CTK Insights

01 Apr

Doodling on April 1st

I came across the following problem several days ago but hesitated to write about it until April 1st. It is simple, practically trivial, and still, after doodling with it for some time, I was left with an open question. If it appears too trivial, even unworthy of mention, do please make an allowance for the date, selected in the hope of forgiveness.

Here's the problem:

Given that (1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)=\cos\alpha\cos\beta\cos\gamma. Find an alternative expression for

\qquad(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma).

Do give it a thought before reading further.

To be on safe side, assume that none of the factors (1-\sin\alpha), (1-\sin\beta), (1-\sin\gamma) vanishes, for, otherwise, 0 would be an alternative form. Under this assumption, we may multiply (1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)=\cos\alpha\cos\beta\cos\gamma by (1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma):

\qquad(1-\sin^2\alpha)(1-\sin^2\beta)(1-\sin^2\gamma)=\cos\alpha\cos\beta\cos\gamma(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)

or (\cos^2\alpha)(\cos^2\beta)(\cos^2\gamma)=\cos\alpha\cos\beta\cos\gamma(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma) and, assuming \cos\alpha\cos\beta\cos\gamma\ne 0, we obtain

\qquad(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)=\cos\alpha\cos\beta\cos\gamma.

So, under the assumption \cos\alpha\cos\beta\cos\gamma\ne 0, that product is equal to both (1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma) and (1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma). Clearly the angles must be special.

First, the problem is generalized in an obvious way: Assuming \displaystyle\prod_{i=1}^{n}\cos\alpha_i\ne 0, that product is equal to both \displaystyle\prod_{i=1}^{n}(1-\sin\alpha_i) and \displaystyle\prod_{i=1}^{n}(1+\sin\alpha_i).

That looks quite right, but for what n? For n=1 there is no product but the question still makes sense: if 1+\sin\alpha=\cos\alpha and \cos\alpha\ne 0, what is 1-\sin\alpha? As before, 1+\sin\alpha=\cos\alpha=1-\sin\alpha. The two equations give \sin\alpha=0 and \cos\alpha=1.

Next, consider the case n=2: \cos\alpha\cos\beta\ne 0 and (1+\sin\alpha)(1+\sin\beta)=\cos\alpha\cos\beta. In this case, (1-\sin\alpha)(1-\sin\beta)=\cos\alpha\cos\beta. The difference of the two equations 2(\sin\alpha+\sin\beta)=0 and we get \displaystyle\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=0, meaning \alpha+\beta=2\pi k, k an integer, or \alpha-\beta=\pi (2k+1). Thus we see that, for example, \alpha=-\beta is a suitable pair of angles. The given equation becomes 1-\sin^2\alpha=\cos^2\alpha which is true.

For n=3, we obtain the equation

\qquad\sin\alpha+\sin\beta+\sin\gamma+\sin\alpha\sin\beta\sin\gamma=0.

With the choce of, say, \alpha=0, we reduce the problem to the case of n=2. Are there other solutions that bind all three variables? There bound to be. For example, suppose \alpha+\beta+\gamma=\pi. Then \sin 2\alpha+\sin 2\beta+\sin 2\gamma=4\sin\alpha\sin\beta\sin\gamma, meaning that, for \alpha+\beta+\gamma=2\pi, \displaystyle\sin \alpha+\sin \beta+\sin\gamma=4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}, which converts our equation to

\qquad 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}+\sin\alpha\sin\beta\sin\gamma=0

or, via the double argument formulas, and assuming that none of the sines vanishes,

\qquad\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}=-\frac{1}{2}.

A solution is plausible; I'll have to look into that later on.

Now, for the general case, \displaystyle\prod_{i=1}^{n}(1+\sin\alpha_i)=\prod_{i=1}^{n}\cos\alpha_i, with the latter not being 0.

Immediately, there are special values of \alpha\text{'s}, for which our derivation will work. For example we may split all factors into single angle expressions or pair up some angles and solve equations with one or two variables. Let's call such sets of angles reducible. In general, for a given n, the set \{\alpha_i\},\,i=1,\ldots,n is said to be reducible if it may be split into two non-empty sets, say, I_1 and I_2 such that I_1\cup I_2=\{1,\ldots,n\} and \displaystyle\prod_{i\in I_1}(1+\sin\alpha_i)=\prod_{i\in I_1}\cos\alpha_i=\prod_{i\in I_1}(1-\sin\alpha_i) and \displaystyle\prod_{i\in I_2}(1+\sin\alpha_i)=\prod_{i\in I_2}\cos\alpha_i=\prod_{i\in I_2}(1-\sin\alpha_i).

The original problem did not require solving equations, however, the question for which angles the stipulation of the problem made sense, was reasonable and natural. So lets call such sets of angles "solutions." In this terminology, we just defined "reducible solutions." Thus the questions pops up, "Are there irreducible solutions?"

For n=2, \alpha=\beta=0 should be considered "reducible" but, say, \displaystyle\alpha=-\beta=\frac{\pi}{4} is "irreducible". Thus the question of the existence of the irreducible solutions for n\gt 2 appears quite legitimate: "Are there irreducible solutions?" If that is too difficult, we may try answering an easier question: "Assuming all solutions are reducible, in how many ways the problem may be reduced?"

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