已知π⼀4<α<3π⼀4,0<β<π⼀4,cos(π⼀4+α)=-3⼀5,sina(3π⼀4+β)=5⼀13,求sina(α+β)的值

2025年05月09日 09:09
有2个网友回答
网友(1):

π/4<α<3π/4 ,cos(π/4+α)=-3/5 …… sin(π/4+α)=4/5

0<β<π/4, sin(3π/4+β)=5/13…… cos(3π/4+β)= - 12/13

sina(α+β)= - sin(α+β+π)=- sin((π/4+α)+(3π/4+β))

= - (sin(π/4+α)*cos(3π/4+β) + cos(π/4+α)*sin(3π/4+β) )

=63/65

网友(2):

π/4<α<3π/4 cos(π/4+α)=-3/5 就有 sin(π/4+α)=4/5

0<β<π/4 sin(3π/4+β)=5/13 cos(3π/4+β)= - 12/13

sina(α+β)= - sin(α+β+π)=- sin((π/4+α)+(3π/4+β))

= - (sin(π/4+α)*cos(3π/4+β) + cos(π/4+α)*sin(3π/4+β) )

=自己算吧