सिद्ध कीजिए - `cos 6x = 32 cos^(6)x - 48 cos^(4)x + 18 cos^(2)x-1`.

Explanation

`L.H.S. = cos 6x`
`=cos 3(2x)`
`= 4 cos^(3)2x-3cos 2x " "[because cos 3A=4 cos^(3)A-3cos A]`
`= 4[(cos^(2)x-1)^(3)-3(2cos^(2)x-1)] " " (therefore cos 2x = 2 cos^(2) x - 1)`
`= 4[(2cos^(2)x)^(3)-(1)^(3)-3(2cos^(2)x)^(2)+3(2 cos^(2)x)]-6 cos^(2)x+3`
`=4[8cos^(6)x-1-12 cos^(4)x+6 cos^(2)x]-6 cos^(2)x + 3`
`= 32 cos^(6)x-4-48cos^(4)x+24 cos^(2)x-6cos^(2)x+3`
`= 32 cos^(6)x-48 cos^(4)x + 18 cos^(2)x - 1 = R.H.S. " "` इति सिद्धम्