0
$\begingroup$

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!

  • 1
    It isn't necessarily. A correct expression is $\sqrt{a}|\cos s|$.2012-11-10

2 Answers 2

3

$\sqrt{a-t^2}=\sqrt{a-a\sin^2s}=\sqrt a\sqrt{1-\sin^2s}=\sqrt a\cos s$ as $\cos s>0$ as $\mid s\mid <\frac \pi 2$

  • 0
    @lab: My objection was not that it is false, but rather that it is better to be clear where you use assumptions. If you only write $P = Q = R = S$ it is not clear where you used those assumptions, and if you have used them at all. Anyway, it's fine now.2012-11-11
3

Hint

Use $\cos^2 (x) + \sin^2 (x) = 1 \forall x \in \mathbb R$.