckily here, the signs match. – Blue May 07 '20 at 19:27

1 Answers1

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Hint:

In polar coordinates, $\theta\mapsto\frac\pi 2-\theta$ is the composition of a symmetry w.r.t. the abscissæ axis by a rotation of $\frac\pi 2$.

Other interpretation: Points on the unit circle with polar angles $\theta$ and $\frac\pi 2-\theta$ are symmetric w.r.t. the first bissectrix.

Bernard
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  • That sounds interesting, but I don't have any exposure to polar coordinates yet. I was hoping for something even more direct, using basic stuff about right-angled triangles, albeit one in the 2nd, 3rd, or 4th quadrant. – Alec May 04 '20 at 12:40