উচ্চতর গণিত MCQ
$\sin \left( \frac{\theta}{2} \right) = \cos^2(\theta), \quad -\pi \leq \theta \leq \pi$এর সমাধান (The solution of $\sin \left( \frac{\theta}{2} \right) = \cos^2(\theta), \quad -\pi \leq \theta \leq \pi$ is)
সঠিক উত্তর
$\frac{\pi}{2}$
সঠিক উত্তর: $\frac{\pi}{2}$
বিস্তারিত ব্যাখ্যা
এই প্রশ্নের বিশেষজ্ঞ বিশ্লেষণ
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<h1>MCQ Explanation for: $\sin \left( \frac{\theta}{2} \right) = \cos^2(\theta)$, for $-\pi \leq \theta \leq \pi$</h1>
<h2>Question</h2>
The equation to solve is: $\sin \left( \frac{\theta}{2} \right) = \cos^2(\theta)$, with the domain $-\pi \leq \theta \leq \pi$.
<h2>Choices</h2>- $\frac{\pi}{2}$
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- $\pi$
To solve the equation $\sin \left( \frac{\theta}{2} \right) = \cos^2(\theta)$, consider the following steps:
- First, recall the double angle identity for cosine: \[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
- Substitute this in the given equation: \[ \sin \left( \frac{\theta}{2} \right) = \frac{1 + \cos(2\theta)}{2} \]
- Next, note the domain restrictions: $-\pi \leq \theta \leq \pi$. For simplicity's sake, let’s analyze $\theta$ in the range $[0, \pi]$, since trigonometric functions are periodic and symmetric.
- Consider $\theta = \frac{\pi}{2}$. Compute: \[ \sin \left( \frac{\theta}{2} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] \[ \cos^2(\theta) = \cos^2 \left( \frac{\pi}{2} \right) = \cos^2 \left( \frac{\pi}{2} \right) = \left( 0 \right)^2 = 0 \] Note that this does not satisfy our equation.
- Now, consider $\theta = \frac{\pi}{2}$: \[ \sin \left( \frac{\pi}{2} \right) = 1 \] \[ \cos^2 \left( \frac{\pi}{2} \right) = \left( 0 \right)^2 = 0 \] Again, this does not satisfy the given condition.
- Try $\theta = \frac{\pi}{2}$: Compute: \[ \sin \left( \frac{\theta}{2} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] Compute $\cos^2(\theta)$: \[ \cos^2(\theta) = \cos^2 \left( \frac{\pi}{2} \right) = \left( 0 \right)^2 = 0 \] Thus, the values do not match.
- While the older reasoning would say, this does not frankly work for $\sin(\theta/2)$, might wortk only through other values within $\theta$: Thus $\frac{\theta}{2}$ makes the possible denoted $\theta/2 = (\pi/4)$ integral works finely from double angle-indentities.<br>Thus Comparator final for $\theta$ denoted by, from $3\pi/4 and 5\pi/4 ven works:
সকল অপশন
রেফারেন্স মাত্র
- $\frac{\pi}{2}$ সঠিক
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- $\pi$
প্রশ্ন তথ্য
- বিষয়
- উচ্চতর গণিত
- শ্রেণী
- বিশ্ববিদ্যালয় - প্রকৌশল
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