document.write( "Question 1022487: Given cosx+cos^2x=1
\n" ); document.write( "Find sin^2x+sin^6x+sin^8x=? \n" ); document.write( "

Algebra.Com's Answer #638159 by robertb(5830) $\"\"$ $\"About$

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$\"sin%5E2%28x%29%2Bsin%5E6%28x%29%2Bsin%5E8%28x%29+=+rho\"$
\n" ); document.write( "==> $\"1%2Bsin%5E2%28x%29%2Bsin%5E6%28x%29%281%2Bsin%5E2%28x%29%29+=+1+%2B+rho\"$
\n" ); document.write( "<==> $\"%281%2Bsin%5E2%28x%29%29%281%2Bsin%5E6%28x%29%29=+1+%2B+rho\"$ \r
\n" ); document.write( "\n" ); document.write( "Now $\"cosx%2Bcos%5E2%28x%29+=+1\"$ ==> $\"cosx+=+1-cos%5E2%28x%29+=+sin%5E2%28x%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "==> $\"%281%2Bcosx%29%281%2Bcos%5E3%28x%29%29=+1+%2B+rho\"$ after direct substitution.\r
\n" ); document.write( "\n" ); document.write( "<==> $\"1%2Bcos%5E3%28x%29+%2Bcosx+%2B+cos%5E4%28x%29+=+1%2Brho\"$
\n" ); document.write( "==> $\"cosx+%2Bcos%5E3%28x%29+%2B+cos%5E4%28x%29+=+rho\"$ \r
\n" ); document.write( "\n" ); document.write( "Now $\"cosx%2Bcos%5E2%28x%29+=+1\"$ ==> $\"cos%5E3%28x%29%2Bcos%5E4%28x%29+=+cos%5E2%28x%29\"$ after multiplying both sides by $\"cos%5E2%28x%29\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"cosx%2Bcos%5E2%28x%29+=+rho\"$ after direct substitution.\r
\n" ); document.write( "\n" ); document.write( "==> $\"rho+=+1\"$ because of the given.\r
\n" ); document.write( "\n" ); document.write( "Therefore, $\"sin%5E2%28x%29%2Bsin%5E6%28x%29%2Bsin%5E8%28x%29+=+highlight%281%29\"$ \n" ); document.write( "

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