Question

Determine the range of values of $\theta \in[0,2 \pi]$ for which the point $(\cos \theta, \sin \theta)$ lies inside the triangle formed by the lines $x+y=2; x-y=1$ & $6 x+2 y-\sqrt{10}=0$.

Answer 1

$\left(0, \arctan\left(\frac{1}{3}\right)+\frac{\pi}{3}\right)$

Answer 2

The problem asks for the range of values of $\theta \in[0,2 \pi]$ for which the point $P(\cos \theta, \sin \theta)$ lies inside the triangle formed by the lines:

$L_1: x+y-2=0$ $L_2: x-y-1=0$ $L_3: 6x+2y-\sqrt{10}=0$

For a point to lie inside a triangle, it must satisfy certain conditions with respect to each line. The standard method is to check the sign of the expression $Ax+By+C$ for the point with respect to each line, ensuring that it is the same as the sign for the third vertex (the vertex not lying on that line).

Let's find the vertices of the triangle:

1. Vertex A ($L_1 \cap L_2$): $x+y=2$ $x-y=1$ Adding the equations gives $2x=3 \Rightarrow x=3/2$. Subtracting the second from the first gives $2y=1 \Rightarrow y=1/2$. So, $A=(3/2, 1/2)$.

2. Vertex B ($L_1 \cap L_3$): $x+y=2 \Rightarrow y=2-x$ $6x+2y=\sqrt{10}$ Substitute $y$: $6x+2(2-x)=\sqrt{10} \Rightarrow 6x+4-2x=\sqrt{10} \Rightarrow 4x=\sqrt{10}-4 \Rightarrow x=(\sqrt{10}-4)/4$. $y=2-(\sqrt{10}-4)/4 = (8-\sqrt{10}+4)/4 = (12-\sqrt{10})/4$. So, $B=((\sqrt{10}-4)/4, (12-\sqrt{10})/4)$.

3. Vertex C ($L_2 \cap L_3$): $x-y=1 \Rightarrow y=x-1$ $6x+2y=\sqrt{10}$ Substitute $y$: $6x+2(x-1)=\sqrt{10} \Rightarrow 6x+2x-2=\sqrt{10} \Rightarrow 8x=\sqrt{10}+2 \Rightarrow x=(\sqrt{10}+2)/8$. $y=(\sqrt{10}+2)/8 - 1 = (\sqrt{10}-6)/8$. So, $C=((\sqrt{10}+2)/8, (\sqrt{10}-6)/8)$.

Now, we check the conditions for $P(\cos \theta, \sin \theta)$ to be inside the triangle:

Condition 1: For $L_1: x+y-2=0$. The third vertex is $C=((\sqrt{10}+2)/8, (\sqrt{10}-6)/8)$. Evaluate $L_1(C) = (\sqrt{10}+2)/8 + (\sqrt{10}-6)/8 - 2 = (2\sqrt{10}-4)/8 - 2 = (\sqrt{10}-2)/4 - 2 = (\sqrt{10}-10)/4$. Since $\sqrt{10} \approx 3.16$, $L_1(C) < 0$. So, for $P(\cos \theta, \sin \theta)$ to be on the same side of $L_1$ as $C$, we must have: $\cos \theta + \sin \theta - 2 < 0$. The maximum value of $\cos \theta + \sin \theta$ is $\sqrt{1^2+1^2} = \sqrt{2}$. So, $\cos \theta + \sin \theta - 2 \le \sqrt{2}-2 \approx 1.414-2 = -0.586 < 0$. This condition is always true for any point on the unit circle.

