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Question
In ΔABC, A + B + C = π show that
cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C
Solution
L.H.S. = cos 2A + cos 2B + cos 2C
= `2.cos((2"A" + 2"B")/2).cos((2"A"- 2"B")/2) + cos2"C"`
= 2 . cos (A + B) . cos ( A – B) + 2 cos2 C – 1
In ΔABC, A + B + C = π
∴ A + B = π − C
∴ cos(A + B) = cos(π − C)
∴ cos (A + B) = − cos C ....(i)
∴ L.H.S. = – 2 . cos C . cos (A − B) + 2 cos2 C − 1 .....[from (i)]
= – 1 – 2 . cos C . [cos (A – B) – cos C]
= – 1 – 2 . cos C . [cos (A – B) + cos (A + B)] ....[from (i)]
= –1 – 2 . cos C . (2 cos A · cos B)
= – 1 – 4 cos A · cos B · cos C
= R.H.S.
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