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 April 30th, 2007, 12:05 AM #1 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Trigonometric equation / Differential inequality (12/18/06) A solution received for the trigonometric equation (characters edited): "Find all the solutions of the equation cos4(x) + sin4(x) = 1/4. cos(4x) â‰¡ 2cos2(2x) _ 1 â‰¡ 2(2cos2x _ 1)2 _ 1 â‰¡ 4(2cos4x _ 2cos2x + 1) _ 3 â‰¡ 4(cos4x + (cos2x _ 1)2) _ 3 â‰¡ 4(cos4x + sin4x) _ 3 (Or this result can be quoted as a standard identity.) Hence the equation to be solved is equivalent to cos(4x) = -2. Hence x =1/4 Cos-1(-2) (where Cos-1 is multi-valued). You probably would prefer this in explicit complex form, involving a logarithm. cos x â‰¡ (eix + e-ix)/2, where i = sqrt(-1) and eiÏ€ = -1 Solving the quadratic equation obtainable by rearranging the identity and taking natural logarithms (or using a standard logarithmic formula for Cos-1), leads to 1/4 Cos-1(-2) = (m Ï€ +/- i ln(2 + sqrt(3)))/4, where m is any odd integer. The plus sign can be replaced by a minus sign without affecting the result. I chose plus so that the logarithm is positive." [Skipjack] A solution to the differential inequality problem (characters edited): "Let g : R -> R be a real function of class C2 (meaning g is twice differentiable everywhere). Assume also that for all x in R, 0 <= g(x) + gÂ´Â´(x). Show that for all x in R, 0 <= g(Ï€ + x) + g(x). To avoid confusion between the two uses of x, I shall prove 0 <= g(Ï€) + g(0), which implies the general result (by a simple translation). Let f(x) = g(x) + gÂ´Â´(x) Let h(x) = int( f(x) sin x dx, 0..x) = int( (sin x g(x) + sin x gÂ´Â´(x)) dx, 0..x) = [- cos x g(x) + sin x gÂ´(x)]x=x - [- cos x g(x) + sin x gÂ´(x)]x=0 = g(0) - cos x g(x) + sin x gÂ´(x) Now assume 0 <= x <= Ï€, so that 0 <= f(x) sin x and hence 0 <= h(x) also. 0 <= int(h(x) / sin^2x dx, t..n-t) (where 0 < t < Pi/2) =Int (g(0) / sin^2x - (cos x / sin^2x) g(x) + gÂ´(x) / sin x) dx, t..n-t) = [-(cos x / sin x))g(0) + g(x) / sin x]x=Pi-t - [-(cos x / sin x))g(0) + g(x) / sin x]x=t = (cos t g(0) + g(p - t) - (-cos t g(0) + g(t)) / sin t (since sin(p - t) = sin t, etc.) Multiplying by sin t and then taking the limit as t -> 0+ gives 0 <= g(Pi) + g(0)." [Skipjack]

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