In a triangle ABC if Cos C= sinA / (2*sinB) then :?

for any triangle a/sin(A) = b/sin(B) = c/sin(C) [the sine rule]

∴ a = csin(A)/sin(C)

∴ b = csin(B)/sin(C)

for any triangle c² = a² + b² − 2abcos(C) [the cosine rule]

∴ 2abcos(C) = a² + b² − c²

∴ cos(C) = (a² + b² − c²)/(2ab)

∴ cos(C) = ((csin(A)/sin(C))² + (csin(B)/sin(C))² − c²)/(2(csin(A)/sin(C))(csin(B)/sin(C)))

∴ cos(C) = (c²sin²(A)/sin²(C) + c²sin²(B)/sin²(C) − c²)/(2(csin(A)/sin(C))(csin(B)/sin(C)))

∴ cos(C) = c²(sin²(A)/sin²(C) + sin²(B)/sin²(C) − 1)/(2(csin(A)/sin(C))(csin(B)/sin(C)))

∴ cos(C) = sin²(C)c²(sin²(A)/sin²(C) + sin²(B)/sin²(C) − 1)/(2c²sin(A)sin(B))

∴ cos(C) = sin²(C)(sin²(A)/sin²(C) + sin²(B)/sin²(C) − 1)/(2sin(A)sin(B))

∴ cos(C) = sin²(C)(sin²(A) + sin²(B) − sin²(C))/(2sin(A)sin(B)sin²(C))

∴ cos(C) = (sin²(A) + sin²(B) − sin²(C))/(2sin(A)sin(B))

if cos(C) = sin(A)/(2sin(B))

then (sin²(A) + sin²(B) − sin²(C))/(2sin(A)sin(B)) = sin(A)/(2sin(B))

∴ (sin²(A) + sin²(B) − sin²(C))/sin(A) = sin(A)

∴ sin²(A) + sin²(B) − sin²(C) = sin²(A)

∴ sin²(B) − sin²(C) = 0

∴ sin²(B) = sin²(C)

so we approved at ABC is at least an isosceles triangle, which rules out Answer d.

if ABC is an right angle isosceles A must be the 90° hence B and C are 45°

cos(45°) = sin(90°)/(2sin(45°))

∴ 1/√2 = 1/(2(1/√2))

∴ 1/√2 = √2/2

∴ 1/√2 * √2/√2 = √2/2

∴ √2/2 = √2/2

So Answer B is true when A = 90°, B = 45°, C = 45°

but and equilateral A = B = C = 60°

and cos(60°) = sin(60°)/(2sin(60°))

∴ cos(60°) = 1/2

So answer C is true too

and An obtuse isosceles like A = 120°, B = 30°, C=30°

cos(30°) = sin(120°)/(2sin(30°))

∴ √(3)/2 = (√(3)/2)/(2(1/2))

∴ √(3)/2 = √(3)/2

so answer A is true too.

however Answer B is the only one that's always true. [there are right angle triangle and obtuse triangles for which cos(C) = sin(A)/(2sin(B)) is not true] it just B isn't the only answer it's true for