(1)展开,得sinAcosπ/4 + cosAsinπ/4 + sinAcosπ/4 - cosAsinπ/4=2sinAcosπ/4=√2×sinA=√2/3
所以,sinA=1/3.
(2) 原式=[sinAcosπ/4 - cosAsinπ/4]/[2sin^A-2sinAcosA]=[√2/2]/2sinA=3√2/4.
(1)展开,得sinAcosπ/4 + cosAsinπ/4 + sinAcosπ/4 - cosAsinπ/4=2sinAcosπ/4=√2×sinA=√2/3
所以,sinA=1/3.
(2) 原式=[sinAcosπ/4 - cosAsinπ/4]/[2sin^A-2sinAcosA]=[√2/2]/2sinA=3√2/4.