三角恒等变换习题
cos(x-y)=√5/5，cos2x=√10/10，并且x，y均为锐角，且x<y,则x+y=？A.π/6B.π/4C.3π/4D.5π/6...
cos(x-y)=√5/5 ，cos2x=√10/10 ，并且x，y均为锐角，且x<y,则x+y=？
A.π/6 B.π/4 C.3π/4 D.5π/6 展开
A.π/6 B.π/4 C.3π/4 D.5π/6 展开
1个回答
展开全部
【解】C
cos(x-y)=√5/5且x<y ==> sin(x-y)=-2√5/5
cos2x=√10/10 ==> sin2x=3√10/10
∵x+y=2x-(x-y)
∴cos(x+y)=cos[2x-(x-y)]
=cos2xcos(x-y)+sin2xsin(x-y)
=√10/10×√5/5+3√10/10×(-2√5/5)
=-√2/2
∵x,y都是锐角
∴(x+y)=3π/4
cos(x-y)=√5/5且x<y ==> sin(x-y)=-2√5/5
cos2x=√10/10 ==> sin2x=3√10/10
∵x+y=2x-(x-y)
∴cos(x+y)=cos[2x-(x-y)]
=cos2xcos(x-y)+sin2xsin(x-y)
=√10/10×√5/5+3√10/10×(-2√5/5)
=-√2/2
∵x,y都是锐角
∴(x+y)=3π/4
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
