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If $\sin \alpha+\cos \alpha=\mathrm{p}$, then what is $\cos ^{2}(2 \alpha)$ equal to?
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The correct answer is:
$\mathrm{P}^{2}\left(2-\mathrm{p}^{2}\right)$
\sin \alpha+\cos \alpha=p \\ & \Rightarrow(\sin \alpha+\cos \alpha)^{2}=p^{2} \\ & \Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=p^{2} \\ & \Rightarrow 1+\sin 2 \alpha=p^{2} \\ & \Rightarrow \sin 2 \alpha=\mathrm{p}^{2}-1 \\ & \cos ^{2} 2 \alpha=1-\sin ^{2} 2 \alpha=1-\left(\mathrm{p}^{2}-1\right)^{2} \\ &=1-\left(\mathrm{p}^{4}+1-2 \mathrm{p}^{2}\right)=-\mathrm{p}^{4}+2 \mathrm{p}^{2} \\ &=\mathrm{p}^{2}\left(2-\mathrm{p}^{2}\right) \\
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