遇到做不来的数学题了.在△ABC中a、b、c分别为角A、B、C的对边,设f(x)=a平方x平方-(a平方-b平方)x-4c平方.①若f(1)=0且B-C=π/3,求角C.②若f(2)=0,求角C的取值范围.
遇到做不来的数学题了.在△ABC中a、b、c分别为角A、B、C的对边,设f(x)=a平方x平方-(a平方-b平方)x-4c平方.①若f(1)=0且B-C=π/3,求角C.②若f(2)=0,求角C的取值范围.
遇到做不来的数学题了.
在△ABC中a、b、c分别为角A、B、C的对边,设f(x)=a平方x平方-(a平方-b平方)x-4c平方.①若f(1)=0且B-C=π/3,求角C.②若f(2)=0,求角C的取值范围.
遇到做不来的数学题了.在△ABC中a、b、c分别为角A、B、C的对边,设f(x)=a平方x平方-(a平方-b平方)x-4c平方.①若f(1)=0且B-C=π/3,求角C.②若f(2)=0,求角C的取值范围.
f(x) = a² x² - (a² - b² )x - 4c²
(1) f(1) =0 ==> a² -(a² -b² ) -4c² =0 ==> b² /c² =4 ==> b =2C
==> sinB = 2sinC
B-C=π/3 ==> B = C+π/3
==> sinB = sin(C+π/3)==>2sinC =sinC* cos(π/3) + cosC*sin(π/3)
==> tanC = √3/3 ==> C = π/6
(2) f(2)=0 ==> 4a² -2(a²-b²) - 4c² = 0 ==> a² + b² = 2c²;可知C为锐角(c小于直角边长)
∵ a² + b² - c² = 2ab*cosC,等式化为:
2ab*cosC = c² ==> cosC = c²/(2ab) = sin²C/(2sinA*sinB)
==> cosC = sin²C/[cos(A-B) - cos(A+B)]
= sin²C/[cos(A-B) + cosC]
整理得:
2cos²C + cos(A-B)*cosC -1 =0
cosC = 1/4*[-cos(A-B)±√(cos²(A-B)+8)]
由于C为锐角,cosC> 0,因此:
取 cosC = 1/4*[-cos(A-B)+√(cos²(A-B)+8)]
-1 < cos(A-B) ≤ 1 ==> 1/2 < cosC < 1
==> 0 < C < π/3
题目不清楚......