求助一道高一的三角函数化简题在三角形ABC中,已知角A为锐角,且f(A)=[{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} / sin^2(π/2-A/2)-sin^2(π-A/2)]+cos^2(A)(1)求f(A)的最大值(2)若A+B=7π/12 ,f(A)=1 ,BC=2 ,求三角形ABC的三个
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 23:35:31
![求助一道高一的三角函数化简题在三角形ABC中,已知角A为锐角,且f(A)=[{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} / sin^2(π/2-A/2)-sin^2(π-A/2)]+cos^2(A)(1)求f(A)的最大值(2)若A+B=7π/12 ,f(A)=1 ,BC=2 ,求三角形ABC的三个](/uploads/image/z/7109439-15-9.jpg?t=%E6%B1%82%E5%8A%A9%E4%B8%80%E9%81%93%E9%AB%98%E4%B8%80%E7%9A%84%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%8C%96%E7%AE%80%E9%A2%98%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5%E8%A7%92A%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94f%28A%29%3D%5B%7B%5Bcos%28%CF%80-2A%29-1%5Dsin%28%CF%80%2BA%2F2%29sin%28%CF%80%2F2-A%2F2%29%7D+%2F+sin%5E2%28%CF%80%2F2-A%2F2%29-sin%5E2%28%CF%80-A%2F2%29%5D%2Bcos%5E2%28A%29%281%29%E6%B1%82f%28A%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%282%29%E8%8B%A5A%2BB%3D7%CF%80%2F12+%2Cf%28A%29%3D1+%2CBC%3D2+%2C%E6%B1%82%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA)
求助一道高一的三角函数化简题在三角形ABC中,已知角A为锐角,且f(A)=[{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} / sin^2(π/2-A/2)-sin^2(π-A/2)]+cos^2(A)(1)求f(A)的最大值(2)若A+B=7π/12 ,f(A)=1 ,BC=2 ,求三角形ABC的三个
求助一道高一的三角函数化简题
在三角形ABC中,已知角A为锐角,且f(A)=[{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} / sin^2(π/2-A/2)-sin^2(π-A/2)]+cos^2(A)
(1)求f(A)的最大值
(2)若A+B=7π/12 ,f(A)=1 ,BC=2 ,求三角形ABC的三个内角.
求助一道高一的三角函数化简题在三角形ABC中,已知角A为锐角,且f(A)=[{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} / sin^2(π/2-A/2)-sin^2(π-A/2)]+cos^2(A)(1)求f(A)的最大值(2)若A+B=7π/12 ,f(A)=1 ,BC=2 ,求三角形ABC的三个
令f(A) = I / J + cos^2(A) ,
其中I = [{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} ,
J = sin^2(π/2-A/2)-sin^2(π-A/2)].
分别化简两式:
由于cos(π-2A)-1 = -cos2A - 1 = -[cos2A + 1] = -2(cosA)^2 ,
sin(π+A/2)sin(π/2-A/2) = -sin(A/2)·cos(A/2) = -(sinA)/2
故 I = (sinA)·(cosA)^2 =
sin^2(π/2-A/2) = [cos(A/2)]^2 ,sin^2(π-A/2) = sin^2(A/2)
所以sin^2(π/2-A/2)-sin^2(π-A/2)] = [cos(A/2)]^2 - [sin(A/2)]^2 = cosA,
所以f(A) = I / J + cos^2(A)
= 【sin(2A)】/2 + 【1 + cos(2A)】/2
= (1/2)【sin(2A) + cos(2A)】 + (1/2)
= (1/2)【√2·sin(2A + π/4)】 + (1/2)
因为0