1. can anyone help me with this question??

i tried to solve it like;

where;
then so on...

but it doesn't work..

would appreciate any help. thank you.

2.

3. Try (1-sin^2)^5 - sin^10 = 1. You now have a fifth degree polynomial in sin^2. Good luck!!

4. it looks like a de'moivres theorem kind of problem, but not sure how to apply it here

5. It's quite obvious what the answer is, but I get stuck here.

6. oo thank you for the replies and To:ArcaneMathematician; we continue to arrange them using trigonometric identities, but playing with say does not seem easy.

7. Originally Posted by mathman
Try (1-sin^2)^5 - sin^10 = 1. You now have a fifth degree polynomial in sin^2. Good luck!!
And if you recognize that x= 0 is a solution you can reduce this to a quartic. From there is looks a bit ugly.

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