Question

cos2A+cos2B+cos2C=-4cosAcosBcosC-1​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

cos2A+cos2B+cos2C=-4cosAcosBcosC-1

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

⇒ -4cos(A)cos(B)cos(C) -1

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

LETS START

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.

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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Answer 2

Answer:

LETS START

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