Problem about sin&cos law

(sinA cosB) / (cosA sinB) = (4c - b) / b
(a / b) (cosB / cosA) = (4c - b) / b ...... By Sine law , a / b = sinA / sinB

2015-03-02 22:33:31 補充：
Proof 1 :(sinA cosB) / (cosA sinB) = (4c - b) / b
(sinA / sinB) (cosB / cosA) = (4c - b) / b , by Sine law :
(a / b) (cosB / cosA) = (4c - b) / b
a cosB = (4c - b) cosA , by Cosine law :
a (a² + c² - b²) / (2ac) = (4c - b) (b² + c² - a²) / (2bc)
b (a² + c² - b²) = (4c - b) (b² + c² - a²)
ba² + bc² - b³ = 4cb² + 4c³ - 4ca² - b³ - bc² + ba²
2bc² = 4cb² + 4c³ - 4ca²
bc = 2b² + 2c² - 2a²
a² = b² + c² - ½ bc
Prove 2 :(sinA cosB) / (cosA sinB) = (4c - b) / b
(sinA cosB) / (cosA sinB) + 1 = (4c - b) / b + 1
(sinA cosB + cosA sinB) / (cosA sinB) = 4c / b
sin(A+B) / (cosA sinB) = 4c / b
sinC / (cosA sinB) = 4sinC / sinB ...... by Sine law
cosA = 1/4 for sinC ≠ 0 and sinB ≠ 0
(b² + c² - a²) / (2bc) = 1/4 ...... by Cosine law
a² = b² + c² - ½ bc
Prove 3 :(sinA cosB) / (cosA sinB) = (4c - b) / b
a cosB / (b cosA) = (4c - b) / b ...... by Sine law
a cosB / (b cosA) + 1 = (4c - b) / b + 1
(a cosB + b cosA) / (b cosA) = 4c / b
c / (b cosA) = 4c / b
cosA = 1/4 for c ≠ 0
(b² + c² - a²) / (2bc) = 1/4 ...... by Cosine law
a² = b² + c² - ½ bc