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I am working for an exam so I am unable to transcript in this exact moment the problem of which I will put the image below, my question is why are we defining the domain of theta? is it to keep the value under the root positive? yet I sill don't see how currently. (attempted transcription: The image shows the problem:

Use the substitution $x = 3 \sin\theta, 0 < \theta < π/2$, to write $\sqrt{9-x^2}$ as a trigonometric function of $\theta$

Thanks very much!

The image shows the problem:Use the substitution x = 3 sin 0, 0 < 0 < π/2, to write sqrt(9-x^2) as a trigonometric function of 0

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    $\begingroup$ They could have given you a different domain of $\theta$. It's not that important: give the main question a go first. It's good to always check if you're taking a square root of a positive thing, and handling absolute values and so and so on, but you won't know if you'll need that or not until you try the question! $\endgroup$
    – FShrike
    Commented Aug 8 at 9:41

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As @FShrike mentioned, the domain of $\theta$ is not particularly important; the chosen domain simply helps express the final solution without needing to consider the potential cases. They selected for you a specific case.

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  • $\begingroup$ what if they did not specify the domain what owed be the resulting possible cases and how to solve them? $\endgroup$
    – Federico Ruck
    Commented Aug 8 at 10:21
  • $\begingroup$ @FedericoRuck Note $\sqrt{9-x^2}$ makes sense for $-3\le x\le 3$. So $3\sin\theta$ for any value of $\theta$ "works". But there is some subtlety - because your answer might be different! Meaning no offence to Abdellah, I'm surprised you've accepted the answer. Have you managed to solve the question yet? $\endgroup$
    – FShrike
    Commented Aug 8 at 10:27
  • $\begingroup$ hi what do you mean? $\endgroup$
    – Federico Ruck
    Commented Aug 8 at 11:04
  • $\begingroup$ After the change of variable, you obtain $\sqrt{cos(\theta)^2}$. Therefore, you need to analyze the sign of the cosine function. The cases are $\theta \in [2k\pi -\frac{\pi}{2}, 2k\pi + \frac{\pi}{2}]$ (cos $\geq 0$) and $\theta \in [2k\pi + \frac{\pi}{2}, 2k\pi + \frac{3\pi}{2}]$ (cos $\leq 0$). $\endgroup$
    – Abdellah
    Commented Aug 8 at 11:04
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Let $x=3\sin\theta$, we have

$$ \begin{align} \sqrt{9-x^2} &= \sqrt{9 - 9\sin^2\theta},\\ &= \sqrt{9}\sqrt{1-\sin^2\theta},\\ &= 3 \sqrt{\cos^2\theta},\\ &= 3 |\cos\theta|. \end{align} $$ Because $\cos\theta \geq 0$ when $0 < \theta < \frac{\pi}{2}$, we can discard the absolute value; hence,

$$ \sqrt{9-x^2} = 3\cos\theta. $$

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