已知cos(a+π⼀4)=3⼀5,x⼀2<=a<=3π⼀2,求1:cos(2a+π⼀4)的值。

2024-11-23 09:22:52
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回答(1):

π/2≤a≤3π/2,
cos(a+π/4)=3/5,就是在第四象限;
3π/2<a+π/4<2π,3π<2a+π/2<4π,π<2a+π/2<2π,
3π-π/4<2a+π/4<4π-π/4,
cos(a+π/4)=3/5,
sin(a+π/4)= -4/5,
cos(2a+π/2)=9/25-16/25=-7/25;
sin(2a+π/2)=-24/25
cos(2a+π/4)=cos(2a+π/2-π/4)=(-7/25+24/25)/√2=17/25/√2