If \( \sin \theta=\frac{-1}{\sqrt{7}} \) where \( t>0 \) and \( \theta \in\left[90^{\circ}: 270^{\circ}\right] \) determine, using a diagram, an expression in terms of \( t \) for \( \sin \left(\theta+30^{\circ}\right) \).
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To find \(\sin(\theta + 30^\circ)\), we can use the sine addition formula: \[ \sin(\theta + 30^\circ) = \sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ \] From the given, \(\sin \theta = \frac{-1}{\sqrt{7}}\). Now, since \(\theta\) is in the range \([90^{\circ}, 270^{\circ}]\), we know \(\theta\) is in either the second or the third quadrant, where cosine is negative in the second quadrant and negative in the third quadrant. To find \(\cos \theta\), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for \(\sin \theta\): \[ \left(\frac{-1}{\sqrt{7}}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{1}{7} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{7} = \frac{6}{7} \] Taking the square root (and remembering that \(\cos \theta\) is negative in this interval): \[ \cos \theta = -\sqrt{\frac{6}{7}} = -\frac{\sqrt{6}}{\sqrt{7}} = -\frac{\sqrt{42}}{7} \] Now substituting \(\sin \theta\) and \(\cos \theta\) into the sine addition formula: Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\): \[ \sin(\theta + 30^\circ) = \left(-\frac{1}{\sqrt{7}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{42}}{7}\right) \left(\frac{1}{2}\right) \] Calculating this step-by-step: 1. \(-\frac{1}{\sqrt{7}} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2\sqrt{7}} = -\frac{\sqrt{21}}{14}\) 2. \(-\frac{\sqrt{42}}{7} \cdot \frac{1}{2} = -\frac{\sqrt{42}}{14}\) Putting it all together: \[ \sin(\theta + 30^\circ) = -\frac{\sqrt{21}}{14} - \frac{\sqrt{42}}{14} = -\frac{\sqrt{21} + \sqrt{42}}{14} \] Thus, the expression for \(\sin(\theta + 30^\circ)\) is: \[ \sin(\theta + 30^\circ) = -\frac{\sqrt{21} + \sqrt{42}}{14} \]
