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sin a + sin b + sin c ???
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(a) (i) Prove that x^2 + y^2 + z^2 >= xy + yz + zx for any real numbers x, y, zLet a, b, c (measured in radians) be the interior angles of a triangle(ii) By using (a)(i), prove that 1/a^2 + 1/b^2 + 1/c^2 >= 27/(pi^2)(b) By using AM >= GM, prove that sin a + sin b + sin c <= 3 root(3) / 2[Hint:... 顯示更多 (a) (i) Prove that x^2 + y^2 + z^2 >= xy + yz + zx for any real numbers x, y, z Let a, b, c (measured in radians) be the interior angles of a triangle (ii) By using (a)(i), prove that 1/a^2 + 1/b^2 + 1/c^2 >= 27/(pi^2) (b) By using AM >= GM, prove that sin a + sin b + sin c <= 3 root(3) / 2 [Hint: cos[(a-b)/2] <= 1 This is a Pure Maths question. I don't know how to do part (b). Please help~ 更新: To takkinl: The method is amazing. I have never thought of adding a new unknown. But as winwincy92 mentioned, inequality AM >= GM is needed to be used. And I still cannot use the inequality to solve part (b). Can you help me?
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sin a + sin b + sin c ???
發問:
(a) (i) Prove that x^2 + y^2 + z^2 >= xy + yz + zx for any real numbers x, y, zLet a, b, c (measured in radians) be the interior angles of a triangle(ii) By using (a)(i), prove that 1/a^2 + 1/b^2 + 1/c^2 >= 27/(pi^2)(b) By using AM >= GM, prove that sin a + sin b + sin c <= 3 root(3) / 2[Hint:... 顯示更多 (a) (i) Prove that x^2 + y^2 + z^2 >= xy + yz + zx for any real numbers x, y, z Let a, b, c (measured in radians) be the interior angles of a triangle (ii) By using (a)(i), prove that 1/a^2 + 1/b^2 + 1/c^2 >= 27/(pi^2) (b) By using AM >= GM, prove that sin a + sin b + sin c <= 3 root(3) / 2 [Hint: cos[(a-b)/2] <= 1 This is a Pure Maths question. I don't know how to do part (b). Please help~ 更新: To takkinl: The method is amazing. I have never thought of adding a new unknown. But as winwincy92 mentioned, inequality AM >= GM is needed to be used. And I still cannot use the inequality to solve part (b). Can you help me?
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sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2] ≦ 2sin[(A+B)/2] similarly sinC + sinD ≦ 2sin[(C+D)/2] sinA + sinB + sinC + sinD ≦ 2sin[(A+B)/2] + 2sin[(C+D)/2] 2sin[(A+B)/2] + 2sin[(C+D)/2] = 4sin[(A+B+C+D)/4]cos[A+B-C-D)/4] ≦ 4sin[(A+B+C+D)/4] ∴ sinA + sinB + sinC + sinD ≦ 4sin[(A+B+C+D)/4] let D = (A+B+C)/3 sinA + sinB + sinC + sinD = sinA + sinB + sinC + sin[(A+B+C)/3] ≦ 4sin[(A+B+C+((A+B+C)/3)/4] = 4sin[(A+B+C)/3] ∴ sinA + sinB + sinC ≦ 3sin[(A+B+C)/3] = 3sin[(pi)/3] = 3sqrt(3)/2 2009-12-04 21:14:47 補充： http://www.enotes.com/math/q-and-a/by-considering-arithmetic-mean-geometric-mean-119089 我晤識用AM>=GM。 2009-12-05 11:18:20 補充： sinA + sinB + sinC = 4(cosA/2)(cosB/2)(cosC/2) 4(cosA/2)(cosB/2)(cosC/2) ≦ 4{(1/3)[(cosA/2) + (cosB/2) + (cosC/2)]}^3 2009-12-05 11:18:25 補充： when A/2 = B/2 =C/2 , 4{(1/3)[(cosA/2) + (cosB/2) + (cosC/2)]}^3 attains max value, and GM = AM ∴ 4(cosA/2)(cosB/2)(cosC/2) ≦ 4{[cos(pi)/6 + (cos(pi)/6 + (cos(pi)/6]/3}^3 = 4(sqrt(3)/2)^3 = 3sqrt3/2 節錄某網站 neela君其他解答:
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