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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 $\sqrt{a}|\cos s|$.2012-11-10

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$\sqrt{a-t^2}=\sqrt{a-a\sin^2s}=\sqrt a\sqrt{1-\sin^2s}=\sqrt a\cos s$ 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!2012-11-10
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    @Lukas Arvidsson, the factoring is allowed for real numbers only. $\sqrt{-1}\sqrt{-1}\ne \sqrt{(-1)(-1)}=\sqrt 1$2012-11-10
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    $\sqrt{\cdot}$ is only defined for positive real numbers, so factoring out negative parts is not only not allowed, but also not defined.2012-11-10
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    @Stefan So do you mean that the solution above is not allowed?2012-11-10
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    http://en.wikipedia.org/wiki/Imaginary_number#Multiplication_of_square_roots2012-11-10
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    No, it is fine if $a \geq 0$.2012-11-10
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    In my Analysis I course we defined $\sqrt{\cdot}$ only on the positive real numbers.2012-11-10
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    $\sqrt{1 - \sin^2(s)} = \cos s$ only if $\cos s > 0$. Without explicitly mentioning it in your analysis, I would deduct points in an exam (for math students, at least).2012-11-10
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    @TMM, have you seen $\mid s\mid <\frac{\pi}2$ in the question and the edited answer?2012-11-11
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    @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
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Hint

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