Prove - (a^2)cos2B + (b^2) cos2A + 2ab cos(A - B) = c^2
Answers
1) By multiple angle trigonometric identity cos(2B) = cos²B - sin²B and cos(2A) = cos²A - sin²A
Cos(A-B) = cos(A)cos(B) + sin(A)sin(B) [compound angle identity]
2) So, a²cos(2B) + b²cos(2A) + 2abcos(A-B) = a²cos²B - a²sin²B + b²cos²A - b²sin²A + 2ab[cos(A)cos(B) + sin(A)sin(B)]
3) [a²cos²B + 2abcos(B)cos(A) + b²cos²A] - [b²sin²A - 2absin(A)sin(B) + a²sin²B]
= {acos(B) + bcos(A)}² - {bsin(A) - asin(B)}²
[Application of the algebraic identities (a+b)² and (a-b)²]
4)acos(B) + bcos(A) = c [Projection law of triangle] and asin(B) = bsin(A) [By sine law of triangle]
Hence, bsin(A) - asin(B) = 0 and the first part in the above is = c²
Substituting these in (3), we have
a²cos(2B) + b²cos(2A) + 2abcos(A-B) = c² - 0
Thus, a²cos(2B) + b²cos(2A) + 2abcos(A-B) = c² [Proved]
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