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

Question

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

UpStudy AI · Accepted Answer

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 $$\sin \alpha = \frac{\sqrt{11}}{6}$$.
$$ \cos 2P = \frac{5}{6} $$, $$ \cos(\theta + 45^\circ) = \frac{\sqrt{7} - 1}{4} $$, and $$ \sin 2\theta = \frac{\sqrt{7}}{2} $$.

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