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prove that cos(6x)=32cos^6(x)-48cos^4(x)+18cos^2(x) -1

prove that cos(6x)=32cos^6(x)-48cos^4(x)+18cos^2(x) -1

Grade:12th pass

4 Answers

noogler
489 Points
5 years ago
cos6x=cos[2(3x)]=2cos2(3x)-1                                                   cos2x=2cos2x-1
                         =2[cos(3x)]2-1                                                 cos3x=4cos3x-3cosx
                         =2[4cos3x-3cosx]2-1                                       (a-b)2=a2+b2-2ab
                         =2[16cos6x+9cos2x-24cos4x]-1
                         =32cos6x-48cos4x+18cos2x-1
Aman
13 Points
3 years ago
cos6x=cos[2(3x)]=2cos2(3x)-1 cos2x=2cos2x-1 =2[cos(3x)]2-1 cos3x=4cos3x-3cosx =2[4cos3x-3cosx]2-1 (a-b)2=a2+b2-2ab =2[16cos6x+9cos2x-24cos4x]-1 =32cos6x-48cos4x+18cos2x-1
kirti
11 Points
3 years ago
R.H.S. ==>  cos(6x) = cos 2(3x) = 2cos2(3x) – 1
L.H.S. ==> 32cos6x – 48cos4x + 18cos2x – 1
             => 2(16cos6x – 24cos4x + 9cos2x) – 1
             => 2(4cos3x – 3cosx)2 – 1
             => 2cos2(3x) – 1
L.H.S. = R.H.S.
 
Hence, Proved
rishiraj
15 Points
one year ago
cos6x=cos3(2x)        {cos3a=4cos3a-3cosa)   (2x)=a
   4cos3(2x)-3cos(2x)
   4(cos2x)2-3cos(2x)
   4(cos2x)3-3  6cos2x-1 
   [4(2cos2x-1)3-6cos2x+1]
   [4(2cos2x-1)3-6cos2x+3]
   4[8cos6x-2+3×2cos2x-3×4cos4x-6cos2x+3
   32cos6x-4+24cos2x-48cos4x-6cos2x+3
   32cos2x-48cos4x+18cos2x-1

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