1.In triangle ABC, it is given that (sinAcosB)/(cosAsinB) = (4c-b)/(b).

Using the Sine law and Cosine law, prove that a^2=b^2+c^2-(1/2)bc.

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• 5 年前
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(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

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