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

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बायाँ पक्ष `=cos6x`
`=cos 3(2x)=cos3theta=cos(2theta+theta)`
`=cos 2 theta cos theta - sin 2 theta sin theta" "("माना "theta = 2x)`
`=(2cos^(2)theta-1)cos theta - 2 sin theta cos theta sin theta`
`=2cos^(3)theta-cos theta-2cos theta(1-cos^(2)theta)`
`=2cos^(3)theta-cos theta-2cos theta+2cos^(3)theta`
`=4cos^(3)theta-3cos theta =4cos^(3)2x-3 cos 2x" "(because theta=2x)`
`=4(2cos^(2)x-1)^(3)-3(2cos^(2)x-1)`
`=4(8cos^(6)x-12 cos^(4)x+6cos^(2)x-1)-(6cos^(2)x-3)`
`=32cos^(6)x-48cos^(4)x+18cos^(2)x-1=` दायाँ पक्ष

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