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Challenging!
發問:
1. a) Given that a is a root of the equation x^2-2x-2=0. Show that a^3 = 6a+4. b) Using (a), find the value of a^3+b^3. ( I just know how to find it by not using a)) 更新: 2. If in triangle ABC, (ab^2)cosA = (bc^2)cosB = (ca^2)cosC. Prove that the triangle is equilateral. 3. By using Sine and Cosine Formular, prove that a(bsinC - ccosB) = b^2-c^2
Let me finish Q2 and Q3 for you: 2) Looking into ab2cos A = bc2cos B and applying the sine formula, we have: c/sin C = b/sin B c2/sin2 C = b2/sin2 B c2 = b2sin2 C/sin2 B Also, a/b = sin A/sin B ∴ab2cos A = bc2cos B ab2cos A = bcos B(b2sin2 C/sin2 B) acos A = bsin2 Ccos B/sin2 B a/b = (sin2 Ccos B)/(sin2 Bcos A) sin A/sin B = (sin2 Ccos B)/(sin2 Bcos A) sin A cos A sin B = sin2 C cos B sin A(1/2)[sin (B+A) + sin (B-A)] = sin C(1/2)[sin (C+B) + sin (C-B)] sin Asin (π-C) + sin Asin (B-A) = sin Csin (π-A) + sin Csin (C-B) sin Asin C + sin Asin (B-A) = sin Csin A + sin Csin (C-B) sin Asin (B-A) = sin Csin (C-B) -2{cos[A+(B-A)]/2}{cos[A-(B-A)]/2} = -2{cos[C+(C-B)]/2}{cos[C-(C-B)]/2} cos(B/2)cos[A-(B/2)] = cos(B/2)cos[C-(B/2)] ∴cos(B/2) = 0 (rejected since B < π) or cos[A-(B/2)] = cos[C-(B/2)] For 0 < A,B,C < π, A-(B/2) = C-(B/2) ∴ A = C Redoing the same deductions for another pair of equality, we can obtain A = B or C = B and hence △ABC is an equilateral triangle. 3) Let me start with a(b cosC - c cosB) first and explain in the latter part. a(b cosC - c cosB) = ab cos C - ac cos B In △ABC, by the cosine formula: b2 = a2 + c2 - 2ac cos B ... (1) c2 = a2 + b2 - 2ab cos C ... (2) (1) - (2) yields: b2 - c2 = c2 - 2ac cos B - b2 + 2ab cos C 2(b2 - c2) = 2ab cos C - 2ac cos B ∴ab cos C - ac cos B = b2 - c2 (You may use a simple right-angled triangle with sides 3,4 and 5 to check against my result and the one you stated in the question.)
其他解答:
1. a) Given that a is a root of the equation x^2-2x-2=0. Show that a^3 = 6a+4. b) Using (a), find the value of a^3+b^3. ( I just know how to find it by not using a)) (a) a^2-2a-2=0 a^3 =(2a+2)a =2a^2+2a =2(2a+2)+2a =6a+4 (b) I think b is another root of the equation x^2-2x-2=0 b^3=6b+4 a^3+b^3 =6a+4+6b+4 =6(a+b)+8 =6(2)+8 (sum of the roots) =20
Challenging!
發問:
1. a) Given that a is a root of the equation x^2-2x-2=0. Show that a^3 = 6a+4. b) Using (a), find the value of a^3+b^3. ( I just know how to find it by not using a)) 更新: 2. If in triangle ABC, (ab^2)cosA = (bc^2)cosB = (ca^2)cosC. Prove that the triangle is equilateral. 3. By using Sine and Cosine Formular, prove that a(bsinC - ccosB) = b^2-c^2
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最佳解答:Let me finish Q2 and Q3 for you: 2) Looking into ab2cos A = bc2cos B and applying the sine formula, we have: c/sin C = b/sin B c2/sin2 C = b2/sin2 B c2 = b2sin2 C/sin2 B Also, a/b = sin A/sin B ∴ab2cos A = bc2cos B ab2cos A = bcos B(b2sin2 C/sin2 B) acos A = bsin2 Ccos B/sin2 B a/b = (sin2 Ccos B)/(sin2 Bcos A) sin A/sin B = (sin2 Ccos B)/(sin2 Bcos A) sin A cos A sin B = sin2 C cos B sin A(1/2)[sin (B+A) + sin (B-A)] = sin C(1/2)[sin (C+B) + sin (C-B)] sin Asin (π-C) + sin Asin (B-A) = sin Csin (π-A) + sin Csin (C-B) sin Asin C + sin Asin (B-A) = sin Csin A + sin Csin (C-B) sin Asin (B-A) = sin Csin (C-B) -2{cos[A+(B-A)]/2}{cos[A-(B-A)]/2} = -2{cos[C+(C-B)]/2}{cos[C-(C-B)]/2} cos(B/2)cos[A-(B/2)] = cos(B/2)cos[C-(B/2)] ∴cos(B/2) = 0 (rejected since B < π) or cos[A-(B/2)] = cos[C-(B/2)] For 0 < A,B,C < π, A-(B/2) = C-(B/2) ∴ A = C Redoing the same deductions for another pair of equality, we can obtain A = B or C = B and hence △ABC is an equilateral triangle. 3) Let me start with a(b cosC - c cosB) first and explain in the latter part. a(b cosC - c cosB) = ab cos C - ac cos B In △ABC, by the cosine formula: b2 = a2 + c2 - 2ac cos B ... (1) c2 = a2 + b2 - 2ab cos C ... (2) (1) - (2) yields: b2 - c2 = c2 - 2ac cos B - b2 + 2ab cos C 2(b2 - c2) = 2ab cos C - 2ac cos B ∴ab cos C - ac cos B = b2 - c2 (You may use a simple right-angled triangle with sides 3,4 and 5 to check against my result and the one you stated in the question.)
其他解答:
1. a) Given that a is a root of the equation x^2-2x-2=0. Show that a^3 = 6a+4. b) Using (a), find the value of a^3+b^3. ( I just know how to find it by not using a)) (a) a^2-2a-2=0 a^3 =(2a+2)a =2a^2+2a =2(2a+2)+2a =6a+4 (b) I think b is another root of the equation x^2-2x-2=0 b^3=6b+4 a^3+b^3 =6a+4+6b+4 =6(a+b)+8 =6(2)+8 (sum of the roots) =20
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