Find the exact value of [tex]\sin 2 \theta[/tex] given that [tex]\sin \theta = \frac{9}{41}[/tex] and [tex]\theta[/tex] is in quadrant I.



Answer :

Given that \(\sin \theta = \frac{9}{41}\) and \(\theta\) is in the first quadrant, we need to find the exact value of \(\sin 2\theta\).

1. Recollect the identity for \(\sin 2\theta\):
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]

2. Find \(\cos \theta\) using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Given \(\sin \theta = \frac{9}{41}\), we substitute and solve for \(\cos \theta\):
[tex]\[ \left(\frac{9}{41}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left(\frac{81}{1681}\right) + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{81}{1681} \][/tex]
[tex]\[ \cos^2 \theta = \frac{1681}{1681} - \frac{81}{1681} \][/tex]
[tex]\[ \cos^2 \theta = \frac{1600}{1681} \][/tex]
Since \(\theta\) is in the first quadrant where cosine is positive:
[tex]\[ \cos \theta = \sqrt{\frac{1600}{1681}} = \frac{40}{41} \][/tex]

3. Apply the double-angle formula using \(\sin \theta = \frac{9}{41}\) and \(\cos \theta = \frac{40}{41}\):
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
[tex]\[ \sin 2\theta = 2 \left(\frac{9}{41}\right) \left(\frac{40}{41}\right) \][/tex]
[tex]\[ \sin 2\theta = 2 \left(\frac{360}{1681}\right) \][/tex]
[tex]\[ \sin 2\theta = \frac{720}{1681} \][/tex]

Thus, the exact value of [tex]\(\sin 2\theta\)[/tex] is [tex]\(\frac{720}{1681}\)[/tex].

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