Let A = {theta in R mid cos^2(sin theta) + si | vedclass

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(A) For set $A$,we have $\cos^2(\sin 	heta) + \sin^2(\cos 	heta) = 1$. Using the identity $\sin^2 x + \cos^2 x = 1$,this equation holds if and only

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the empty set

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(A) For set $A$,we have $\cos^2(\sin 	heta) + \sin^2(\cos 	heta) = 1$. Using the identity $\sin^2 x + \cos^2 x = 1$,this equation holds if and only if $\sin^2 	heta = \cos^2 	heta$,which implies $	an^2 	heta = 1$,so $	an 	heta = \pm 1$. Thus,$	heta = n\pi \pm \frac{\pi}{4}$. For set $B$,we have $\cos(\sin 	heta) \sin(\cos 	heta) = 0$. This implies $\cos(\sin 	heta) = 0$ or $\sin(\cos 	heta) = 0$. Since $-1 \le \sin 	heta \le 1$,$\cos(\sin 	heta)$ can never be $0$ because $\cos x = 0$ at $x = \pm \frac{\pi}{2} \approx \pm 1.57$,which is outside the range $[-1, 1]$. Similarly,$\sin(\cos 	heta) = 0$ implies $\cos 	heta = n\pi$. Since $-1 \le \cos 	heta \le 1$,the only possible value is $\cos 	heta = 0$. Thus,$	heta = (2n+1)\frac{\pi}{2}$. Comparing the sets,$A = \{n\pi \pm \frac{\pi}{4}\}$ and $B = \{(2n+1)\frac{\pi}{2}\}$,we see that there is no common value. Therefore,$A \cap B = \emptyset$.

Let $A = \{\theta \in R \mid \cos^2(\sin \theta) + \sin^2(\cos \theta) = 1\}$ and $B = \{\theta \in R \mid \cos(\sin \theta) \sin(\cos \theta) = 0\}$. Then,$A \cap B$ is:

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