Consider the following trigonometric identity, valid for any set of angles $u,v,t$:

$$\cos t⋅\cos u⋅\cos v =\frac14\left[\cos(t + u + v)+\cos(t + u - v)+\cos(u+v-t)+\cos(v + t - u)\right]$$

This identity and its derivation have previously appeared as a question on this site (though the version quoted in the question is in fact missing a term).

But the solutions to the linked question, while valid, were all algebraic in nature. Is there a geometric proof/demonstration of the above identity? I would also be satisfied with a proof-without-words of a special case e.g. $u=u+v$, $u+v+t=\pi$, $u+v+t=2\pi$, etc.

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    $\begingroup$ the LHS could be seen as the volume of a box in $\Bbb R^3$. Maybe from this point of view we can see what "means" the RHS. $\endgroup$ – Masacroso Nov 24 '17 at 17:44
  • $\begingroup$ As a starting point, my trigonograph for sine and cosine of $\alpha+\beta+\gamma$ may be helpful. $\endgroup$ – Blue Nov 24 '17 at 18:08
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    $\begingroup$ This is the formula for $\cos(u)\cos(v)$ to sum applied twice, is there even a geometric interpretation limited to two variables $u,v$ ? Maybe we could extrapolate from there. $\endgroup$ – zwim Nov 24 '17 at 18:10
  • $\begingroup$ @Blue Very nice. Indeed, my identity follows (algebraically) from yours for $\cos(\alpha+\beta+\gamma)$ if we sum all four sign variations. So it could be that there's an easy geometric interpretation of that version but not mine. $\endgroup$ – Semiclassical Nov 24 '17 at 18:12
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The key idea is let $P_1$ and $P_2$ be points on a circle $C$ of radius $r$ with center $O$, $P_3:=(P_1+P_2)/2$ be the midpoint, and $P_4$ on circle $C$ be on the ray from center $O$ through point $P_3$. Given a Cartesian coordinate system express $$ P_1=(r\cos(u+v),r\sin(u+v)), P_2=(r\cos(u-v),r\sin(u-v)),$$ $$P_3=\cos(v)P_4,\; P_4=(r\cos u,r\sin u),.$$ The reason that $P_3=\cos(v)P_4$ is that they are on the same ray from the origin and by rotating the points and/or the coordinate system making $u=0$ gives $P_3=(r\cos(v),0)$ and $P_4=(r,0)$.

The next steps can be done completely geometrically, but I write them algebraically solely for convenience.Thus, setting $r=1$ and using the first coordinates of the points we must have $$(\cos(u+v)+\cos(u-v))/2=\cos(v)\cos(u)$$ and similarly for the second coordinates $$(\sin(u+v)+\sin(u-v))/2=\cos(v)\sin(u).$$

Now apply the first result with $u$ replaced by $u+t$ and then also $u-t$ giving

$$(\cos(u+t+v)+\cos(u+t-v))/2=\cos(v)\cos(u+t)$$ and $$(\cos(u-t+v)+\cos(u-t-v))/2=\cos(v)\cos(u-t).$$ Averaging the two equations results in the desired trigonometry identity.


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