设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a
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![设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a](/uploads/image/z/3857906-2-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%28x%2B2%CF%80%2F3%29%2B2cos%26sup2%3Bx%2F2%2Cx%E5%B1%9E%E4%BA%8ER.%E8%AE%B0%E2%96%B3ABC%E5%86%85%E8%A7%92A+B+C%E7%9A%84%E5%AF%B9%E8%BE%B9%E9%95%BF%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E8%8B%A5f%28B%29%3D1%2Cb%3D1%2Cc%3D%E6%A0%B9%E5%8F%B7%E4%B8%89%2C%E6%B1%82a)
设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a
设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R
.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a
设函数f(x)=cos(x+2π/3)+2cos²x/2,x属于R.记△ABC内角A B C的对边长分别为a,b,c,若f(B)=1,b=1,c=根号三,求a
f(B)=cos(B+2π/3)+2cos²B/2=1推出cos(B+2π/3)= -cosB所以B+2π/3+B=π推出B=π/6
cosB=(a²+c²-b²)/2*a*c=(a²+3-1)/2*a*根号3=根号3/2推出a=1或2~