Hi can anyone please explain to me how you develop this? I am currently stuck...
When $t =\sqrt{a} \sin s$ how is $\sqrt{a-t^2} = \sqrt{a}\cos s$?
$|s| < \frac {\pi}{2}$
Thank you very much!
Hi can anyone please explain to me how you develop this? I am currently stuck...
When $t =\sqrt{a} \sin s$ how is $\sqrt{a-t^2} = \sqrt{a}\cos s$?
$|s| < \frac {\pi}{2}$
Thank you very much!
$\sqrt{a-t^2}=\sqrt{a-a\sin^2s}=\sqrt a\sqrt{1-\sin^2s}=\sqrt a\cos s$ as $\cos s>0$ as $\mid s\mid <\frac \pi 2$
Hint
Use $\cos^2 (x) + \sin^2 (x) = 1 \forall x \in \mathbb R$.