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) \( \sin 2 \beta \)

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) \(\sin 2\beta=-\frac{240}{289}\)