急求高手解答三角题!已知锐角△ABC中,角A,B,C的对边分别为a,b,c,且(Sin^2A+Sin^2C-Sin^2B)tanB=√3SinASinC.(Ⅰ)如果b=2,△ABC的面积为√3,求a、c的值.(Ⅱ)求函数f(x)=cos^2(x+B/2)+√3Sinxcosx-1/2的单调区间.
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![急求高手解答三角题!已知锐角△ABC中,角A,B,C的对边分别为a,b,c,且(Sin^2A+Sin^2C-Sin^2B)tanB=√3SinASinC.(Ⅰ)如果b=2,△ABC的面积为√3,求a、c的值.(Ⅱ)求函数f(x)=cos^2(x+B/2)+√3Sinxcosx-1/2的单调区间.](/uploads/image/z/13693054-22-4.jpg?t=%E6%80%A5%E6%B1%82%E9%AB%98%E6%89%8B%E8%A7%A3%E7%AD%94%E4%B8%89%E8%A7%92%E9%A2%98%21%E5%B7%B2%E7%9F%A5%E9%94%90%E8%A7%92%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E4%B8%94%EF%BC%88Sin%5E2A%2BSin%5E2C-Sin%5E2B%29tanB%3D%E2%88%9A3SinASinC.%28%E2%85%A0%29%E5%A6%82%E6%9E%9Cb%3D2%2C%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BA%E2%88%9A3%2C%E6%B1%82a%E3%80%81c%E7%9A%84%E5%80%BC.%EF%BC%88%E2%85%A1%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2%28x%2BB%2F2%29%2B%E2%88%9A3Sinxcosx-1%2F2%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4.)
急求高手解答三角题!已知锐角△ABC中,角A,B,C的对边分别为a,b,c,且(Sin^2A+Sin^2C-Sin^2B)tanB=√3SinASinC.(Ⅰ)如果b=2,△ABC的面积为√3,求a、c的值.(Ⅱ)求函数f(x)=cos^2(x+B/2)+√3Sinxcosx-1/2的单调区间.
急求高手解答三角题!
已知锐角△ABC中,角A,B,C的对边分别为a,b,c,且(Sin^2A+Sin^2C-Sin^2B)tanB=√3SinASinC.
(Ⅰ)如果b=2,△ABC的面积为√3,求a、c的值.
(Ⅱ)求函数f(x)=cos^2(x+B/2)+√3Sinxcosx-1/2的单调区间.
最好有详细过程!谢谢了!
急求高手解答三角题!已知锐角△ABC中,角A,B,C的对边分别为a,b,c,且(Sin^2A+Sin^2C-Sin^2B)tanB=√3SinASinC.(Ⅰ)如果b=2,△ABC的面积为√3,求a、c的值.(Ⅱ)求函数f(x)=cos^2(x+B/2)+√3Sinxcosx-1/2的单调区间.
由正弦定理,转化原式为:
(a^2+c^2-b^2)tanB=√3ac
再据余弦定理:
2accosB*tanB=√3ac
sinB=√3/2
B=π/3或2π/3
A) B=π/3
设c=x,可列方程:[x/2+√(4-3x^3/4)]*0.5*√3x/2=√3
16/x^2+x^2=8
x=2=c
0.5acsinB=√3,a=2
B) B=2π/3,无解
所以,a=c=2
(2)
cos^2(x+B/2)=2cos(2x+B)-1=2cos2xcosB-2sin2xsinB-1
=cos2x-√3sin2x-1
f(x)=cos2x-3/2
-π/2+kπ