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(i) Using the fact that $\sin x \lt x ~\forall~ x \gt 0$, prove that for all $x \gt 0$,

$\cos x > 1- \dfrac{x^2}{2}$

(ii) Using (i) or otherwise, prove that for all $x \gt 0$,

$\sin x > x - \dfrac{x^3}{6}$

For part (i) I did the following,

Let $f(x)= \cos x, g(x)= 1- \dfrac{x^2}{2}$

After differentiation, I've got

$\sin x < x$

Subbing $x=1$, I've got

$0.84 < 1$ (proven)

Am I doing it right for part (i), if so, How should I proceed for part (ii)?

Thanks

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2 Answers 2

There is another way to prove.

$\cos x =1-2\sin^2(\frac{x}{2})>1-\frac{x^2}{2}$

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The inequality required by the OP has $1-x^2/2$ on the rhs. So this does not work. –  1015 Feb 21 '13 at 14:34
    
Oh, you're right. I edited it. –  Guillermo Feb 21 '13 at 14:35
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Don't differentiate here, integrate!

If $f,g$ are continuous on $[0,x]$ and if $f<g$ on $(0,x)$ for some $x>0$, then $$ \int_0^xf<\int_0^x g. $$

This solves both of your questions.

For the first one, $$ 1-\cos x=\int_0^x \sin t dt< \int_0^xtdt=\frac{x^2}{2}. $$

And for the second one, $$ x-\frac{x^3}{6}=\int_0^x (1-\frac{t^2}{2})dt<\int_0^x\cos tdt=\sin x $$

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