QUESTION 5 5.1 Prove, without the use of a calculator, that, $$ \cos 75^{\circ}+\cos 15^{\circ}=\frac{\sqrt{6}}{2} $$ 5.2 Determine the general solution of: $$ 1+4 \sin ^{2} x-5 \sin x+\cos 2 x=0 $$ 5.3 Prove the identity $$ \frac{\sin 2 A}{1+\cos 2 A}=\tan A $$ 5.4 Simplify $$ \frac{\sin \left(450^{\circ}-x\right) \tan \left(x-180^{\circ}\right) \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} \sin (-x)} $$

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Verify the identity by following steps:
- step0: Verify:
  $$\cos\left(75\right)+\cos\left(15\right)=\frac{\sqrt{6}}{2}$$
- step1: Verify the identity:
  $$\textrm{false}$$
Solve the equation $$ 1 + 4\sin^2(x) - 5\sin(x) + \cos(2x) = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$1+4\sin^{2}\left(x\right)-5\sin\left(x\right)+\cos\left(2x\right)=0$$
- step1: Rewrite the expression:
  $$1+4\sin^{2}\left(x\right)-5\sin\left(x\right)+2\cos^{2}\left(x\right)-1=0$$
- step2: Expand the expression:
  $$4\sin^{2}\left(x\right)-5\sin\left(x\right)+2\cos^{2}\left(x\right)=0$$
- step3: Rewrite the expression:
  $$2\sin^{2}\left(x\right)-5\sin\left(x\right)+2=0$$
- step4: Factor the expression:
  $$\left(\sin\left(x\right)-2\right)\left(2\sin\left(x\right)-1\right)=0$$
- step5: Separate into possible cases:
  $$\begin{align}&\sin\left(x\right)-2=0\&2\sin\left(x\right)-1=0\end{align}$$
- step6: Solve the equation:
  $$\begin{align}&x \notin \mathbb{R}\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step7: Find the union:
  $$x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Determine whether the expression $$ \sin(2A)/(1 + \cos(2A)) = \tan(A) $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\frac{\sin\left(2A\right)}{\left(1+\cos\left(2A\right)\right)}=\tan\left(A\right)$$
- step1: Choose a side to work on:
  $$\tan\left(A\right)=\tan\left(A\right)$$
- step2: Verify the identity:
  $$\textrm{true}$$
Let's go through each part of the problem step by step.

### 5.1 Prove that $$\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$$

To prove this, we can use the cosine addition formula:
$$
\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)
$$
Let $$A = 75^{\circ}$$ and $$B = 15^{\circ}$$:
- $$A + B = 90^{\circ}$$
- $$A - B = 60^{\circ}$$

Now substituting into the formula:
$$
\cos 75^{\circ} + \cos 15^{\circ} = 2 \cos\left(\frac{90^{\circ}}{2}\right) \cos\left(\frac{60^{\circ}}{2}\right) = 2 \cos(45^{\circ}) \cos(30^{\circ})
$$
We know:
- $$\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$$
- $$\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$$

Thus:
$$
\cos 75^{\circ} + \cos 15^{\circ} = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}
$$
This proves that $$\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$$.

### 5.2 Determine the general solution of $$1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$$

Using the identity $$\cos 2x = 1 - 2\sin^2 x$$, we can rewrite the equation:
$$
1 + 4 \sin^2 x - 5 \sin x + (1 - 2 \sin^2 x) = 0
$$
This simplifies to:
$$
2 + 2 \sin^2 x - 5 \sin x = 0
$$
Dividing through by 2:
$$
1 + \sin^2 x - \frac{5}{2} \sin x = 0
$$
Rearranging gives:
$$
\sin^2 x - \frac{5}{2} \sin x + 1 = 0
$$
Let $$y = \sin x$$:
$$
y^2 - \frac{5}{2}y + 1 = 0
$$
Using the quadratic formula:
$$
y = \frac{\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4 \cdot 1}}{2}
$$
Calculating the discriminant:
$$
\left(\frac{5}{2}\right)^2 - 4 = \frac{25}{4} - 4 = \frac{25}{4} - \frac{16}{4} = \frac{9}{4}
$$
Thus:
$$
y = \frac{\frac{5}{2} \pm \frac{3}{2}}{2} = \frac{8/2}{2} \text{ or } \frac{2/2}{2} = 2 \text{ or } \frac{1}{2}
$$
Since $$\sin x$$ must be between -1 and 1, we discard $$y = 2$$ and keep $$y = \frac{1}{2}$$.

The general solution for $$\sin x = \frac{1}{2}$$ is:
$$
x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}
$$

### 5.3 Prove the identity $$\frac{\sin 2A}{1 + \cos 2A} = \tan A$$

Using the double angle formulas:
$$
\sin 2A = 2 \sin A \cos A \quad \text{and} \quad \cos 2A = 2 \cos^2 A - 1
$$
Thus:
$$
1 + \cos 2A = 1 + (2 \cos^2 A - 1) = 2 \cos^2 A
$$
Now substituting into the left side:
$$
\frac{\sin 2A}{1 + \cos 2A} = \frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A} = \tan A
$$
This proves the identity.

### 5.4 Simplify $$\frac{\sin(450^{\circ} - x) \tan(x - 180^{\circ}) \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} \sin(-x)}$$

1. **Simplifying $$\sin(450^{\circ} - x)$$**:
   $$
   \sin(450^{\circ} - x) = \sin(90^{\circ} - x) = \cos x
   $$

2. **Simplifying $$\tan(x - 180^{\circ})$$**:
   $$
   \tan(x - 180^{\circ}) = \tan x
   $$

3. **Simplifying $$\sin(-x)$$**:
   $$
   \sin(-x) = -\sin x
   $$

Now substituting these into the expression:
$$
\frac{\cos x \tan x \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} (-\sin x)} = -\frac{\cos x \cdot \frac{\sin x}{\cos x} \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}} = -\frac{\sin x \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}}
$$

Thus, the simplified expression is:
$$
-\frac{\sin x \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}}
$$

This concludes the solutions for all parts of the problem.
1. $$\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$$ 2. The general solution is $$x = \frac{\pi}{6} + 2k\pi$$ or $$x = \frac{5\pi}{6} + 2k\pi$$, where $$k$$ is an integer. 3. The identity $$\frac{\sin 2A}{1 + \cos 2A} = \tan A$$ is proven. 4. The simplified expression is $$-\frac{\sin x \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}}$$.

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