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With usual notations prove that 2(bc cosA + ac cosB + ab cosC) = a2 + b2 + c2 .

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Question

With usual notations prove that 2(bc cosA + ac cosB + ab cosC) = a2 + b2 + c2 .

Sum
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Solution

L.H.S. = 2(bc cosA + ac cosB + ab cosC)
= 2bc cosA + 2ac cosB + 2ab cosC

`= 2bc((b^2 + c^2 - a^2)/(2bc)) + 2ac((c^2 + a^2 - b^2)/(2ca)) + 2ab((a^2 + b^2 - c^2)/(2ab))`  ...[By cosine rule]

= b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2
= a2 + b2 + c2
= R.H.S.

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Chapter 3: Trigonometric Functions - Exercise 3.2 [Page 88]

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