怎么做，求解！
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(sinA)^2 = 1/2 * {cos[(π/3 + B) - (π/3 -B)] - cos[(π/3 + B) + (π/3 - B)]} + (sinB)^2
= 1/2 * [cos(2B) - cos(2π/3)] + (sinB)^2
= 1/2 * [1 - 2(sinB)^2 + cos(π/3)] + (sinB)^2
= 1/2 - (sinB)^2 + 1/4 + (sinB)^2
= 3/4
所以，sinA = √3/2，则 A = 60°
AB' * AC' = b * c * cosA = bc * 1/2 = 12
所以，b*c = 24
又因为 b^2 + c^2 - 2bc*cosA = b^2 + c^2 - bc = a^2
则 b^2 + c^2 - 2bc = (b - c)^2 = a^2 - bc = 4*7 - 24 = 4
所以，c - b = 2, c = b + 2
因此，b = 4, c = 6
= 1/2 * [cos(2B) - cos(2π/3)] + (sinB)^2
= 1/2 * [1 - 2(sinB)^2 + cos(π/3)] + (sinB)^2
= 1/2 - (sinB)^2 + 1/4 + (sinB)^2
= 3/4
所以，sinA = √3/2，则 A = 60°
AB' * AC' = b * c * cosA = bc * 1/2 = 12
所以，b*c = 24
又因为 b^2 + c^2 - 2bc*cosA = b^2 + c^2 - bc = a^2
则 b^2 + c^2 - 2bc = (b - c)^2 = a^2 - bc = 4*7 - 24 = 4
所以，c - b = 2, c = b + 2
因此，b = 4, c = 6
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