\(sina=\frac{4}{5}\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{3}{5}\)
\(\Rightarrow sin2a=2sina.cosa=\frac{24}{25}\)
\(cos2a=2cos^2a-1=-\frac{7}{25}\)
\(0< \frac{a}{2}< \frac{\pi}{4}\Rightarrow\left\{{}\begin{matrix}sin\frac{a}{2}>0\\cos\frac{a}{2}>0\end{matrix}\right.\)
\(\frac{3}{5}=cosa=2cos^2\frac{a}{2}-1\Rightarrow cos\frac{a}{2}=\sqrt{\frac{\frac{3}{5}+1}{2}}=\frac{2\sqrt{5}}{5}\)
\(sin\frac{a}{2}=\sqrt{1-cos^2\frac{a}{2}}=\frac{\sqrt{5}}{5}\)
Vậy: \(sin\frac{5a}{2}=sin\left(2a+\frac{a}{2}\right)=sin2a.cos\frac{a}{2}+cos2a.sin\frac{a}{2}\) (thay số bấm máy)
