Question

已知函数f(x)=根号2(sinx-cosx),求(1)函数f(x)的值域(2)若函数的图象过(a,6/5),兀/4<a<3/4兀,求f(兀/4+a)=

Answer 1

f(x)=2sin(x-45) 值域为[-2,2]
-兀/2+2k兀<=x-兀/4<=兀/2+2k兀
-兀/4+2k兀<=x<=3兀/4+2k兀
6/5=2sin(a-兀/4)
sin(a-兀/4)=3/5
兀/4<a<3/4兀
兀/2<兀/4+a<兀
0<a-兀/4<兀/2在-兀/4+2k兀<=x<=3兀/4+2k兀内
f(兀/4+a)=2sin(a+兀/4-兀/4)=2（sin(a-兀/4)√2/2+cos(a-兀/4)√2/2)=3√2/5+4√2/5

Answer 2

值域[-2,2] f(兀/4+a)= (3√2＋4√2)/5

追问

要过程

追答

过程：f(x)=根号2(sinx-cosx)
f(x)=2(√2/2sinx-√2/2cosx)
=2sin(x-兀/4)..
x取任何数...所以sin的值域是[-1,1]...所以就是[-2,2]
f(a)=2sin(a-兀/4)=6/5
cos(a-兀/4)=4/5
f(兀/4+a)= 2sin(兀/4+a-兀/4)=2sina=2sin(a-兀/4)cos兀/4+2cos(a-兀/4)sin兀/4= (3√2＋4√2)/5

Answer 3

1.
f(x)=2×[cos45sinx-cosxsin45)
f(x)=2×sin(x-45)
=2sin(x-π/4) 方式1
y=f(x)∈[-2,2]
2.
将（a,6/5)代入上面方式1，得sin(a-π/4)=3/5
将y=f(π/4+a)=2sin(a+π/4-π/4)=2sin(a-π/4+π/4)=2[sin(a-π/4)cosπ/4+cos(a-π/4)sinπ/4]
y=f(π/4+a)=2[√2/2sin(a-π/4)+√2 /2×cos(a-π/4)]=√2×(3/5+4/5)=7√2/5