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Little Lemma: $\frac{x^{2n}}{(2n)!}>\frac{x^{2n+2}}{(2n+2)!}$ where $x \in (0,\frac{\pi}{2})$ and $n \in N$ By multiplying we get: $(2n+1)(2n+2)x^{2n}>x^{2n+2}$ $(2n+1)(2n+2)>x^2$ which is obvious because $x \in (0,\frac{\pi}{2})$ and $n \in N$ Coming back to our problem: $cos(x) = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{6!}+...<1-\frac{x^2}{2}+\frac{x^4}{16}$ $-\frac{x^4}{48}-\frac{x^6}{6!}+...<0$ which is true using our lemma

Since $\sin x<x<\frac{3}{2}x\ \left(0<x<\frac{\pi}{2}\right),$ we have $\int_0^x \sin x\ dx<\int_0^x \frac{3}{2}x\ dx \Longleftrightarrow \left[-\cos x\right]_0^x <\left[\frac{3}{4}x^2\right]_0^x \Longleftrightarrow 1-\cos x<\frac{3}{4}x^2.$ $\therefore \int_0^x (1-\cos x)\ dx<\int_0^x\frac{3}{4}x^2\ dx \Longleftrightarrow \left[x-\sin x\right]_0^x <\left[\frac{1}{4}x^3\right]_0^x$ $\Longleftrightarrow x-\sin x<\frac{1}{4}x^3.$ $\therefore \int_0^x (x-\sin x)dx<\int_0^x \frac{1}{4}x^3\ dx \Longleftrightarrow \left[\frac{1}{2}x^2+\cos x\right]_0^x <\left[\frac{1}{16}x^4\right]_0^x$ $\Longleftrightarrow \frac{1}{2}x^2+\cos x-1<\frac{1}{16}x^4.$ $\therefore \cos x<1-\frac{1}{2}x^2+\frac{1}{16}x^4\ \left(0<x<\frac{\pi}{2}\right)\ Q.E.D.$

yo yo yo man i apreceate yours solution it a very good one but we can apply the development limited

lastquincy wrote: yo yo yo man i apreceate yours solution it a very good one but we can apply the development limited Thanks. but, what is the development limited? kunny

is that a common problem solving technique (in upper math...kindof...) to take an expression, then make an integral from 0 to x of both sides, then equate , then repeat? or is it a really original way to attack a problem?? it seems like it would only work (practicably) on sin/cosine/(?tangent?etc.) or natural logs (maybe?). Anyway, is it useful or does it only work on this problem.

orl wrote: Prove the trigonometric inequality $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{16},$ when $\dsp x \in \left(0, \frac{\pi}{2} \right).$

Using 4-derrivative is also avallable

me@home wrote: is that a common problem solving technique (in upper math...kindof...) to take an expression, then make an integral from 0 to x of both sides, then equate , then repeat? or is it a really original way to attack a problem?? it seems like it would only work (practicably) on sin/cosine/(?tangent?etc.) or natural logs (maybe?). Anyway, is it useful or does it only work on this problem. How about this one?

ok, so I guess it is fairly common when you're stuck. also what do people have to say about using calculus to solve some AIME problems without really knowing the "proper" solution?