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aiven \( \sin \beta=\frac{8}{c 7} \) where \( 90^{\circ} \leq \beta \leq 270^{\circ} \) dotormine witrad using a calechator a) \( \cos 2 \beta \) b) \( \sin 2 \beta \)

Ask by Newton Phillips. in South Africa
Mar 14,2025

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a) \( \cos 2\beta = \frac{161}{289} \) b) \( \sin 2\beta = -\frac{240}{289} \)

Solution

We are given \[ \sin \beta=\frac{8}{17} \] with \[ 90^\circ\le\beta\le270^\circ. \] Because \(\sin\beta>0\) and the angle is between \(90^\circ\) and \(270^\circ\), it follows that \(\beta\) is in the second quadrant. (In the third quadrant, sine is negative.) In the second quadrant the cosine is negative. We now use the following steps. **Step 1. Find \(\cos\beta\).** Using the Pythagorean identity \[ \sin^2\beta+\cos^2\beta=1, \] we have \[ \cos^2\beta=1-\sin^2\beta=1-\left(\frac{8}{17}\right)^2=1-\frac{64}{289}=\frac{289-64}{289}=\frac{225}{289}. \] Taking the square root, and choosing the negative sign (since cosine is negative in the second quadrant), \[ \cos\beta=-\frac{15}{17}. \] **Step 2. Compute \(\cos 2\beta\).** One version of the double-angle formula is \[ \cos2\beta=1-2\sin^2\beta. \] Substitute \(\sin\beta=\frac{8}{17}\): \[ \cos2\beta=1-2\left(\frac{8}{17}\right)^2=1-2\frac{64}{289}=1-\frac{128}{289}. \] Express 1 as \(\frac{289}{289}\): \[ \cos2\beta=\frac{289-128}{289}=\frac{161}{289}. \] **Step 3. Compute \(\sin 2\beta\).** The double-angle formula for sine is \[ \sin2\beta=2\sin\beta\cos\beta. \] Substitute the known values: \[ \sin2\beta=2\cdot\frac{8}{17}\cdot\left(-\frac{15}{17}\right)= -\frac{240}{289}. \] Thus, the answers are: a) \(\cos 2\beta=\frac{161}{289}\) b) \(\sin 2\beta=-\frac{240}{289}\)

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The Deep Dive

To determine \( \cos 2\beta \) and \( \sin 2\beta \) from the equation \( \sin \beta = \frac{8}{c 7} \), where \( 90^{\circ} \leq \beta \leq 270^{\circ} \), first we need to find the value of \( \beta \). Using a calculator, compute \( \beta = \arcsin\left(\frac{8}{c 7}\right) \). Given the range of \( \beta \), we can determine that it will be in the second or third quadrant, where sine is positive for the second quadrant and negative for the third (but here, we only consider the context that implies \( \sin \) must still yield a valid output based on the calculator's range). Once you've found \( \beta \), apply the double angle formulas: - \( \cos 2\beta = 1 - 2\sin^2\beta \) - \( \sin 2\beta = 2\sin\beta \cos\beta \) From sorting those values out through calculation, you’ll find the numbers to plug into those formulas to get your final answers. Don’t forget the relation \( \cos^2\beta + \sin^2\beta = 1 \) to determine \( \cos \beta \) as well! Time to crunch that calculator!

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