Question

在三角形ABC中，内角ABC的对边分别为abc。已知A=B,2b=根号3c.（1）求COSC的值（2）求COS(2C+4分之派)的值

Answer 1

1. 余弦定理
cosC=(a^2+b^2-c^2)/2ab 因为A=B，所以a=b,
=(2b^2-c^2)/2b^2 又因为2b=根号3c，所以4b^2=3c^2
=(2b^2-4/3b^2) /2b^2
=1/3
2. cos2C=2cos^2C-1=-7/9 cosC>0 0°<C<90° 0°<2C<180°
sin2C=4√2/9
COS(2C+4分之派)
=cos2Ccos4分之派-sin2Csin4分之派
=-7/9*√2/2-4√2/9*√2/2
=-(7√2+8)/18

Answer 2

A=B
三角形为等腰三角形a=b
根据余弦定理：
cosC=(a²+b²-c²)/(2ab) = (b²+b²-c²)/(2b²) = (2b²-c²)/(2b²) = 1-2(c/2b)² = 1 - 2×(1/√3)² = 1/3
sinC=√(1-cos²C) = 2√2/3
cos2C=2cos²C-1=-7/9
sin2C=2sinCcosC=4√2/9
cos(2C+π/4) = cos2Ccosπ/4-sin2Csinπ/4 = -7/9×√2/2-4√2/9×√2/2 = - (7√2+8)/18