sinα+sinβ=2分之根号2，求cosα+cosβ取值范围

sinα+sinβ=2分之根号2，
令cosα+cosβ=k
分别平方并相加
由sin²x+cos²x=1
所以2+2(cosαcosβ+sinαsinβ)=1/2+k²
cos(α-β)=cosαcosβ+sinαsinβ=(k²-3/2)/2
-1<=cos(α-β)<=1
-1<=(k²-3/2)/2<=1
-1/2<=k²<=7/2
即0<=k²<=7/2
所以-√14/2<=cosα+cosβ<=√14/2

f(x)=5sinxcosx-5√3cos²x+5√3/2
=5sin(2x)/2-5√3cos(2x)/2
=5[sin(2x)/2-√3cos(2x)/2]
=5[sin(2x)cos(π/3)-cos(2x)sin(π/3)]
=5sin(2x-π/3)

T=2π/2=π
f(x)的最小正周期是π

f(x)=5sin(2x-π/3)=5sin[2(x-π/6)]的图象即把f(x)=sinx的图象横坐标缩小1倍，纵坐标扩大4倍，向右平移π/6个单位得到

f(x)=5sinx的单调递减区间是[2kπ+π/2，2kπ+3π/2]
f(x)=5sin(2x)的单调递减区间是[kπ+π/4，kπ+3π/4]
f(x)=5sin[2(x-π/6)]的单调递减区间是[kπ+5π/12，kπ+11π/12]

f(x)的单调递减区间是[kπ+5π/12，kπ+11π/12]，k∈Z

f(x)=5sin(2x-π/3)
当2x-π/3=2kπ+π/2，即x=kπ+5π/12时，f(x)有最大值5，k∈Z
当2x-π/3=2kπ+3π/2，即x=kπ+11π/12时，f(x)有最小值-5，k∈Z

... 你怎么看的啊 我在这只能说你看错了