设向量m=(√3sinπ/4,1),n=(cosx/4,cos^2x/4)(1)若m·n=1,求cos(2π/3-x)的值;(2)记f(x)=m·n,在△ABC中,角ABC的对边是abc,且满足(2a-c)cosB=bcosC,求函数f(A)的取值范围
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![设向量m=(√3sinπ/4,1),n=(cosx/4,cos^2x/4)(1)若m·n=1,求cos(2π/3-x)的值;(2)记f(x)=m·n,在△ABC中,角ABC的对边是abc,且满足(2a-c)cosB=bcosC,求函数f(A)的取值范围](/uploads/image/z/2522771-35-1.jpg?t=%E8%AE%BE%E5%90%91%E9%87%8Fm%3D%EF%BC%88%E2%88%9A3sin%CF%80%2F4%2C1%EF%BC%89%2Cn%3D%28cosx%2F4%2Ccos%5E2x%2F4%EF%BC%89%EF%BC%881%EF%BC%89%E8%8B%A5m%C2%B7n%3D1%2C%E6%B1%82cos%EF%BC%882%CF%80%2F3-x%EF%BC%89%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%B0f%EF%BC%88x%EF%BC%89%3Dm%C2%B7n%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92ABC%E7%9A%84%E5%AF%B9%E8%BE%B9%E6%98%AFabc%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%282a-c%EF%BC%89cosB%3DbcosC%2C%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88A%EF%BC%89%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
设向量m=(√3sinπ/4,1),n=(cosx/4,cos^2x/4)(1)若m·n=1,求cos(2π/3-x)的值;(2)记f(x)=m·n,在△ABC中,角ABC的对边是abc,且满足(2a-c)cosB=bcosC,求函数f(A)的取值范围
设向量m=(√3sinπ/4,1),n=(cosx/4,cos^2x/4)(1)若m·n=1,求cos(2π/3-x)的值;
(2)记f(x)=m·n,在△ABC中,角ABC的对边是abc,且满足(2a-c)cosB=bcosC,求函数f(A)的取值范围
设向量m=(√3sinπ/4,1),n=(cosx/4,cos^2x/4)(1)若m·n=1,求cos(2π/3-x)的值;(2)记f(x)=m·n,在△ABC中,角ABC的对边是abc,且满足(2a-c)cosB=bcosC,求函数f(A)的取值范围
M是SIN四分之X吧?
(1)由M·N=1得 √3/2sinx/2+1/2cosx/2+1/2=1
化简 sin(x/2+派/6)=1/2
所以 cos(pi/3-x/2)=1/2
cos(2pi/3-x0=2cos^2(pi/3-x/2)-1=-1/2
(2)易得f(x)=sin(x/2+pi/6)+1/2
(2a-c)cosB=bcosC由正弦定理得2sinAcosB=sinBcosC+sinCcosB
即2sinAcosB=sin(B+C)=sin(pi-A)=sinA
所以cosB=1/2 B=pi/3
所以 A属于(0,2pi/3)
f(x)属于(1,3/2)