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Size:  10.8 KBNot exactly sure how to get rid of the power of 8 using the identies, splitting it up individually into cos doesn't seem like it'll work
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    (Original post by Purple K)
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Size:  10.8 KBNot exactly sure how to get rid of the power of 8 using the identies, splitting it up individually into cos doesn't seem like it'll work
    \sin^8 = (1-\cos^2 x)^4 then let u = \cos x and get a polynomial in u.

    What'd I do is by inspection x = \pi and then prove that (3 +\cos x)^2 is always bigger than 4 - 2\sin^8 x other than the touching at x = \pi.
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    Looking at it again, this is very easy.  2 \leq 3 + \cos  x \leq 4 whilst 2 \leq 4 - 2\sin^8 x \leq 4
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    Ahh okay so the one solution being pi
    Then just proving that the cos graph is always bigger and not touching the sin graph again
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    (Original post by Purple K)
    Ahh okay so the one solution being pi
    Then just proving that the cos graph is always bigger and not touching the sin graph again
    Yep!
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Size:  279.1 KBThe main issue as I cannot use a calculator is factorising this or using another method to prove that they only meet at pi
 
 
 
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