230问答网 > 已知cos(a+π⼀4)=3⼀5, π⼀2≤a≤3π⼀2, 求cos(2a+3π⼀4)的值

已知cos(a+π⼀4)=3⼀5, π⼀2≤a≤3π⼀2, 求cos(2a+3π⼀4)的值

2024-12-28 14:23:16

推荐回答（2个）

回答1：

因为cos（a+π/4）=3/5>0,
π/2<=a<=3π/2,
3π/4=所以sin(a+π/4)=-4/5

sin(2a+π/2)
=2sin(a+π/4)cos(a+π/4)
=2*(-4/5)*3/5)
=-24/25,

cos(2a+π/2)
=2cos(a+π/4)^2-1
=2*9/25-1
=-7/25

cos(2a+3π/4)
=cos[(2a+π/2)+π/4]
=cos(2a+π/2)cosπ/4-sin(2a+π/2)sinπ/4
=(-7/25)*(√2/2)-(-24/25)*(√2/2)
=-7√2/50+24√2/2
=17√2/50

回答2：

cos(a+π/4)=3/5 cos(2a+π/2)=2*(3/5)^2-1= -7/25 (II,III象限) sin(2a+π/2)=+/-24/25
cos(2a+3π/4)= cos(2a+π/2)cos(π/4)- sin(2a+π/2)sin(π/4)
=-7/25*根号2/2+/-24/25*根号2/2
-17根号2/50 -31根号2/50