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Question: If \(A + B + C = \pi,\) then \(\cos{}2A + \cos{}2B + \cos{}2C =\)...

If A+B+C=π,A + B + C = \pi, then cos⁡2A+cos⁡2B+cos⁡2C=\cos{}2A + \cos{}2B + \cos{}2C =

A

1+4cos⁡Acos⁡Bsin⁡C1 + 4\cos A\cos B\sin C

B

−1+4sin⁡Asin⁡Bcos⁡C- 1 + 4\sin A\sin B\cos C

C

−1−4cos⁡A⥂⥂cos⁡Bcos⁡C- 1 - 4\cos A ⥂ ⥂ \cos B\cos C

D

None of these

Answer

−1−4cos⁡A⥂⥂cos⁡Bcos⁡C- 1 - 4\cos A ⥂ ⥂ \cos B\cos C

Explanation

Solution

L.H.S. =2cos⁡(A+B)cos⁡(A−B)+(2cos⁡2C−1)= 2\cos(A + B)\cos(A - B) + (2\cos^{2}C - 1)

=−1−2cos⁡Ccos⁡(A−B)+2cos⁡2C= - 1 - 2\cos C\cos(A - B) + 2\cos^{2}C

=−1−2cos⁡C[cos⁡(A−B)+cos⁡(A+B)]= - 1 - 2\cos C\lbrack\cos(A - B) + \cos(A + B)\rbrack

=−1−4cos⁡Acos⁡Bcos⁡C= - 1 - 4\cos A\cos B\cos C.