Question

Cho \(\left\{{}\begin{matrix}0< \alpha< \dfrac{\pi}{4}\\sin\alpha+cos\alpha=\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)

Tính \(P=sin\alpha-cos\alpha\)

Answer 1

\(\sin a\cdot\cos a=\dfrac{\left(sina+cosa\right)^2-1}{2}=\dfrac{\dfrac{5}{4}-1}{2}=\dfrac{1}{8}\)

\(\left(sina-cosa\right)^2=1-2\cdot sina\cdot cosa=1-\dfrac{1}{4}=\dfrac{3}{4}\)

nên \(\left[{}\begin{matrix}sina-cosa=\dfrac{3}{4}\\sina-cosa=-\dfrac{3}{4}\end{matrix}\right.\)