2004 HSC Mathematics Extension 2 Paper (1 Viewer)

Jonomyster

New Member
There are several useful hints that you can look for:
1.) These part (iii) style of questions nearly always use the previous part, so it is useful to think about how you can incorporate part (ii) into part (iii)
2.) On part (iii), there are 3 terms on LHS, but 3^2 = 9 and there is a 9 on RHS (think of n^2 from part(ii)). This is another hint to substitute
3.) In (iii), note that there is 1/sin^2(x), 1/cos^2(x) and 1/tan^2(x), this is similar to the 1/a1, 1/a2, 1/a3 from part (ii), a clue you should substitute them

Using these clues you can rearrange part (ii) into:

$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} \geq \dfrac{9}{a_1 + a_2 + a_3} \text{ Then try the substitution:}$
$\text{cosec}^2(x) + \text{sec}^2(x) + \text{cot}^2(x) \geq \dfrac{9}{\sin^2(x) + \cos^2(x) + \tan^2(x)}$

Then we solve from here:

$\dfrac{9}{\sin^2(x) + \cos^2(x) + \tan^2(x)}$
$= \dfrac{9\cos^2(x)}{\sin^2(x)\cos^2(x) + \cos^4(x) + \sin^2(x)}$
$= \dfrac{9\cos^2(x)}{\cos^2(x) (\cos^2(x) + \sin^2(x)) + \sin^2(x)}$
$= \dfrac{9\cos^2(x)}{\cos^2(x) + \sin^2(x)}$
$= 9\cos^2(x)$