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If `sinx+sin^2x+sin^3x=1` then find the value of `cos^6x-4cos^4x+8cos^2x`A. 2B. 1C. 3D. 4 |
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Answer» Correct Answer - D 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.` |
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