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While solving this Question, I could derive the following:

As $\displaystyle 2\cos A\cos B=\cos(A-B)+\cos(A+B)$

substituting $A+B=90^\circ\iff B=90^\circ-A$ we get $\displaystyle 2\cos A\cos(90^\circ-A)=\cos(2A-90^\circ)$

$$\implies \cos A\cos(90^\circ-A)=\cos(2A-90^\circ)\cos60^\circ\ \ \ \ (1A)$$

Putting $2A-90^\circ=2C,$

$$\cos(45^\circ-C)\cos(45^\circ+C)=\cos2C\cos60^\circ\ \ \ \ (1B)$$

Clearly, this $(1A,$ or $1B)$ is one of the solutions of $$\cos A\cos B=\cos C\cos D$$

Is there any other or more generic solution(s)?

Community
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lab bhattacharjee
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    A start would be to consider all linear functions $A(s)$, $B(s)$, $C(s)$ and $D(s)$ such that $$\cos A(s) \cos B(s) = \cos C(s) \cos D(s).$$ Expanding in $\cos s$ and $\sin s$ will yield polynomial expressions, and we'd be looking for polynomial identities. – abnry Oct 23 '13 at 19:05

2 Answers2

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Here is a "general" but useless solution, anyhow unless I am missing something simple, it shows that the general solution cannot be too simple.

Let $A,B \in \mathbb{R}$. Then, $\left| \cos(A)\cos(B)\right| \leq 1$ and

$$\left| \cos(A)\cos(B)\right| =\left| \cos(C)\cos(D)\right| \leq \left| \cos(C)\right| \,.$$

Let $C \in \{ x \in \mathbb R | \left| \cos(A)\cos(B)\right| \leq \left| \cos(C) \right| \leq 1 \,\}$. Then, if $\cos(C)=0$, then $D$ can be anything, otherwise, $D$ is one of the solutions to the consistent equation

$$ \cos(D) =\frac{\cos(A) \cos(B)}{\cos(C) }$$

I should also point that if you pick $\cos(A)=0$, then $\cos(C)=0$ or $\cos(D)=0$, but you have complete freedom over the other two letters...

N. S.
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Suppose cos(a)cos(b) = p where |p| $\le$ 1. There are many ways to factor p, say into q, r such that qr = p. If |p| = 1, q and r have to be chosen to be $\pm$ 1 which is to say that c, d = $ \pm \pi/2$ .

If p is strictly less than one, there are many, many choices of qr = p; you need only to restrict yourself to |q|, |r| < 1. That gives cos(c) = q, cos(d) = r and c = arccos(q), d = arccos(r). Of course both c and d can be altered with $\pm 2\pi$.

Given a and b we cannot conclude anything more about c and d.

Betty Mock
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