If $2\cos\alpha_1=a+ \frac1a$ ,$2\cos\alpha_2 =b+ \frac1 b$ , etc... then show that $abc +\frac{ 1}{ abc} + ... = 2 \cos(\alpha_1 + \alpha_2 + ...)$

Question

If $2\cos\alpha_1=a+\frac1a$,$2\cos\alpha_2 =b+\frac1b$, etc... then show that $abc +\frac{1}{abc} + ... = 2 \cos(\alpha_1 + \alpha_2 + ...)$

I don't get how to solve this problem. I tried the following strategy :

$$(\cos\alpha_1+i\sin\alpha_1)(\cos\alpha_2+i\sin\alpha_2)...(\cos\alpha_n+i\sin\alpha_n)=\cos(\alpha_1+\alpha_2+...+\alpha_n)+i\sin(\alpha_1+\alpha_2+...+\alpha_n)=\cos\alpha_1\cos\alpha_2...\cos\alpha_n(1+is_1+i^2s_2+\cdots),$$ where $$s_1=\Sigma \tan\alpha_1,s_2=\Sigma\tan\alpha_1\tan\alpha_2,\cdots,s_k=\Sigma\tan\alpha_1\tan\alpha_2\cdots\tan\alpha_k.$$ Thus, $$\cos(\alpha_1+\cdots+\alpha_n)=\cos\alpha_1\cos\alpha_2\cdots\cos\alpha_n(1-s_2+s_4-\cdots)$$

But this doesn't reduce to something favourable for solving the problem. Also, I think there might be a loophole in my strategy as in the question it's not specified clearly whether they consider a finite number of terms or not as they write $\alpha_1+\alpha_2+\cdots$, while I assumed a finite number of $\alpha_i$'s i.e $\alpha_1+\cdots+\alpha_n$ in my strategy( Is the question at all valid?). I dont quite get it...

Answer 1

For a better homogeneity, instead of $a,b,c...$, I take $a_1,a_2,a_3...$.

• In a first step, I will assume all the $a_k > 0$.

Let $f$ be the function defined by

$$f(x)=\frac12 \left(x+\frac{1}{x}\right)$$

allowing the hypotheses to be written in the following way :

$$\text{for} \ k=1,2\cdots n, \ \ \cos(\alpha_k)=f(a_k)\tag{1}$$

But, for $x>0$, function $f$ takes values always $\ge 1$ with equality iff $x=1$.

As a consequence, as the values of cosine function are between $-1$ and $+1$, conditions (1) amounts to

$$\text{for} \ k=1,2\cdots n, \ \ \cos(\alpha_k)=1$$

Therefore $\alpha_k = 0 \ (\text{mod} \ 2 \pi)$.

For the general case with numbers $n=3$ for example, with the following interpretation of the question (not sure it is the right one...) ; let :

$$A:=a_1a_2\cdots a_n\tag{2}$$

$$\underbrace{A+\frac{1}{A}}_{1+\tfrac{1}{1}}=2 \cos(0+0+0+\cdots)\tag{3}$$

which is evidently true.

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• Let us assume some of the $a_k$s are negative. Let us begin for example for didactical reasons with the case $a_1<0$, all other ones $>0$. As, for negative values for $x$, (due to the fact that $f$ is an odd function), we have the reverse property $f(x)\le -1$ with equality iff $x=-1$, one must have $a_1=-1$ ; in this case

$$\cos(\alpha_1)=f(-1)=-1 \iff \alpha_1 = \pi \ (\text{mod} \ 2 \pi)$$

The consequence is that $A$, defined by (2), become equal to $-1$.

Then (3) is changed into :

$$\underbrace{A+\frac{1}{A}}_{-1+\tfrac{ \ \ 1}{-1}}=2 \cos(\pi+0+0+\cdots)\tag{3'}$$

which is again true.

If there are more than one $a_k$ which is $<0$, it turns out that with an odd number $(2p+1)$ of negative values, we fall into case (3'), because $\cos((2p+1)\pi) = -1$, otherwise we are in case (3).

Answer 2

I doubt you need summations to prove this.

Using the following rules; $2\cos{\theta} = z + \frac{1}{z} = e^{i\theta} + e^{-i\theta}$ and $2\sin{\theta} = z - \frac{1}{z} = -ie^{i\theta} + ie^{-i\theta}$

We have $2\cos{\alpha_1} = \left(a + \frac{1}{a} \right)$ and $2\cos{\alpha_2} = \left(b + \frac{1}{b} \right)$. Then $a = e^{i \alpha_1}$ and $b = e^{i \alpha_2}$. Similarly, $2\sin{\alpha_1} = \left(-ia + \frac{i}{a} \right)$ and $2\sin{\alpha_2} = \left(-ib + \frac{i}{b} \right)$.

Then, $$ 4\cos{\alpha_1}\cos{\alpha_2} - 4\sin{\alpha_1}\sin{\alpha_2} = 4\cos{\left(\alpha_1 + \alpha_2 \right)} = \left(a + \frac{1}{a} \right) \cdot \left(b + \frac{1}{b} \right) - \left(-ia + \frac{i}{a} \right)\left(-ib + \frac{i}{b} \right) \\ \qquad \\ = \left(\frac{(a^2 + 1)(b^2 + 1)}{ab} \right) + \left(\frac{(a^2 - 1)(b^2 - 1)}{ab} \right) \\ \qquad \\ = \frac{2\left(a^2 b^2 + 1 \right)}{ab} = 2ab + \frac{2}{ab}$$.

$$4\cos{\left(\alpha_1 + \alpha_2 \right)} = 2ab + \frac{2}{ab} \implies 2\cos{\left(\alpha_1 + \alpha_2 \right)} = ab + \frac{1}{ab}$$

Can you continue and prove for a third iteration, $\left(c + \frac{1}{c}\right)$?