 My Math Forum Trigonometric equation / Differential inequality (12/18/06)

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 April 30th, 2007, 01:05 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‰ˇ 2cos2(2x) _ 1 â‰ˇ 2(2cos2x _ 1)2 _ 1 â‰ˇ 4(2cos4x _ 2cos2x + 1) _ 3 â‰ˇ 4(cos4x + (cos2x _ 1)2) _ 3 â‰ˇ 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‰ˇ (eix + e-ix)/2, where i = sqrt(-1) and eiĎ€ = -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 Ď€ +/- 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(Ď€ + x) + g(x). To avoid confusion between the two uses of x, I shall prove 0 <= g(Ď€) + 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 <= Ď€, 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] Tags differential, equation, inequality, trigonometric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fahad nasir Trigonometry 2 July 21st, 2013 05:40 AM bboy114crew Trigonometry 3 July 30th, 2011 12:07 PM wkhur Calculus 4 August 18th, 2010 07:52 PM Tomate Algebra 4 June 18th, 2010 11:51 AM mikeportnoy Algebra 7 October 4th, 2009 06:47 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      