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


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.

  • $\begingroup$ $\sin(\frac\pi2-\theta)=\pm\cos\theta$ depending on the quadrant your angle is in. $\endgroup$ May 4 '20 at 12:27
  • $\begingroup$ Related (duplicate?): "How to remember a particular class of trig identities.". In particular, see my answer. $\endgroup$
    – Blue
    May 4 '20 at 13:36
  • $\begingroup$ @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. $\endgroup$
    – Alec
    May 6 '20 at 23:25
  • $\begingroup$ @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. $\endgroup$
    – Blue
    May 6 '20 at 23:50
  • $\begingroup$ @Blue - Here's as far as I've gotten: 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? $\endgroup$
    – Alec
    May 7 '20 at 0:04


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.

  • $\begingroup$ 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. $\endgroup$
    – Alec
    May 4 '20 at 12:40

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