The set of all possible values of theta in th | vedclass

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Question

(A) Let the line be $L(x, y) = x+y-1 = 0$. For the points $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (\sin 	heta, \cos 	heta)$ to lie on the same side

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$\left(0, \frac{\pi}{2}\right)$

Vedclass · Answer

(A) Let the line be $L(x, y) = x+y-1 = 0$. For the points $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (\sin 	heta, \cos 	heta)$ to lie on the same side of the line,the product $L(x_1, y_1) \cdot L(x_2, y_2)$ must be greater than $0$. $L(1, 2) = 1+2-1 = 2 > 0$. Therefore,we must have $L(\sin 	heta, \cos 	heta) > 0$. $\Rightarrow \sin 	heta + \cos 	heta - 1 > 0$ $\Rightarrow \sin 	heta + \cos 	heta > 1$ Dividing by $\sqrt{2}$,we get: $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta > \frac{1}{\sqrt{2}}$ $\Rightarrow \sin \left(	heta + \frac{\pi}{4}ight) > \frac{1}{\sqrt{2}}$ Since $	heta \in (0, \pi)$,we have $	heta + \frac{\pi}{4} \in \left(\frac{\pi}{4}, \frac{5\pi}{4}ight)$. In this interval,$\sin \left(	heta + \frac{\pi}{4}ight) > \frac{1}{\sqrt{2}}$ holds when: $\frac{\pi}{4} Subtracting $\frac{\pi}{4}$ from all sides,we get: $0 Thus,the set of values is $\left(0, \frac{\pi}{2}ight)$.

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 . . . . . . .

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