(d) \( 6 \sin 2 \mathrm{P}+\sqrt{11=0} \) with \( 2 \hat{\mathrm{P}} \in\left[270^{\circ} ; 360^{\circ}\right] \), determine the value of the following without the use of a calculator and with the aid of a diagram: \( \begin{array}{ll}\text { (1) } \cos 2 \mathrm{P} & \text { (2) } \\ \text { (e) } \\ \text { (e) } \tan ^{2} \theta=7 \text { with } \theta \in\left(-90^{\circ} ;-180^{\circ}\right) \text {, determine the value of the following without the use } \\ \text { of a calculator and with the aid of a diagram: } \\ \begin{array}{ll}\text { (1) } \cos \left(\theta+45^{\circ}\right) & \text { (2) } \sin 2 \theta\end{array}\end{array} \$ . \begin{array}{l}\text { (3) } \mathrm{P}\end{array} \)

Let's break down the problem step by step. ### Part (d) We start with the equation: \[ 6 \sin 2P + \sqrt{11} = 0 \] This can be rearranged to find \(\sin 2P\): \[ 6 \sin 2P = -\sqrt{11} \] \[ \sin 2P = -\frac{\sqrt{11}}{6} \] Next, we need to determine the angle \(2P\) within the interval \([270^\circ, 360^\circ]\). Since \(\sin\) is negative in the fourth quadrant, we can find the reference angle \(\alpha\) such that: \[ \sin \alpha = \frac{\sqrt{11}}{6} \] To find \(2P\), we can express it as: \[ 2P = 360^\circ - \alpha \] ### Step 1: Find \(\cos 2P\) Using the identity \(\cos^2 x + \sin^2 x = 1\), we can find \(\cos 2P\): \[ \cos^2 2P = 1 - \sin^2 2P \] Calculating \(\sin^2 2P\): \[ \sin^2 2P = \left(-\frac{\sqrt{11}}{6}\right)^2 = \frac{11}{36} \] Now substituting into the identity: \[ \cos^2 2P = 1 - \frac{11}{36} = \frac{36}{36} - \frac{11}{36} = \frac{25}{36} \] Taking the square root: \[ \cos 2P = \pm \frac{5}{6} \] Since \(2P\) is in the fourth quadrant, \(\cos 2P\) is positive: \[ \cos 2P = \frac{5}{6} \] ### Step 2: Find \(P\) To find \(P\), we divide by 2: \[ P = \frac{2P}{2} = \frac{360^\circ - \alpha}{2} \] ### Part (e) Given: \[ \tan^2 \theta = 7 \] Taking the square root: \[ \tan \theta = \pm \sqrt{7} \] Since \(\theta\) is in the interval \((-90^\circ, -180^\circ)\), \(\tan \theta\) is positive in the third quadrant: \[ \tan \theta = -\sqrt{7} \] ### Step 1: Find \(\cos(\theta + 45^\circ)\) Using the angle addition formula: \[ \cos(\theta + 45^\circ) = \cos \theta \cos 45^\circ - \sin \theta \sin 45^\circ \] Where \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\). To find \(\cos \theta\) and \(\sin \theta\): \[ \sin \theta = -\sqrt{7}/\sqrt{8} = -\frac{\sqrt{7}}{2\sqrt{2}} = -\frac{\sqrt{14}}{4} \] \[ \cos \theta = -\frac{1}{\sqrt{8}} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4} \] Now substituting: \[ \cos(\theta + 45^\circ) = \left(-\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{14}}{4}\right)\left(\frac{\sqrt{2}}{2}\right) \] \[ = -\frac{2}{8} + \frac{\sqrt{28}}{8} = -\frac{1}{4} + \frac{\sqrt{7}}{4} = \frac{\sqrt{7} - 1}{4} \] ### Step 2: Find \(\sin 2\theta\) Using the double angle formula: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] Substituting the values: \[ \sin 2\theta = 2 \left(-\frac{\sqrt{14}}{4}\right)\left(-\frac{\sqrt{2}}{4}\right) = \frac{\sqrt{28}}{8} = \frac{\sqrt{7}}{2} \] ### Summary of Results 1. \(\cos 2P = \frac{5}{6}\) 2. \(\cos(\theta + 45^\circ) = \frac{\sqrt{7} - 1}{4}\) 3. \(\sin 2\theta = \frac{\sqrt{7}}{2}\) 4. \(P\) can be calculated as \(P = \frac{360^\circ - \alpha}{2}\) where \(\alpha\) is the reference angle corresponding to \(\sin \alpha = \frac{\sqrt{11}}{6}\).