If A+B+C=π, prove that cos2A+cos2B+cos2C=-1-4cosAcosBcosC
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From question, cos2A+cos2B+cos2C ⇒ 2cos(A+B)cos(A-B)+cos2C (∵ cos(C)+cos(D)=2cos(C/2+D/2)cos(C/2-D/2)
⇒ 2cos(π-C)cos(A-B)+cos2C (∵ A+B=π-C)
⇒ -2cos(C)cos(A-B)+2 cos²C-1 (∵cosc2C=2cos²C-1)
⇒ 2cos(C)[cos(C)-cos(A-B)] -1
⇒ 2cos(C)[cos(π-(A+B)-cos(A-B)]-1 (∵C=π-(A+B))
⇒ 2cos(C)[-cos(A+B)-cos(A-B)]-1
⇒ -2cos(C)[cos(A+B)+cos(A-B)] -1
⇒ -2cos(C)[2cos(A)cos(B)] -1 (∵cos(A+B)+cos(A-B)
=2cos(A)cos(B)
⇒ -4cos(A)cos(B)cos(C) -1
∴Hence proved.
From question, cos2A+cos2B+cos2C ⇒ 2cos(A+B)cos(A-B)+cos2C (∵ cos(C)+cos(D)=2cos(C/2+D/2)cos(C/2-D/2)
⇒ 2cos(π-C)cos(A-B)+cos2C (∵ A+B=π-C)
⇒ -2cos(C)cos(A-B)+2 cos²C-1 (∵cosc2C=2cos²C-1)
⇒ 2cos(C)[cos(C)-cos(A-B)] -1
⇒ 2cos(C)[cos(π-(A+B)-cos(A-B)]-1 (∵C=π-(A+B))
⇒ 2cos(C)[-cos(A+B)-cos(A-B)]-1
⇒ -2cos(C)[cos(A+B)+cos(A-B)] -1
⇒ -2cos(C)[2cos(A)cos(B)] -1 (∵cos(A+B)+cos(A-B)
=2cos(A)cos(B)
⇒ -4cos(A)cos(B)cos(C) -1
∴Hence proved.
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