SOLUTION: Given cosx+cos^2x=1 Find sin^2x+sin^6x+sin^8x=?

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Question 1022487: Given cosx+cos^2x=1
Find sin^2x+sin^6x+sin^8x=?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
sin%5E2%28x%29%2Bsin%5E6%28x%29%2Bsin%5E8%28x%29+=+rho
==> 1%2Bsin%5E2%28x%29%2Bsin%5E6%28x%29%281%2Bsin%5E2%28x%29%29+=+1+%2B+rho
<==> %281%2Bsin%5E2%28x%29%29%281%2Bsin%5E6%28x%29%29=+1+%2B+rho
Now cosx%2Bcos%5E2%28x%29+=+1==> cosx+=+1-cos%5E2%28x%29+=+sin%5E2%28x%29.
==> %281%2Bcosx%29%281%2Bcos%5E3%28x%29%29=+1+%2B+rho after direct substitution.
<==> 1%2Bcos%5E3%28x%29+%2Bcosx+%2B+cos%5E4%28x%29+=+1%2Brho
==> cosx+%2Bcos%5E3%28x%29+%2B+cos%5E4%28x%29+=+rho
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
==> cosx%2Bcos%5E2%28x%29+=+rho after direct substitution.
==> rho+=+1 because of the given.
Therefore, sin%5E2%28x%29%2Bsin%5E6%28x%29%2Bsin%5E8%28x%29+=+highlight%281%29