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Question
Prove the following:
If A + B + C = `(3pi)/2`, then cos 2A + cos 2B + cos 2C = 1 − 4 sin A sin B sin C
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Solution
In ΔABC,
A + B + C = `(3pi)/2`
∴ A + B = `(3pi)/2 - "C"`
∴ cos (A + B) = `cos((3pi)/2 - "C")`
= – sin C ...(i)
L.H.S. = cos 2A + cos 2B + cos 2C
= `2cos((2"A" + 2"B")/2)cos((2"A" - 2"B")/2) + cos2"C"`
= 2 cos (A + B) cos (A – B) + cos 2C
= 2 (– sin C) cos (A – B) + 1 – 2 sin2C …[From (i)]
= 1 – 2 sin C [cos (A – B) + sin C]
= 1 – 2 sin C {cos (A – B) – cos (A + B)} …[From (i)]
= `1 - 2 sin "C" xx 2sin(("A" - "B" + "A" + "B")/2)sin(("A" + "B" - "A" + "B")/2)`
= 1 – 2 sin C (2 sin A sin B)
= 1 – 4 sin A sin B sin C
= R.H.S.
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