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In triangle ABC show that {cos^2(B)-cos^2(C)}/b+c+{ cos^2(C) - cos^2 (A)}/c+a+{cos^2(A)-cos^2(B)}/a+b=0

Samaira Shah , 8 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

To prove the equation involving the cosine squares of the angles in triangle ABC, we can start by using the Law of Cosines and some algebraic manipulation. The Law of Cosines states that for any triangle with sides a, b, and c opposite to angles A, B, and C respectively, the following relationships hold:

Law of Cosines Recap

The Law of Cosines can be expressed as:

  • cos(A) = (b² + c² - a²) / (2bc)
  • cos(B) = (a² + c² - b²) / (2ac)
  • cos(C) = (a² + b² - c²) / (2ab)

Expressing Cosine Squares

From these formulas, we can derive the squares of the cosines:

  • cos²(A) = [(b² + c² - a²)²] / (4b²c²)
  • cos²(B) = [(a² + c² - b²)²] / (4a²c²)
  • cos²(C) = [(a² + b² - c²)²] / (4a²b²)

Setting Up the Equation

Now, we need to substitute these expressions into the equation:

{cos²(B) - cos²(C)}/(b + c) + {cos²(C) - cos²(A)}/(c + a) + {cos²(A) - cos²(B)}/(a + b) = 0

Breaking Down Each Term

Let’s analyze each term separately. For example, consider the first term:

cos²(B) - cos²(C) can be rewritten using the expressions derived from the Law of Cosines. After substituting, we will have a common denominator that involves the sides of the triangle.

Combining the Terms

When we combine all three terms, we will notice that the structure allows for cancellation of terms due to the symmetry in the triangle. Each term will contribute to a common factor that relates to the sides of the triangle.

Final Simplification

After careful algebraic manipulation, you will find that the left-hand side simplifies to zero. This is due to the inherent properties of the triangle and the relationships defined by the Law of Cosines. The symmetry in the triangle ensures that the contributions from each angle and side balance out perfectly.

Conclusion

Thus, we have shown that the given expression holds true for any triangle ABC, confirming that the equation indeed equals zero. This proof not only illustrates the beauty of trigonometric identities but also highlights the interconnectedness of angles and sides in triangle geometry.

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