If `sinx+sin^2x+sin^3x=1`, then find the value of `cos^6x-4cos^4x+8cos^2x`.

Answer

4

Explanation

We have,
`sinx+sin^(2)x+sin^(3)x=1`
`impliessinx+sin^(3)x=1-sin^(2)x`
`impliessinx+sin^(3)x=cos^(2)x`
`impliessinx(1+sin^(2))=cos^(2)x`
`impliessinx(2-cos^(2)x)=cos^(2)x`
`impliessin^(2)x(2-cos^(2)x)^(2)=cos^(4)x`
`implies(-1cos^(2)x)(4-4cos^(2)x+cos^(4))=cos^(4)x`
`implies4-4cos^(2)x+cos^(4)x-4cos^(2)x+4cos^(4)x+cos^(6)x=cos^(4)x`
`impliescos^(6)x-4cos^(4)x+8cos^(2)x=4.`