\( \theta_{1}=\theta_{2}=\theta_{2}=2 \) If \( 1+\sin \theta=9 \cos \theta \) and \( 0^{\circ}<\theta<90^{\circ} \), then analyse and assess the value of \( 5(1-\cos \theta) \). \( 1+\sin \theta=9 \cos \theta \)
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To analyze the equation \( 1+\sin \theta=9 \cos \theta \), we can rearrange it to \( \sin \theta = 9 \cos \theta - 1 \). To further explore \( 5(1-\cos \theta) \), we start by using the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), and substituting for \(\sin \theta\). First, substitute \( \sin \theta \): \[ (9 \cos \theta - 1)^2 + \cos^2 \theta = 1 \] Expanding gives: \[ 81 \cos^2 \theta - 18 \cos \theta + 1 + \cos^2 \theta = 1 \] This simplifies to: \[ 82 \cos^2 \theta - 18 \cos \theta = 0 \] Factoring out \( \cos \theta \): \[ \cos \theta (82 \cos \theta - 18) = 0 \] Since \( \cos \theta \neq 0 \) for \( 0^{\circ}<\theta<90^{\circ} \), we set: \[ 82 \cos \theta - 18 = 0 \implies \cos \theta = \frac{18}{82} = \frac{9}{41} \] Next, we need \( \sin \theta \): \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{9}{41}\right)^2 \] Calculating \( \left(\frac{9}{41}\right)^2 = \frac{81}{1681} \): \[ \sin^2 \theta = \frac{1681 - 81}{1681} = \frac{1600}{1681} \] Thus, \[ \sin \theta = \frac{40}{41} \] Now, substituting \( \cos \theta \) into \( 5(1-\cos \theta) \): \[ 5(1 - \cos \theta) = 5 \left( 1 - \frac{9}{41} \right) = 5 \left( \frac{41 - 9}{41} \right) = 5 \left( \frac{32}{41} \right) = \frac{160}{41} \] Thus, the value of \( 5(1-\cos \theta) \) is \( \frac{160}{41} \).
