Problem
Solve the system:
$$
\begin{align}
Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1982-t_0 \right ) \right )&=1.5 \\
Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1984-t_0 \right ) \right )&=1 \\
Y+Z*\cos\left ( \frac{\pi}{4}\left ( 1985-t_0 \right ) \right )&=0.6464
\end{align}
$$
Progress / thoughts
Using some basic trig identities, we observe that
$\cos{\left( \frac{\pi}{4}(1984-t_0)\right)} = \cos{\left(\frac{\pi}{4}\cdot 1984\right)}\cdot \cos{\left(\frac{\pi}{4}\cdot t_0\right)} + \sin{\left(\frac{\pi}{4}\cdot 1984\right)}\cdot \sin{\left(\frac{\pi}{4}\cdot t_0\right)}$.
From here, we can also use that $$\cos{\left(\frac{\pi}{4}\cdot 1984\right)} = \cos{\left(\pi \cdot 496 \right)} = \cos{\left(2\pi \cdot 248 \right)} = 1$$ and likewise for
$$\sin{\left(\frac{\pi}{4}\cdot 1984\right)} = \sin{\left(\pi \cdot 496 \right)} = \sin{\left(2\pi \cdot 248 \right)} = 0$$
Thus $$\cos{\left( \frac{\pi}{4}(1984-t_0)\right)} = 1\cdot \cos{\left(\frac{\pi}{4}\cdot t_0\right)} + 0 \cdot \sin{\left(\frac{\pi}{4}\cdot t_0\right)} = \cos{\left(\frac{\pi}{4}\cdot t_0\right)}$$
But even with this simplification, the problems just blows up in my face as soon as I start to try any substitution between the equations.
Question
Assume I can solve this with substitution, but the answer ends up as something ridiculous like $$L \in \left \{ Y=0.6464+\frac{3\sqrt{2}}{4}-\frac{\sqrt{2}}{2},\, \, \, \, t_0 =1.99+4 n,\, \, \, Z= \frac{\left ( 0.6464+\frac{3\sqrt{2}}{4}-\frac{\sqrt{2}}{2}\right )-1.5}{\sin\left ( \frac{\pi}{4}*\left ( -1.99+4 n \right ) \right )} \right \}$$ and may contain careless errors.
Is there a method that would make the answer "prettier"?
Best Answer
To keep track of things, help spot errors, and avoid having to re-do everything if (when?) values change, it's often better to obscure distracting specific values (especially messy decimals that may not be exact). So, let's consider the system in this form: $$\begin{align} y + z \cos(\theta-2\alpha) &= a \tag{1}\\ y + z \cos(\theta-2\beta) &= b \tag{2}\\ y + z \cos(\theta-2\gamma) &= c \tag{3} \end{align}$$ (The $2\alpha$, etc, anticipate having to divide angles by $2$ and so avoids having to write visually-cluttering fractions.)
We can easily solve the linear system $(1)$ and $(2)$ for $y$ and $z$. Simplifying gives $$(y,z) = \left(-\frac{a \cos(\theta-2\beta)-b \cos(\theta-2\alpha)}{2\sin(\alpha-\beta)\sin(\theta-\alpha-\beta)},\; \frac{a - b}{2\sin(\alpha-\beta)\sin(\theta-\alpha-\beta)}\;\right) \tag{4}$$ Substituting this into $(3)$, expanding the trig functions, and dividing-through by $\cos\theta$ leaves a linear equation in $\tan\theta$ with this "simplified" solution: $$\tan\theta = \frac{ \sum_{cyc} a \sin(\beta-\gamma) \sin(\beta+\gamma)}{ \sum_{cyc} a \sin(\beta-\gamma) \cos(\beta+\gamma)} \tag{5}$$ It's a little unwieldy if written-out, but it's clearly symmetric in the parameters, which gives us some confidence that we haven't completely messed things up. The form suggests defining, say, $$a' := a\sin(\beta-\gamma) \qquad b' := b\sin(\gamma-\alpha) \qquad c' := c \sin(\alpha-\beta)$$ so that we can write $$\tan\theta = \frac{\sum_{cyc}a'\sin(\beta+\gamma)}{\sum_{cyc}a'\cos(\beta+\gamma)} \tag{5'}$$
Substituting back into $(4)$ and simplifying takes some persistence. I won't show a step-by-step, but here are the results:
$$\begin{align} y &= -\frac{a' \cos(\beta-\gamma) + b' \cos(\gamma-\alpha) + c' \cos(\alpha-\beta)}{2 \sin(\alpha-\beta) \sin(\beta-\gamma) \sin(\gamma-\alpha)} \\[6pt] z &= \frac{\sqrt{a'^2 + b'^2 + c'^2 + 2 a' b' \cos(\alpha-\beta) + 2 b' c' \cos(\beta-\gamma) + 2 c' a' \cos(\gamma-\alpha) }}{2 \sin(\alpha-\beta)\sin(\beta-\gamma)\sin(\gamma-\alpha)} \end{align} \tag{6}$$
Here, the form suggests that there's something interesting about the geometry of the problem that led to the system. Knowing that geometry would likely help in confirming validity (and/or spotting errors), and may actually provide a more-direct path to the solution. Be that as it may ...
Applying these formulas to the problem at hand is left as an exercise to the reader.