设函数f(x)=cos(2x+π/3)+sin^2x设函数f(x)=cos(2x+π/3)+sin^2 X设A,B,C为△ABC的三个内角,若cosB=1/3,f(c/2)=1/4,且C为锐角,求sinA最好30分钟内给答案,设函数f(x)=cos(2x+π/3)+sin^2 X设A,C为△ABC的三个内角,若cosB=1/
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![设函数f(x)=cos(2x+π/3)+sin^2x设函数f(x)=cos(2x+π/3)+sin^2 X设A,B,C为△ABC的三个内角,若cosB=1/3,f(c/2)=1/4,且C为锐角,求sinA最好30分钟内给答案,设函数f(x)=cos(2x+π/3)+sin^2 X设A,C为△ABC的三个内角,若cosB=1/](/uploads/image/z/1063676-20-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2x%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2+X%E8%AE%BEA%2CB%2CC%E4%B8%BA%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E8%8B%A5cosB%3D1%2F3%2Cf%28c%2F2%29%3D1%2F4%2C%E4%B8%94C%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82sinA%E6%9C%80%E5%A5%BD30%E5%88%86%E9%92%9F%E5%86%85%E7%BB%99%E7%AD%94%E6%A1%88%2C%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2+X%E8%AE%BEA%EF%BC%8CC%E4%B8%BA%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%EF%BC%8C%E8%8B%A5cosB%3D1%2F)
设函数f(x)=cos(2x+π/3)+sin^2x设函数f(x)=cos(2x+π/3)+sin^2 X设A,B,C为△ABC的三个内角,若cosB=1/3,f(c/2)=1/4,且C为锐角,求sinA最好30分钟内给答案,设函数f(x)=cos(2x+π/3)+sin^2 X设A,C为△ABC的三个内角,若cosB=1/
设函数f(x)=cos(2x+π/3)+sin^2x
设函数f(x)=cos(2x+π/3)+sin^2 X
设A,B,C为△ABC的三个内角,若cosB=1/3,f(c/2)=1/4,且C为锐角,求sinA
最好30分钟内给答案,
设函数f(x)=cos(2x+π/3)+sin^2 X
设A,C为△ABC的三个内角,若cosB=1/3,f(c/2)=-1/4,且C为锐角,求sinA
是写错了,麻烦帮我做下
设函数f(x)=cos(2x+π/3)+sin^2x设函数f(x)=cos(2x+π/3)+sin^2 X设A,B,C为△ABC的三个内角,若cosB=1/3,f(c/2)=1/4,且C为锐角,求sinA最好30分钟内给答案,设函数f(x)=cos(2x+π/3)+sin^2 X设A,C为△ABC的三个内角,若cosB=1/
f(x)=cos(2x+π/3)+sin^2 X=1/2cos2x-根号3/2sin2x+(1-cos2x)/2=1/2-根号3/2sin2x
因为f(c/2)=-1/4,所以sinC=根号3/2,cosC=1/2
又因为cosB=1/3,所以sinB=2倍根号2/3
那么sinA=sin(π-B-C)=sin(B+C)=sinBcosC+sinCcosB=(2倍根号2+根号3)/6
主要是2倍角的公式运用