Animated Solution for Mathematics - Straight Lines: The set of all possible values of $\theta$ in the interval $(0,\pi )$ for which the points $(1,2)$ and $(\sin \theta ,\cos \theta )$ lie on the same side of the line $x+y=1$ is:

Imagine you are standing on a vast, flat coordinate plane. You have a straight line, defined by the equation $x+y=1$, cutting across the landscape.

This line is a boundary, a divider that splits the entire plane into two distinct regions. On one side, the expression $x+y-1$ is positive; on the other, it is negative.

Our mission is to find the set of angles $\theta$ such that a point $P_2(\sin \theta ,\cos \theta )$—which is dancing along the unit circle—stays on the same side of this line as our fixed point $P_1(1,2)$.

How do we know if two points are on the same side of a line? We use the power of the line function $L(x,y)=x+y-1$.

If we plug the coordinates of a point into this function, the sign of the result tells us which side of the line the point resides on. For two points to be on the same side, the sign of $L(P_1)$ must match the sign of $L(P_2)$.

Mathematically, this is most elegantly expressed as $L(P_1)\cdot L(P_2)>0$. Let us test our fixed point $P_1(1,2)$.

Substituting these coordinates, we get:

Since $2>0$, we know that any point on the same side must also yield a positive value when plugged into $L(x,y)$.

Now, we turn our attention to the variable point $P_2(\sin \theta ,\cos \theta )$. For this point to be on the same side as $P_1$, we must satisfy the inequality $L(\sin \theta ,\cos \theta )>0$, which simplifies to:

This is where the magic of trigonometry comes in. We have a sum of sine and cosine, and we want to condense it. We use the harmonic addition identity, multiplying and dividing by $\sqrt{1^2+1^2}=\sqrt{2}$.

This transforms our expression into:

Recognizing that $\frac{1}{\sqrt{2}}$ is $\cos (\frac{\pi }{4})$ and $\sin (\frac{\pi }{4})$, we see the expansion of $\sin (\theta +\frac{\pi }{4})$. Thus, our inequality becomes:

We are almost there, but we must be vigilant. We are given the domain $\theta \in (0,\pi )$.

However, our sine function is operating on the argument $(\theta +\frac{\pi }{4})$. Therefore, the domain for our sine function is $(\frac{\pi }{4},\frac{5\pi }{4})$.

Within this interval, when is the sine of an angle greater than $\frac{1}{\sqrt{2}}$? The sine curve hits $\frac{1}{\sqrt{2}}$ at $\frac{\pi }{4}$ and $\frac{3\pi }{4}$. It stays above this value strictly between these two points.

So, we have:

To find the range of $\theta$, we simply subtract $\frac{\pi }{4}$ from all parts of our inequality. This leaves us with:

This result is beautiful in its simplicity. It tells us that for the point $(\sin \theta ,\cos \theta )$ to lie on the same side of the line as $(1,2)$, the angle $\theta$ must be in the first quadrant.

In this region, both $\sin \theta$ and $\cos \theta$ are positive, pushing the point far enough from the origin to stay on the 'positive' side of the line $x+y=1$. You have successfully navigated the algebra, the trigonometry, and the domain constraints.