Question

cho sin (a-b) = \(\dfrac{2}{5}\) ; sin (a+b) = \(-\dfrac{3}{5}\)

tính sin a.cos b

Answer 1

ta có:\(sin\alpha.cosb=\dfrac{1}{2}\left[sin\left(a-b\right)+sin\left(a+b\right)\right]\)

\(=\dfrac{1}{2}\left[\dfrac{2}{5}+\left(-\dfrac{3}{5}\right)\right]\)

\(=\dfrac{1}{2}.\left(-\dfrac{1}{5}\right)\)

\(=-\dfrac{1}{10}\)

Answer 2

Ta có \(sin\left(a-b\right)+sin\left(a+b\right)=2sin\left(\dfrac{a-b+a+b}{2}\right)cos\left(\dfrac{a+b-\left(a-b\right)}{2}\right)\\ \Rightarrow2sin\left(a\right).cos\left(b\right)=\dfrac{2}{5}+\left(-\dfrac{3}{5}\right)=-\dfrac{1}{5}\\ \Rightarrow sin\left(a\right)cos\left(b\right)=-\dfrac{1}{10}\)