InterviewSolution
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InterviewSolution
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If `sinx+sin^2x+sin^3x=1` then find the value of `cos^6x-4cos^4x+8cos^2x` |
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Answer» We have , `sinx+sin^(2)x+sin^(3)x=1` `rArrsinx+sin^(3)x=1-sin^(2)x` `rArrsinx(1+sin^(2)x)=cos^(2)x` `rArrsinx(1+1-cos^(2)x)=cos^(2)x` `rArrsinx(2-cos^(2)x)=cos^(2)x` Squaring both sides ,we get `sin^(2)x(2-cos^(2)x)^(2)=cos^(4)x` `rArr(1-cos^(2)x)(4+cos^(4)x-4cos^(2)x)=cos^(4)x` `rArr4+cos^(4)x-4cos^(2)x-4cos^(2)x-cos^(6)x+4cos^(4)x=cos^(4)x` `rArrcos^(6)x-4cos^(4)x+8cos^(2)x-4=0` |
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