已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间
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![已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间](/uploads/image/z/7055986-58-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%88cosx%2F2%2Ctan%28x%2F2%2B%CF%80%EF%BC%8F4%29%29%2C%E5%90%91%E9%87%8Fb%3D%EF%BC%88%E2%88%9A2sin%28x%2F2%2B%CF%80%2F4%29%2Ctan%28x%2F2-%CF%80%2F4%29%29%2C%E4%BB%A4f%EF%BC%88x%EF%BC%89%3Dab%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%2C%E5%B9%B6%E5%86%99%E5%87%BAf%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4)
已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间
已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab
求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间
已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间
f(x)=2cosx/2×(√2sin(x/2+π/4)+ tan(x/2+π/4)×tan(x/2-π/4))
=√2[sin(x+π/4)+sin(π/4)] + [1+tan(x/2)]/[1-tan(x/2)]×[tan(x/2)-1]/[1+tan(x/2)]
=√2sin(x+π/4)
最大值=√2
最小正周期=2π
sinx的增区间是:-π/2+2kπ≤x≤π/2+2kπ
带入-π/2+2kπ≤x+π/4≤π/2+2kπ
所以增区间-3π/4+2kπ≤x≤π/4+2kπ,在[0,π]上是[0,π/4]
减区间:π/2+2kπ
套定义去做,不要告诉我你三角函数不会...
f(x)=ab=cosx/2*√2sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)
=cosx/2*(sinx/2+cosx/2)+1
=1/2*sinx+1/2*cosx+3/2
=√2/2*sin(x+π/4)+3/2,
因为-1<=sin(x+π/4)<=1,所以
-√2/2+3/2<=f(x)<=√...
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f(x)=ab=cosx/2*√2sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)
=cosx/2*(sinx/2+cosx/2)+1
=1/2*sinx+1/2*cosx+3/2
=√2/2*sin(x+π/4)+3/2,
因为-1<=sin(x+π/4)<=1,所以
-√2/2+3/2<=f(x)<=√2/2+3/2,
故f(x)的最大值为:√2/2+3/2,
最小正周期为:T=2π。
由函数图象,可知:
f(x)在[0,π/4]上递增,在 [π/4,π]上递减。
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