\(\sin\alpha=4\cos\alpha\Leftrightarrow\cos\alpha=\dfrac{1}{4}\sin\alpha\)
Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\sin^2\alpha+\left(\dfrac{1}{4}\sin\alpha\right)^2=1\\ \Leftrightarrow\sin^2\alpha+\dfrac{1}{16}\sin^2\alpha=1\\ \Leftrightarrow\dfrac{17}{16}\sin^2\alpha=1\\ \Leftrightarrow\sin^2\alpha=\dfrac{16}{17}\)
Vì \(\alpha\) là một góc nhọn => \(\alpha>0\) => \(\sin\alpha=\sqrt{\dfrac{16}{17}}=\dfrac{4\sqrt{17}}{17}\)
\(\cos\alpha=\dfrac{1}{4}\sin\alpha=\dfrac{1}{4}.\dfrac{4\sqrt{17}}{17}=\dfrac{\sqrt{17}}{17}\)
Vậy \(P=3\sin\alpha.\cos\alpha=3.\dfrac{4\sqrt{17}}{17}.\dfrac{\sqrt{17}}{17}=\dfrac{12}{17}\)
