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Problem about sin&cos law

發問:

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!

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 - ? 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 - ? 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 - ? bc

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