Question

设三角形ABC的内角A、B、C的对边长分别为abc,且3b^2+3c^2-3a^2=4(根号下2)bc

a)求SinA的值
b)求[2sin(A+π/4)sin(B+C+π/4)]/1-cos2A的值。

Answer 1

余弦定理
CosA = (b^2+c^2-a^2)/(2bc) = 2/3 * 根号(2)
所以(SinA)^2 = 1-(CosA)^2 = 1 - 8/9 = 1/9
所以SinA = 1/3 因为A是 0~π的角,SinA>0

2
B+C = π - A
所以分子变为
2Sin(A + π/4)Sin(π - A + π/4) = 2Sin(A + π/4) Sin(A-π/4)
= - (Cos(A + π/4 + A - π/4) - Cos(A + π/4 - A + π/4) = - (Cos2A -0) = -Cos2A
则原式化为
Cos2A/(Cos2A-1)
而Cos2A = 1-2(SinA)^2 二倍角公式
SinA = 1/3
所以Cos2A = 1 - 2/9 = 7/9
那么原式 = 7/9 / (7/9 -1) = -7/2

Answer 2

根据余弦定理cosa=(b^2+c^2-a^2)/2bc=2sqr(2)/3
因此sina=1/3
（2）sin(b+c+pi/4)=sin(pi-a+pi/4)=-sin(pi/4-a)=-cos(a+pi/4)
原式=-(2sin(a+pi/4)cos(a+pi/4))/(1-cos2a)=-sin(2a+pi/2)/(1-cos2a)=cos2a(cos2a-1)=1-(1/2)sin^2a=1-9/2=-7/2
计算未必对，不过过程就是这样