已知π⼀2<b<a<3π⼀4,cos(a-b)=12⼀13,sin(a+b)=-3⼀5,求cos2a

2025-05-22 05:29:17
推荐回答(2个)
回答(1):

ππ/2所以cos(a+b)=-4/5,sin(a-b)=5/13,
cos2a=cos(a+b+a-b)=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=-4/5*12/13-(-3/5)*5/13=-33/65

回答(2):