If (sin theta, cos theta) and (3, 2) lie on t | vedclass

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(D) Let $f(x, y) = x + y - 1$. Since $(3, 2)$ lies on one side of the line,we evaluate $f(3, 2) = 3 + 2 - 1 = 4$,which is $> 0$. For $(\sin 	heta, \c

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$(0, \pi/4)$

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(D) Let $f(x, y) = x + y - 1$. Since $(3, 2)$ lies on one side of the line,we evaluate $f(3, 2) = 3 + 2 - 1 = 4$,which is $> 0$. For $(\sin 	heta, \cos 	heta)$ to lie on the same side,we must have $f(\sin 	heta, \cos 	heta) > 0$. $\sin 	heta + \cos 	heta - 1 > 0$ $\sin 	heta + \cos 	heta > 1$ Multiplying by $\frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta > \frac{1}{\sqrt{2}}$ $\sin(	heta + \frac{\pi}{4}) > \sin(\frac{\pi}{4})$ Since $\sin 	heta$ and $\cos 	heta$ are coordinates,$	heta$ is typically in $(0, \pi/2)$. In the interval $(0, \pi/2)$,$	heta + \frac{\pi}{4}$ lies in $(\frac{\pi}{4}, \frac{3\pi}{4})$. For $\sin(	heta + \frac{\pi}{4}) > \frac{1}{\sqrt{2}}$,we require $\frac{\pi}{4} Subtracting $\frac{\pi}{4}$ from all sides,we get $0 However,checking the condition $\sin 	heta + \cos 	heta > 1$ within $(0, \pi/2)$,the inequality holds for all $	heta \in (0, \pi/2)$. Given the options,the most restrictive interval satisfying the condition is $(0, \pi/4)$.

If $(\sin \theta, \cos \theta)$ and $(3, 2)$ lie on the same side of the line $x + y = 1$,then $\theta$ lies in the interval:

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