f(x)=sin(π/3+4x)+cos(4x-π/6） 求在【-π/12,11π/48】上的值域

f(x)=sin(π/3+4x)+cos(4x-π/6） 求在【-π/12,11π/48】上的值域
f(x)=sin(π/3+4x)+cos(4x-π/6） 求在【-π/12,11π/48】上的值域

f(x)=sin(π/3+4x)+cos(4x-π/6） 求在【-π/12,11π/48】上的值域
f(x)=sin(π/3+4x)+cos(4x-π/6）
=(√3/2)cos4x+(1/2)sin4x+(√3/2)cos4x+(1/2)sin4x
=√3cos4x+sin4x
=2[(√3/2)cos4x+(1/2)sin4x]
=2sin(4x+π/3)
x∈[-π/12,11π/48]
所以
4x+π/3∈[0,5π/4]
所以
2sin(4x+π/3)∈[-√2,2]
则值域为 [-√2,2]

正解

f(x
=(√3/2)cos4x+(1/2)sin4x+(√3/2)cos4x+(1/2)sin4x
=√3cos4x+sin4x
=2[(√3/2)cos4x+(1/2)sin4x]
=2sin(4x+π/3)
x∈[-π/12,11π/48]
所以
4x+π/3∈[0,5π/4]
所以
2sin(4x+π/3)∈[-√2,2]
则值域为 [-√2,2] pp