已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π÷2)的图像关于点B(-π/4,0)对称,点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,且f(π/2)=1(1)求A ω φ的值(2)若0<θ<π ,且f(θ)=
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![已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π÷2)的图像关于点B(-π/4,0)对称,点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,且f(π/2)=1(1)求A ω φ的值(2)若0<θ<π ,且f(θ)=](/uploads/image/z/1042828-52-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3DAsin%EF%BC%88%CF%89x%2B%CF%86%29%EF%BC%88A%EF%BC%9E0%2C%CF%89%EF%BC%9E0%2C0%EF%BC%9C%CF%86%EF%BC%9C%CF%80%C3%B72%29%E7%9A%84%E5%9B%BE%E5%83%8F%E5%85%B3%E4%BA%8E%E7%82%B9B%EF%BC%88%EF%BC%8D%CF%80%2F4%2C0%EF%BC%89%E5%AF%B9%E7%A7%B0%2C%E7%82%B9B%E5%88%B0%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%E7%9A%84%E5%9B%BE%E5%83%8F%E7%9A%84%E5%AF%B9%E7%A7%B0%E8%BD%B4%E7%9A%84%E6%9C%80%E7%9F%AD%E8%B7%9D%E7%A6%BB%E4%B8%BA%CF%80%2F2+%2C%E4%B8%94f%EF%BC%88%CF%80%2F2%EF%BC%89%3D1%EF%BC%881%EF%BC%89%E6%B1%82A+%CF%89+%CF%86%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E8%8B%A50%EF%BC%9C%CE%B8%EF%BC%9C%CF%80+%2C%E4%B8%94f%EF%BC%88%CE%B8%EF%BC%89%3D)
已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π÷2)的图像关于点B(-π/4,0)对称,点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,且f(π/2)=1(1)求A ω φ的值(2)若0<θ<π ,且f(θ)=
已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π÷2)的图像关于点B(-π/4,0)对称,点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,且f(π/2)=1
(1)求A ω φ的值
(2)若0<θ<π ,且f(θ)=1/3 ,求cos2θ的值
已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π÷2)的图像关于点B(-π/4,0)对称,点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,且f(π/2)=1(1)求A ω φ的值(2)若0<θ<π ,且f(θ)=
(1).由图像关于点B(-π/4,0)对称,可知函的形式为f(x)=Asin【ω(x+π/4)+k*π】;对比原函数f(x)=Asin(ωx+φ),可知w*π/4+k*π=φ;
结合点B到函数y=f(x)的图像的对称轴的最短距离为π/2 ,可得到w=(π/2)/(π/2)=1; 结合ω>0,0<φ<π÷2,及w*π/4+k*π=φ;显然k=0,即φ=π/4;
将w,φ带入f(π/2)=1,即 A*sin(1*π/2+π/4)=1;可得A=2^0.5;
所以f(x)=2^0.5*sin(x+π/4);
(2)f(θ)=2^0.5*sin(θ+π/4)=1/3 所以:θ在第二象限,sin(θ+π/4)=1/3/2^0.5;cos(θ+π/4)=-17^0.5/3/2^0.5,
sin(2θ)=-cos(2θ+π/2)=cos(θ+π/4)^2-sin(θ+π/4)^2=2*2^0.5/3;,2θ在第四象限 所以:cos(2θ)=1/3;
?? 完全堪布动