Math, asked by neerajvermag11, 1 year ago

If A+B+C=π, prove that cos2A+cos2B+cos2C=-1-4cosAcosBcosC

Answers

Answered by ellenmelloharry
42
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. 


neerajvermag11: Bhai
neerajvermag11: If A+B+C=180° , prove that sin²A+sin²B+sin²C=2(1+cosAcosBcosC)
neerajvermag11: Solve this please
ellenmelloharry: Are you taking to me
neerajvermag11: Ya
ellenmelloharry: where can I answer, if you did not post the question
neerajvermag11: I already post
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