已知\(α\in(0,π)\),且\(sinα+cosα=\frac{1}{2}\),求\(cos2α\)的值为____.
\(cos2α\)=\(cos^2\)\(a\)-\(sin^2\)\(a\)=\((sinα+cosα)(cosα-sinα)\)=\(\frac{cosα-sinα}{2}\)
由\((sinα+cosα)^2=\frac{1}{4}\),\(sin^2\)\(α\)+\(cos^2\)\(α\)=1
得\(2sinαcosα\)=-\(\frac{3}{4}\)
所以\((sinα-cosα)^2\)=\(sin^2\)\(α\)+\(cos^2\)\(α\)-\(2sinαcosα\)=\(\frac{7}{4}\)
由\(α\in(0,π)\),得sinα>0
因为\(2sinαcosα\)<0
所以\(cosα\)<0
所以\(cosα-sinα\)=-\(\frac{\sqrt{7}}{2}\)
所以\(cos2α\)=-\(\frac{\sqrt{7}}{4}\)