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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!

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It isn't necessarily. A correct expression is $\sqrt{a}|\cos s|$. – André Nicolas Nov 10 '12 at 18:21
up vote 3 down vote accepted

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

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I forgot that you could factor out $\sqrt{a}$ Thank you! – Lukas Arvidsson Nov 10 '12 at 18:20
@Lukas Arvidsson, the factoring is allowed for real numbers only. $\sqrt{-1}\sqrt{-1}\ne \sqrt{(-1)(-1)}=\sqrt 1$ – lab bhattacharjee Nov 10 '12 at 18:22
$\sqrt{\cdot}$ is only defined for positive real numbers, so factoring out negative parts is not only not allowed, but also not defined. – Stefan Nov 10 '12 at 18:23
@Stefan So do you mean that the solution above is not allowed? – Lukas Arvidsson Nov 10 '12 at 18:26… – lab bhattacharjee Nov 10 '12 at 18:27


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

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Thank you for your quick reply! – Lukas Arvidsson Nov 10 '12 at 18:21
You're welcome! – Stefan Nov 10 '12 at 18:22

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