MHB How is the Domain of a Trigonometric Expression Determined?

[tmt1]

I have the expression

$$\sqrt{ a ^2 - x^2}$$

using trig substitution (with $x = asin\theta$), I get$$\sqrt{ a ^2 - a^2sin^2\theta}$$

which gets simplified to

$ a \sqrt{ cos^2\theta}$ and then $ a cos \theta$

for $$- \frac{\pi}{2} \le 0 \le \frac{\pi}{2} $$

what I don't get is domain of the expression. I understand that $ cos \theta$ must be greater than 0 because of $ a \sqrt{ cos^2\theta}$, but how does that get simplified to $- \frac{\pi}{2} \le 0 \le \frac{\pi}{2} $?

Thanks

 

[I like Serena]

Hi tmt! ;)

I have the expression

$$\sqrt{ a ^2 - x^2}$$

using trig substitution (with $x = asin\theta$), I get

For $x = a\sin\theta$ to be properly defined (bijective), we need to restrict the domain of $\theta$.
Without loss of generality, we can choose $- \frac{\pi}{2} \le 0 \le \frac{\pi}{2} $, which is the domain of $\arcsin$.

$$\sqrt{ a ^2 - a^2sin^2\theta}$$

which gets simplified to

$ a \sqrt{ cos^2\theta}$ and then $ a cos \theta$

for $$- \frac{\pi}{2} \le 0 \le \frac{\pi}{2} $$

what I don't get is domain of the expression. I understand that $ cos \theta$ must be greater than 0 because of $ a \sqrt{ cos^2\theta}$, but how does that get simplified to $- \frac{\pi}{2} \le 0 \le \frac{\pi}{2} $?

Actually, we can't tell if $\cos \theta$ is greater than 0 or not.
We should consider the case that it's not.
Btw, there's also an assumption in there that $a \ge 0$.
Is that given?

Properly we have:
$$\sqrt{ a ^2 - a^2\sin^2\theta}
= \sqrt{a^2(1-\sin^2\theta)}
= \sqrt{a^2 \cos^2\theta}
= |a \cos\theta|
$$