sinα+sinβ=2分之根号2,求cosα+cosβ取值范围答案选项里都有14分之根号14,
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sinα+sinβ=2分之根号2,求cosα+cosβ取值范围答案选项里都有14分之根号14,
sinα+sinβ=2分之根号2,求cosα+cosβ取值范围
答案选项里都有14分之根号14,
sinα+sinβ=2分之根号2,求cosα+cosβ取值范围答案选项里都有14分之根号14,
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
... 你怎么看的啊 我在这只能说你看错了
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...
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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
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