If P = frac{1}{2} sin^2 theta + frac{1}{3} co | vedclass

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(A) Given $P = \frac{1}{2} \sin^2 	heta + \frac{1}{3} \cos^2 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we get: $P = \frac{1}{2

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$\frac{1}{3} \leq P \leq \frac{1}{2}$

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(A) Given $P = \frac{1}{2} \sin^2 	heta + \frac{1}{3} \cos^2 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we get: $P = \frac{1}{2} \sin^2 	heta + \frac{1}{3} (1 - \sin^2 	heta)$ $P = \frac{1}{3} + (\frac{1}{2} - \frac{1}{3}) \sin^2 	heta$ $P = \frac{1}{3} + \frac{1}{6} \sin^2 	heta$. Since $0 \leq \sin^2 	heta \leq 1$, we have: $0 \leq \frac{1}{6} \sin^2 	heta \leq \frac{1}{6}$ Adding $\frac{1}{3}$ to all parts: $\frac{1}{3} \leq \frac{1}{3} + \frac{1}{6} \sin^2 	heta \leq \frac{1}{3} + \frac{1}{6}$ $\frac{1}{3} \leq P \leq \frac{1}{2}$.

If $P = \frac{1}{2} \sin^2 \theta + \frac{1}{3} \cos^2 \theta$, then:

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