Problem about sin&cos law

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.

Thanks!

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  • 5 年 前
    最佳解答

    (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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