Question

Solve $\underset{\theta \to \pi /4}{lim}\frac{2\sqrt{2}-{(\cos \theta +\sin \theta )}^3}{1-\sin 2\theta }$ is $=\frac{a}{\sqrt{2}}.$ Find a

Answer 1

$\cos \theta +\sin \theta =\sqrt{2}\cos (\theta -\pi /4)$
$\theta -\frac{\pi }{4}=t$ $\therefore \theta \frac{\pi }{4}+t$ or
$2\theta =\frac{\pi }{2}+2t$
$\therefore \sin 2\theta =\cos 2t$
When $\theta \to \pi /4,t\to 0$
$\underset{t\to 0}{lim}\frac{2\sqrt{2}-2\sqrt{2}{\cos }^{3}t}{1-\cos 2t}=\underset{t\to 0}{lim}\frac{2\sqrt{2}(1-{\cos }^{3}t)}{2{\sin }^{2}t}$
$=\underset{t\to 0}{lim}\sqrt{2}.\frac{(1-\cos t)(1+\cos t+{\cos }^{2}t)}{(1-\cos t)(1+\cos t)}$
$=\sqrt{2}.\frac{1}{1+1}(1+1+1)=\frac{3}{\sqrt{2}}.$
Ans: A