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Evaluate, without using a calculator: \( X \) \[ \sum_{y=43^{\circ}}^{47^{\circ}} \sin ^{2} y \]

Ask by Cruz Bond. in South Africa
Mar 12,2025

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Answer

The sum \( X = \frac{5}{2} \).

Solution

We start with \[ X=\sum_{y=43^{\circ}}^{47^{\circ}} \sin ^{2} y = \sin^2 43^\circ + \sin^2 44^\circ + \sin^2 45^\circ + \sin^2 46^\circ + \sin^2 47^\circ. \] **Step 1. Use the complementary angle identity** Recall that \[ \sin^2 \theta = \cos^2 (90^\circ - \theta). \] Thus, notice that: \[ \sin^2 43^\circ = \cos^2 47^\circ \quad \text{and} \quad \sin^2 47^\circ = \cos^2 43^\circ. \] Similarly, \[ \sin^2 44^\circ = \cos^2 46^\circ \quad \text{and} \quad \sin^2 46^\circ = \cos^2 44^\circ. \] **Step 2. Group the complementary terms** Grouping the terms, we have: - For \(43^\circ\) and \(47^\circ\): \[ \sin^2 43^\circ + \sin^2 47^\circ = \cos^2 47^\circ + \cos^2 43^\circ. \] But observe that \(\cos^2 47^\circ + \sin^2 47^\circ = 1\) if the argument were the same. Instead, we apply the identity in a different way: since we already have the pair \(\sin^2 43^\circ\) and \(\cos^2 43^\circ\) if we use \(\sin^2 47^\circ = \cos^2 43^\circ\), so \[ \sin^2 43^\circ + \sin^2 47^\circ = \sin^2 43^\circ + \cos^2 43^\circ = 1. \] - For \(44^\circ\) and \(46^\circ\): \[ \sin^2 44^\circ + \sin^2 46^\circ = \sin^2 44^\circ + \cos^2 44^\circ = 1. \] - The remaining term is: \[ \sin^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}. \] **Step 3. Sum up the values** Adding the groups together: \[ X = 1 + 1 + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2}. \] Thus, the evaluated sum is: \[ \frac{5}{2}. \]

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To evaluate the sum \( X = \sum_{y=43^{\circ}}^{47^{\circ}} \sin^2 y \), we will compute \( \sin^2 43^{\circ} \), \( \sin^2 44^{\circ} \), \( \sin^2 45^{\circ} \), \( \sin^2 46^{\circ} \), and \( \sin^2 47^{\circ} \). First, we recall that \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) from the double angle identity. Therefore, we can rewrite each term as follows: \[ \sin^2 43^{\circ} = \frac{1 - \cos(86^{\circ})}{2} \] \[ \sin^2 44^{\circ} = \frac{1 - \cos(88^{\circ})}{2} \] \[ \sin^2 45^{\circ} = \frac{1 - \cos(90^{\circ})}{2} = \frac{1}{2} \] \[ \sin^2 46^{\circ} = \frac{1 - \cos(92^{\circ})}{2} \] \[ \sin^2 47^{\circ} = \frac{1 - \cos(94^{\circ})}{2} \] Now we can compute the total sum: \[ X = \frac{1 - \cos(86^{\circ})}{2} + \frac{1 - \cos(88^{\circ})}{2} + \frac{1}{2} + \frac{1 - \cos(92^{\circ})}{2} + \frac{1 - \cos(94^{\circ})}{2} \] Combining these, we have: \[ X = \frac{5}{2} - \frac{\cos(86^{\circ}) + \cos(88^{\circ}) + \cos(92^{\circ}) + \cos(94^{\circ})}{2} \] Next, we use the fact that \( \cos(90^{\circ} + x) = -\sin(x) \) to simplify: \[ \cos(86^{\circ}) = \sin(4^{\circ},\quad \cos(88^{\circ}) = \sin(2^{\circ}),\quad \cos(92^{\circ}) = \sin(2^{\circ}),\quad \cos(94^{\circ}) = \sin(4^{\circ}) \] These provide us with the equalities: \[ \cos(86^{\circ}) + \cos(94^{\circ}) = \sin(4^{\circ}) + \sin(4^{\circ}) = 2\sin(4^{\circ}) \] \[ \cos(88^{\circ}) + \cos(92^{\circ}) = \sin(2^{\circ}) + \sin(2^{\circ}) = 2\sin(2^{\circ}) \] Thus, we have: \[ X = \frac{5}{2} - \frac{2\sin(4^{\circ}) + 2\sin(2^{\circ})}{2} = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ})) \] Finally, without precise values for \( \sin(2^{\circ}) \) and \( \sin(4^{\circ}) \), we conclude that the sum \( \sum_{y=43^{\circ}}^{47^{\circ}} \sin^2 y \) can be expressed as: \[ X = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ})). \] While we do not have exact numerical answers, we summarize: \[ \boxed{X = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ}))} \]

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