Hard-to-simplify system of trigonometric equations

Question

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"?

Answer 1

Since $$\frac{1982}{4}\pi=247\times 2\pi+\frac 32\pi$$ $$\frac{1984}{4}\pi =248\times 2\pi$$ $$\frac{1985}{4}\pi=248\times 2\pi+\frac{\pi}{4}$$

the system is equivalent to $$ \begin{align} Y+Z\cos\left (\frac 32\pi- \frac{\pi}{4}t_0 \right )&=1.5 \\ Y+Z\cos\left ( -\frac{\pi}{4}t_0 \right )&=1 \\ Y+Z\cos\left ( \frac{\pi}{4}- \frac{\pi}{4}t_0 \right )&=0.6464 \end{align} $$ i.e. $$ Y-Z\sin\left(\frac{\pi}{4}t_0\right)=1.5\tag1$$ $$Y+Z\cos\left ( \frac{\pi}{4}t_0 \right )=1\tag2$$ $$Y+\frac{Z}{\sqrt 2}\left ( \cos\left ( \frac{\pi}{4}t_0 \right )+\sin\left ( \frac{\pi}{4}t_0 \right ) \right )=0.6464\tag3$$

$(2)-(1)$ gives $$Z\left ( \cos\left ( \frac{\pi}{4}t_0 \right )+\sin\left ( \frac{\pi}{4}t_0 \right ) \right )=-0.5\tag4$$

From $(3)(4)$, we get $$Y+\frac{1}{\sqrt 2}(-0.5)=0.6464\implies Y=0.6464+\frac{1}{2\sqrt 2}\tag5$$

From $(5)(1)(2)$, we get $$(1)\implies \sin\left(\frac{\pi}{4}t_0\right)=\frac{\frac{1}{2\sqrt 2}-0.8536}{Z}\tag6$$ $$(2)\implies \cos\left ( \frac{\pi}{4}t_0 \right )=\frac{0.3536-\frac{1}{2\sqrt 2}}{Z}\tag7$$

From $(6)(7)$ and $$\sin^2\left(\frac{\pi}{4}t_0\right)+\cos^2\left ( \frac{\pi}{4}t_0 \right )=1$$ we obtain $$\bigg(\frac{\frac{1}{2\sqrt 2}-0.8536}{Z}\bigg)^2+\bigg(\frac{0.3536-\frac{1}{2\sqrt 2}}{Z}\bigg)^2=1$$ from which we have $$Z=\pm\sqrt{\bigg(\frac{1}{2\sqrt 2}-0.8536\bigg)^2+\bigg(0.3536-\frac{1}{2\sqrt 2}\bigg)^2}=\pm\frac{\sqrt{1724478 - 943125 \sqrt 2}}{1250}$$

Hence, we get $$(Y,Z)=\bigg(0.6464+\frac{1}{2\sqrt 2},\pm\frac{\sqrt{1724478 - 943125 \sqrt 2}}{1250}\bigg)$$

---

If $0.6464$ is meant to be an approximation of $\frac{4-\sqrt 2}{4}$, then we get $Y=1$.

So, from $(2)$, we get $$Z\cos\left ( \frac{\pi}{4}t_0 \right )=0$$ from which we have $$\cos\left ( \frac{\pi}{4}t_0 \right )=0$$

It follows that $$\frac{\pi}{4}t_0=\frac{\pi}{2}+n\pi\iff t_0=4n+2$$ where $n$ is an integer.

Also, we get $$(1)\implies Z=-\frac{1}{2\sin\left(\frac{\pi}{4}t_0\right)}=-\frac{1}{2\sin\left(\frac{\pi}{2}+n\pi\right)}=\begin{cases}\frac 12&\text{if $n$ is odd}\\\\-\frac 12&\text{if $n$ is even}\end{cases}$$

In conclusion, if $0.6464$ is meant to be an approximation of $\frac{4-\sqrt 2}{4}$, then the answer is $$(Y,Z,t_0)=\bigg(1,\frac 12,4(2k+1)+2\bigg),\bigg(1,-\frac 12,4(2k)+2\bigg),$$ i.e. $$\color{red}{(Y,Z,t_0)=\bigg(1,\frac 12,8k+6\bigg),\bigg(1,-\frac 12,8k+2\bigg)}$$ where $k$ is an integer.

Answer 2

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.