Find $p,q$ to the expression $A = p (\cos^{8}x-\sin^{8}x) + 4(\cos^{6}x-2\sin^{6}x) + q\sin^{4}x$ does not depend upon $x$
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What's the source of this problem, please? – Gerry Myerson May 27 '14 at 07:53
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I think differencing will help. – pointer May 27 '14 at 07:54
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I would start by the change of variable $\sin^2x=z^2, cos^2x=1-z^2$ to turn the expression into a polynomial, and cancel the non-constant terms. – May 27 '14 at 08:05
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Hint: If $A$ doesn't depend on $x$, you should get the same value for $x = 0, \frac{\pi}2, \pi$ etc. Substitute and you should get simple equations. Of course, this does not prove the statement, it just assumes there exists such values... – Macavity May 27 '14 at 08:11
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It's a difficult problem – Jung Hyun Ran May 27 '14 at 08:29
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$$A=p((1-\sin^2x)^4- \sin^8x) + 4((1-\sin^2x)^3 - 2\sin^6x) + q\sin^4x=\\ =p(\sin^8x - 4\sin^6x + 6\sin^4x - 4\sin^2 x + 1 - \sin^8x) + 4(-\sin^6x + 3\sin^4x -3\sin^2x+1-2\sin^6x) + q\sin^4x = \\ =(-4p-12)\sin^6x + (6p+12+q)\sin^4x + (-4p-12)\sin^2x + p+4.$$
5xum
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Differentiate and equate to zero, solve for p,q. But that works only if the p,q are constants. Else apply $(a^2-b^2)$ formula repeatedly to simplify.
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