两道与三角函数有关的题目1.设a为常数,且a>1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为多少.2.已知sin(x+y)=1 求证 tan(2x+y)+tan y=0第一题是选择题、 A 2a+1 B 2a-1 C-2a-1 D a^2
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![两道与三角函数有关的题目1.设a为常数,且a>1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为多少.2.已知sin(x+y)=1 求证 tan(2x+y)+tan y=0第一题是选择题、 A 2a+1 B 2a-1 C-2a-1 D a^2](/uploads/image/z/8653624-16-4.jpg?t=%E4%B8%A4%E9%81%93%E4%B8%8E%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E6%9C%89%E5%85%B3%E7%9A%84%E9%A2%98%E7%9B%AE1.%E8%AE%BEa%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E4%B8%94a%EF%BC%9E1%2C0%E2%89%A4x%E2%89%A42%CF%80+%E5%88%99%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2X%2B2asinX-1%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91.2.%E5%B7%B2%E7%9F%A5sin%28x%2By%29%3D1+%E6%B1%82%E8%AF%81+tan%282x%2By%29%2Btan+y%3D0%E7%AC%AC%E4%B8%80%E9%A2%98%E6%98%AF%E9%80%89%E6%8B%A9%E9%A2%98%E3%80%81+A+2a%2B1+B+2a-1+C-2a-1+D+a%5E2)
两道与三角函数有关的题目1.设a为常数,且a>1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为多少.2.已知sin(x+y)=1 求证 tan(2x+y)+tan y=0第一题是选择题、 A 2a+1 B 2a-1 C-2a-1 D a^2
两道与三角函数有关的题目
1.设a为常数,且a>1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为多少.
2.已知sin(x+y)=1 求证 tan(2x+y)+tan y=0
第一题是选择题、 A 2a+1 B 2a-1 C-2a-1 D a^2
两道与三角函数有关的题目1.设a为常数,且a>1,0≤x≤2π 则函数f(x)=cos^2X+2asinX-1的最大值为多少.2.已知sin(x+y)=1 求证 tan(2x+y)+tan y=0第一题是选择题、 A 2a+1 B 2a-1 C-2a-1 D a^2
(1)f(x)=cos^2X+2asinX-1
=-sinX^2+2asinX
=-(sinX-a)^2+a^2
0≤x≤2π ,-1≤sinX≤1
a>1,sinX=1时,f(x)最大值为a^2
选D
(2)证明:sin(x+y)=1
x+y=π/2+2kπ
2x+y=π/2+2kπ+x
tan(2x+y)+tan y
= tan(π/2+2kπ+x)+tan(π/2+2kπ-x)
=tan(π/2+x)+tan(π/2-x)
=-cotx+cotx
=0
1. B 2a-1 .证明如下:
f(x)=cos²x+2asinx-1=1-sin²x+2asinx-1=2asinx-sin²x.
f(x)-(2a-1)=2asinx-sin²x-(2a-1)=2a(sinx-1)-(sin²x-1)
=(sinx-1)[2a-(sinx+1)]≤0.
故...
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1. B 2a-1 .证明如下:
f(x)=cos²x+2asinx-1=1-sin²x+2asinx-1=2asinx-sin²x.
f(x)-(2a-1)=2asinx-sin²x-(2a-1)=2a(sinx-1)-(sin²x-1)
=(sinx-1)[2a-(sinx+1)]≤0.
故f(x)≤2a-1,得f(x)max=2a-1.
2.已知sin(x+y)=1 ,得2(x+y)=mπ(m=4n+1).则2x+y=2(x+y)-y=mπ -y.
故:tan(2x+y)+tan y=tan(mπ -y)+tany=-tany+tany=0.
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