3.在△ABC中,已知sin(A-B)cos B+cos(A-B)sin B≥1,则△ABC是( )
A.锐角三角形
B.钝角三角形
C.直角三角形
D.等腰非直角三角形
五、反思小结,观点提炼
布置作业
1.课本P131练习第1,2,3,4,5,6,7题.
2.课本P137习题3.1A组第1,6,7,8,13题.
参考答案
三、运用规律,解决问题
【例1】解:由sin α=-3/5,α是第四象限角,得cos α=√(1"-" sin^2 α)=√(1"-(-" 3/5 ")" ^2 )=4/5.
∴tan α=sinα/cosα=-3/4.
于是有sin(π/4-α)=sinπ/4cos α-cosπ/4sin α=√2/2×4/5-√2/2×(-3/5)=(7√2)/10,
cos(π/4+α)=cosπ/4cos α-sinπ/4sin α=√2/2×4/5-√2/2×(-3/5)=(7√2)/10,
tan(α-π/4)=(tanα"-" tan π/4)/(1+tanαtan π/4)=(tanα"-" 1)/(1+tanα)=("-" 3/4 "-" 1)/(1+"(-" 3/4 ")" )=-7.
【例2】解:(1)sin 72°cos 42°-cos 72°sin 42°=sin (72°-42°)=sin 30°=1/2.
(2)cos 20°cos 70°-sin 20°sin 70°=cos (20°+70°)=cos 90°=0.
(3)(1+tan15"°" )/(1"-" tan15"°" )=(tan45"°" +tan15"°" )/(1"-" tan45"°" tan15"°" )=tan (45°+15°)=tan 60°=√3.