\(\begin{array}{l} \Rightarrow ab = 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}.2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\ = \left( {2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha + \beta }}{2}} \right).2{\cos ^2}\frac{{\alpha - \beta }}{2}\\ = 2\sin \left( {\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\{a^2} + {b^2}\\ = 4{\cos ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\ + 4{\sin ^2}\frac{{\alpha + \beta }}{2}{\cos ^2}\frac{{\alpha - \beta }}{2}\\ = 4{\cos ^2}\frac{{\alpha - \beta }}{2}\left( {{{\cos }^2}\frac{{\alpha + \beta }}{2} + {{\sin }^2}\frac{{\alpha + \beta }}{2}} \right)\\ = 4{\cos ^2}\frac{{\alpha - \beta }}{2}\\ \Rightarrow {\cos ^2}\frac{{\alpha - \beta }}{2} = \frac{{{a^2} + {b^2}}}{4}\\ \Rightarrow ab = 2\sin \left( {\alpha + \beta } \right){\cos ^2}\frac{{\alpha - \beta }}{2}\\ = 2\sin \left( {\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{4}\\ = \sin \left( {\alpha + \beta } \right).\frac{{{a^2} + {b^2}}}{2}\\ \Rightarrow \sin \left( {\alpha + \beta } \right) = \frac{{2ab}}{{{a^2} + {b^2}}}\end{array}\) Đề bài Biết cosα +cosβ =a; sinα+sinβ =b (a,b là hằng số và a2+ b2 0) Hãy tính sin(α + β ) theo a và b. Lời giải chi tiết Ta có: \(\left. \matrix{ \(\begin{array}{l} Cách khác: 
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