Condition 2: For $L_2: x-y-1=0$. The third vertex is $B=((\sqrt{10}-4)/4, (12-\sqrt{10})/4)$. Evaluate $L_2(B) = (\sqrt{10}-4)/4 - (12-\sqrt{10})/4 - 1 = (2\sqrt{10}-16)/4 - 1 = (\sqrt{10}-8)/2 - 1 = (\sqrt{10}-10)/2$. Since $\sqrt{10} \approx 3.16$, $L_2(B) < 0$. So, for $P(\cos \theta, \sin \theta)$ to be on the same side of $L_2$ as $B$, we must have: $\cos \theta - \sin \theta - 1 < 0 \Rightarrow \cos \theta - \sin \theta < 1$. We can write $\cos \theta - \sin \theta = \sqrt{2} \cos(\theta + \pi/4)$. So, $\sqrt{2} \cos(\theta + \pi/4) < 1 \Rightarrow \cos(\theta + \pi/4) < 1/\sqrt{2}$. Let $\alpha = \theta + \pi/4$. Since $\theta \in [0, 2\pi]$, $\alpha \in [\pi/4, 2\pi + \pi/4] = [\pi/4, 9\pi/4]$. For $\cos \alpha < 1/\sqrt{2}$ in the interval $[\pi/4, 9\pi/4]$, we have: $\pi/4 < \alpha < 7\pi/4$. Substituting back $\alpha = \theta + \pi/4$: $\pi/4 < \theta + \pi/4 < 7\pi/4$ $0 < \theta < 3\pi/2$.

Condition 3: For $L_3: 6x+2y-\sqrt{10}=0$. The third vertex is $A=(3/2, 1/2)$. Evaluate $L_3(A) = 6(3/2) + 2(1/2) - \sqrt{10} = 9+1-\sqrt{10} = 10-\sqrt{10}$. Since $\sqrt{10} \approx 3.16$, $L_3(A) > 0$. So, for $P(\cos \theta, \sin \theta)$ to be on the same side of $L_3$ as $A$, we must have: $6\cos \theta + 2\sin \theta - \sqrt{10} > 0 \Rightarrow 6\cos \theta + 2\sin \theta > \sqrt{10}$. We can write $6\cos \theta + 2\sin \theta = \sqrt{6^2+2^2} \cos(\theta - \phi) = \sqrt{36+4} \cos(\theta - \phi) = \sqrt{40} \cos(\theta - \phi) = 2\sqrt{10} \cos(\theta - \phi)$, where $\tan \phi = 2/6 = 1/3$. So, $2\sqrt{10} \cos(\theta - \phi) > \sqrt{10} \Rightarrow \cos(\theta - \phi) > 1/2$. Let $\beta = \theta - \phi$. We need $\cos \beta > 1/2$. We know $\phi = \arctan(1/3)$. Since $0 < 1/3 < 1/\sqrt{3}$, we have $0 < \phi < \pi/6$. So, $\phi \approx 0.322$ radians ($18.43^\circ$). Since $\theta \in [0, 2\pi]$, $\beta = \theta - \phi \in [-\phi, 2\pi-\phi]$. For $\cos \beta > 1/2$, we have $\beta \in (-\pi/3, \pi/3)$ (modulo $2\pi$). So, $-\pi/3 < \theta - \phi < \pi/3$. $\phi - \pi/3 < \theta < \phi + \pi/3$. Numerically, $\pi/3 \approx 1.047$ radians. So, $0.322 - 1.047 < \theta < 0.322 + 1.047$. $-0.725 < \theta < 1.369$. Considering $\theta \in [0, 2\pi]$, this interval becomes: $0 \le \theta < \phi + \pi/3$.

Combining the conditions:

1. $0 < \theta < 3\pi/2$
2. $0 \le \theta < \phi + \pi/3$

We need to find the intersection of these two ranges. The final range for $\theta$ is $0 < \theta < \min(3\pi/2, \phi + \pi/3)$. We know $\phi < \pi/6$. So $\phi + \pi/3 < \pi/6 + \pi/3 = \pi/2$. Since $\pi/2 < 3\pi/2$, the intersection is $0 < \theta < \phi + \pi/3$.

The range of values for $\theta$ is $(0, \arctan(1/3) + \pi/3)$.