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Step-by-Step Solution
Step 1: Express the line and define the sign check
The given line is
$x + y = 1$. We can rewrite it in the form
$f(x, y) = x + y - 1 = 0$.
Two points lie on the same side of this line if
$f(\text{first point}) \cdot f(\text{second point}) > 0$.
Step 2: Evaluate $f(x, y)$ for the given points
1) For the point $(1, 2)$:
$f(1, 2) = 1 + 2 - 1 = 2$.
2) For the point $(\sin \theta, \cos \theta)$:
$f(\sin \theta, \cos \theta) = \sin \theta + \cos \theta - 1$.
Step 3: Impose the condition for lying on the same side
We require
$f(1,2) \cdot f(\sin \theta, \cos \theta) > 0$.
Since $f(1,2) = 2$, this inequality becomes:
$2 \left[\sin \theta + \cos \theta - 1 \right] > 0$
$\quad \Rightarrow \quad \sin \theta + \cos \theta > 1$.
Step 4: Use a trigonometric identity to simplify
Recall the identity:
$\sin \theta + \cos \theta = \sqrt{2} \,\sin\Bigl(\theta + \frac{\pi}{4}\Bigr).$
So the inequality
$\sin \theta + \cos \theta > 1$
becomes
$\sqrt{2}\,\sin\Bigl(\theta + \frac{\pi}{4}\Bigr) > 1,$
which simplifies to
$\sin\Bigl(\theta + \frac{\pi}{4}\Bigr) > \frac{1}{\sqrt{2}}.$
Step 5: Solve the resulting trigonometric inequality
We know
$\sin\Bigl(\theta + \frac{\pi}{4}\Bigr) > \frac{1}{\sqrt{2}}$
means
$\theta + \frac{\pi}{4} \in \Bigl(\frac{\pi}{4}, \frac{3\pi}{4}\Bigr)$
because on the interval $(0, \pi)$ (extended appropriately to fit the shift by $\frac{\pi}{4}$),
$\sin x > \frac{1}{\sqrt{2}}$ corresponds to $x \in \Bigl(\frac{\pi}{4}, \frac{3\pi}{4}\Bigr)$.
Hence,
$\theta \in \Bigl(0, \frac{\pi}{2}\Bigr).$
Final Answer
Therefore, the set of all possible values of $\theta$ in the interval $(0, \pi)$ that satisfy the condition is
$\boxed{\bigl(0, \tfrac{\pi}{2}\bigr)}.$
