# Expressing in different form and finding a maximum? See picture below (7 marks)

Apr 11, 2017

(a)

$\sqrt{15} \sin \left(2 x\right) + \sqrt{5} \cos \left(2 x\right)$

$= \sqrt{20} \left(\frac{\sqrt{15}}{\sqrt{20}} \sin \left(2 x\right) + \frac{\sqrt{5}}{\sqrt{20}} \cos \left(2 x\right)\right)$

[Taking $\frac{\sqrt{15}}{\sqrt{20}} = \cos \alpha \mathmr{and} \frac{\sqrt{5}}{\sqrt{20}} = \sin \alpha$]

$= \sqrt{20} \left(\cos \alpha \sin \left(2 x\right) + \sin \alpha \cos \left(2 x\right)\right)$

$= 2 \sqrt{5} \sin \left(2 x + \alpha\right)$

Where $\alpha = {\tan}^{-} 1 \left(\frac{\sqrt{5}}{\sqrt{15}}\right) = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

(b)

(i)
$f \left(x\right) = \frac{2}{5 + \sqrt{15} \sin \left(2 x\right) + \sqrt{\cos} \left(2 x\right)}$

$\implies f \left(x\right) = \frac{2}{5 + 2 \sqrt{5} \sin \left(2 x + \alpha\right)}$

The value of $f \left(x\right)$ will be maximum when $\sin \left(2 x + \alpha\right)$ is minimum i.e. $\sin \left(2 x + \alpha\right) = - 1$

So

$f {\left(x\right)}_{\text{max}} = \frac{2}{5 - 2 \sqrt{5}}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{2 \left(5 + 2 \sqrt{5}\right)}{\left(5 - 2 \sqrt{5}\right) \left(5 + 2 \sqrt{5}\right)}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{10 + 4 \sqrt{5}}{25 - 20} = 2 + \frac{4}{5} \sqrt{5}$

(ii)Here $f \left(x\right)$ is maximum when

$\sin \left(2 x + \alpha\right) = - 1$

$\implies \sin \left(2 x + \frac{\pi}{6}\right) = \sin \left(\frac{3 \pi}{2}\right)$

$\implies 2 x + \frac{\pi}{6} = \frac{3 \pi}{2}$

$\implies 2 x = \frac{3 \pi}{2} - \frac{\pi}{6} = \frac{4 \pi}{3}$

$\implies x = \frac{2 \pi}{3}$