三角大题 球解析1.已知函数f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2(1)求函数f(x)的最小正周期和值域;(2)若f(a)=3√2/10,求sin2a的值;
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![三角大题 球解析1.已知函数f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2(1)求函数f(x)的最小正周期和值域;(2)若f(a)=3√2/10,求sin2a的值;](/uploads/image/z/15016563-27-3.jpg?t=%E4%B8%89%E8%A7%92%E5%A4%A7%E9%A2%98+%E7%90%83%E8%A7%A3%E6%9E%901.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2%28x%2F2%29-sin%28x%2F2%29cos%28x%2F2%29-1%2F2%281%29%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C%E5%80%BC%E5%9F%9F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5f%28a%29%3D3%E2%88%9A2%2F10%2C%E6%B1%82sin2a%E7%9A%84%E5%80%BC%EF%BC%9B)
三角大题 球解析1.已知函数f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2(1)求函数f(x)的最小正周期和值域;(2)若f(a)=3√2/10,求sin2a的值;
三角大题 球解析
1.已知函数f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2
(1)求函数f(x)的最小正周期和值域;
(2)若f(a)=3√2/10,求sin2a的值;
三角大题 球解析1.已知函数f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2(1)求函数f(x)的最小正周期和值域;(2)若f(a)=3√2/10,求sin2a的值;
(1)
f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2
=1/2(1+cosx)-1/2sinx-1/2
=1/2cosx-1/2sinx
=√2/2(√2/2cosx-√2/2sinx)
=√2/2cos(x+π/4)
f(x)最小正周期T=2π/1=2π
f(x)值域为[-√2/2,√2/2]
(2)若f(a)=3√2/10,
即1/2cosa-1/2sina=3√2/10
那么cosa-sina=3√2/5
两边平方:
cos²a+sin²a-2sinacosa=18/25
2sinacosa=7/25
即sin2a=7/25
f(x)=cos^2(x/2)-sin(x/2)cos(x/2)-1/2 =[cos^2(x/2)-1/2]-sin(x/2)cos(x/2) =1/2[2cos^2(x/2)-1]-1/2 *2sin(x/2)cos(x/2) =1/2 * cosx-1/2 *sinx =√2/2[√2/2 *cosx-√2/2sinx] =√2/2sin(π/4 -x) 最小正周期 2π,值域[-√2/2,√2/2] f(a)=√2/2sin(π/4 -a)=3√2/10 √2/2 *cosa-√2/2sina=3/5 cosa-sina=3√2/5 两边平方 cos^2a+sin^2a-2cosasina=18/25 1-2cosasina=18/25 sin2a=1-18/25 =7/25
sin(π/4 -a)=3/5