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

noogler
489 Points
```										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
```
2 years ago
Aman
13 Points
```										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
```
6 months ago
kirti
11 Points
```										R.H.S. ==>  cos(6x) = cos 2(3x) = 2cos2(3x) – 1L.H.S. ==> 32cos6x – 48cos4x + 18cos2x – 1             => 2(16cos6x – 24cos4x + 9cos2x) – 1             => 2(4cos3x – 3cosx)2 – 1             => 2cos2(3x) – 1L.H.S. = R.H.S. Hence, Proved
```
4 months ago
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