6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.