已知sin（a+π⼀4）=3⼀5，π⼀2＜a≤3π⼀2，求sin（2a+π⼀4）

首先观察sin（a+π/4）=3/5,由于3,4,5是经典勾股数,知道cos是很好求的值.
而要求的值是a+π/4的两倍差π/4,利用和角公式可得.
求解过程:
sin（a+π/4）=3/5=>[cos（a+π/4）]^2=1-(3/5)^2=16/25
得到cos（a+π/4）=4/5或-4/5(以a+π/4的范围判断)
因π/2＜a≤3π/2,则3π/4所以-1<=cos(a+π/4)<√2/2<4/5
cos(a+π/4)=-4/5
sin(2a+π/2)=2sin(a+π/4)cos(a+π/4)=-24/25
cos(2a+π/2)=cos^2(a+π/4)-sin^2(a+π/4)=7/25
sin(2a+π/4)=sin(2a+π/2-π/4)=sin(2a+π/4)*cosπ/4-sinπ/4*cos(2a+π/4)
=-24/25*√2/2-√2/2*7/25=-31√2/50

希望对你有所帮助 还望采纳~~

π/2≤a≤3π/2
cos(a+π/4)=3/5>0
a+π/4>3π/2
3π/2sin(a+π/4)=-√[1-cos²(a+π/4)]=-4/5
sina=[sin(a+π/4)-π/4]=√2/2*[sin(a+π/4)-cos(a+π/4)]=√2/2*(-4/5-3/5)=-7√2/10
cosa=-√(1-sin²a)=-√2/10
cos2a=2cos²a-1=-24/25,sin2a=2sinacosa=7/25
cos(2a+π/4)=√2/2(cos2a-sin2a)=-31√2/50
希望对你能有所帮助。