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I was able to reduce an equation I have to:

$$f(t) = \tan(\mu) \tan(\nu) - C = 0$$

where $\mu, \nu$ are linear functions of t and $C$ is a constant.

  1. Are there any identities for the product of tangents?
  2. Is there a way to solve this equation analytically?
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You should give explicitely $\mu$ and $\nu$ as functions of time. This can simplify a lot. – Jon Feb 9 '12 at 8:40
For 1., sure, there are. I doubt that they'd be helpful here, though. For 2., as mentioned a number of times before, transcendental equations usually don't admit easy analytical solutions; what makes you think it would be different here? – J. M. Feb 9 '12 at 8:48
Special case: $\mu=t$, $\nu=5t$. There's probably an identity for $\tan5t$ in terms of $(\tan t)^5$ which will lead to some polynomial equation of high degree in $\tan t$ and thereby to numerical methods of solution only. – Gerry Myerson Feb 9 '12 at 11:45
@Jon: They're just run-of-the-mill linear functions. ie: – Jay Lemmon Feb 9 '12 at 17:34
@J.M.: If one of $\mu$ or $\nu$ are constants, or $C$ is $0$, it's solvable analytically. So there was enough reason for me to suspect an analytic solution might be possible. – Jay Lemmon Feb 9 '12 at 17:34

Using the identity pointed out by J.M., we can transform the equation into $$\cos(\mu-\nu)-\cos(\mu+\nu) = C(\cos(\mu-\nu)+\cos(\mu+\nu)) \tag1$$ or better yet, $$(1-C)\cos(\mu-\nu) = (1+C) \cos(\mu+\nu) \tag2$$ Here both $\mu-\nu$ and $\mu+\nu$ are linear functions of $t$, which, as I infer from a comment by the OP, means linear and not affine. Assuming neither $\mu-\nu$ nor $\mu+\nu$ happen to be identically zero (which is an easy case), we can make (2) more compact: $$\cos x = A\cos bx \tag3$$ where $A=(1+C)/(1-C)$ and $b=(\mu+\nu)/(\mu-\nu)$, which is a nonzero constant. There are two cases:

$b$ is rational, $b=m/n$. Then both sides of (3) are periodic functions with period $2\pi n$. Let a numerical routine find all roots of (3) on $[0,2\pi n]$, and write down all of them by periodicity. If both $m$ and $n$ are very small, an explicit solution may exist, but otherwise not.

$b$ is irrational. The function $\cos x-A\cos bx$ is not periodic, but it's quasiperiodic. This means that the roots of (3) come arbitrarily close to forming a periodic pattern, but they never do. Here is the graph of $\cos x-\pi \cos ex$, for example:

weird cosines

All I can add is an easy observation: if $A\ll 1$ or $A\gg 1$, the roots of (3) are close to the zeroes of the wave with greater magnitude.

I wonder if (3) has a name...

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interesting, thanks for taking some time to explore the problem :) – Jay Lemmon Jun 11 '13 at 23:40

$$ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}, $$ $$ \tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}, $$ etc.

Each of these involves products of tangents.

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This is one of my own tangent half-angle formulas (or "formulae", if you like), and its in a paper I'm writing on Villarceau's theorem: $$ \begin{align} \cos\gamma & =\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta} \\[10pt] \text{and }\sin\gamma & = \frac{\sin\alpha\sin\beta}{1+\cos\alpha\cos\beta} \\[10pt] \text{if and only if } \tan\frac\gamma2 & = \tan\frac\alpha2\cdot\tan\frac\beta2. \end{align} $$

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