5.2 If $$ \cos 12^{\circ}=m $$, write down the following in terms of m : $$ 5.2 .1 \quad \sin 168^{\circ} $$ 5.2.2 $$ \cos 192^{\circ} $$ 5.2.3 $$ \tan 372^{\circ} $$ Please to 5 Gr. 11/ Mathematics First Push March QUESTION 6 6.1 Simplify: $$ \sin ^{2}\left(180^{\circ}+x\right)-\cos (-x) \cdot \tan \left(90^{\circ}-x\right) \cdot \sin \left(540^{\circ}+x\right) $$ 6.2 Prove the identity: $$ \frac{\sin \left(x-90^{\circ}\right) \cdot \tan (-x)}{\cos \left(450^{\circ}+x\right)}=-1 $$ 6.3 Determine the general solution for: $$ 3 \sin x=2 \cos ^{2} x $$

Question

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Let's solve the problems step by step.

### Problem 5.2

Given that $$ \cos 12^{\circ} = m $$, we need to express the following in terms of $$ m $$.

#### 5.2.1 $$ \sin 168^{\circ} $$

Using the sine angle subtraction identity:
$$
\sin(168^{\circ}) = \sin(180^{\circ} - 12^{\circ}) = \sin(12^{\circ})
$$
Using the identity $$ \sin(12^{\circ}) = \sqrt{1 - \cos^2(12^{\circ})} $$:
$$
\sin(12^{\circ}) = \sqrt{1 - m^2}
$$
Thus,
$$
\sin(168^{\circ}) = \sqrt{1 - m^2}
$$

#### 5.2.2 $$ \cos 192^{\circ} $$

Using the cosine angle addition identity:
$$
\cos(192^{\circ}) = \cos(180^{\circ} + 12^{\circ}) = -\cos(12^{\circ}) = -m
$$

#### 5.2.3 $$ \tan 372^{\circ} $$

First, reduce $$ 372^{\circ} $$ to an equivalent angle within $$ 0^{\circ} $$ to $$ 360^{\circ} $$:
$$
372^{\circ} - 360^{\circ} = 12^{\circ}
$$
Thus,
$$
\tan(372^{\circ}) = \tan(12^{\circ}) = \frac{\sin(12^{\circ})}{\cos(12^{\circ})} = \frac{\sqrt{1 - m^2}}{m}
$$

### Summary of 5.2
1. $$ \sin 168^{\circ} = \sqrt{1 - m^2} $$
2. $$ \cos 192^{\circ} = -m $$
3. $$ \tan 372^{\circ} = \frac{\sqrt{1 - m^2}}{m} $$

---

### Problem 6.1

We need to simplify:
$$
\sin^{2}(180^{\circ} + x) - \cos(-x) \cdot \tan(90^{\circ} - x) \cdot \sin(540^{\circ} + x)
$$

1. **Simplify $$ \sin^{2}(180^{\circ} + x) $$**:
   $$
   \sin(180^{\circ} + x) = -\sin(x) \implies \sin^{2}(180^{\circ} + x) = \sin^{2}(x)
   $$

2. **Simplify $$ \cos(-x) $$**:
   $$
   \cos(-x) = \cos(x)
   $$

3. **Simplify $$ \tan(90^{\circ} - x) $$**:
   $$
   \tan(90^{\circ} - x) = \cot(x) = \frac{\cos(x)}{\sin(x)}
   $$

4. **Simplify $$ \sin(540^{\circ} + x) $$**:
   $$
   540^{\circ} = 360^{\circ} + 180^{\circ} \implies \sin(540^{\circ} + x) = -\sin(x)
   $$

Now substituting these into the expression:
$$
\sin^{2}(x) - \cos(x) \cdot \frac{\cos(x)}{\sin(x)} \cdot (-\sin(x))
$$
This simplifies to:
$$
\sin^{2}(x) + \cos^{2}(x) = 1
$$

### Problem 6.2

We need to prove the identity:
$$
\frac{\sin(x - 90^{\circ}) \cdot \tan(-x)}{\cos(450^{\circ} + x)} = -1
$$

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

2. **Simplify $$ \tan(-x) $$**:
   $$
   \tan(-x) = -\tan(x)
   $$

3. **Simplify $$ \cos(450^{\circ} + x) $$**:
   $$
   450^{\circ} = 360^{\circ} + 90^{\circ} \implies \cos(450^{\circ} + x) = -\sin(x)
   $$

Now substituting these into the expression:
$$
\frac{-\cos(x) \cdot (-\tan(x))}{-\sin(x)} = \frac{\cos(x) \cdot \tan(x)}{\sin(x)} = \frac{\cos(x) \cdot \frac{\sin(x)}{\cos(x)}}{\sin(x)} = 1
$$
Thus, the left-hand side equals $$-1$$.

### Problem 6.3

We need to determine the general solution for:
$$
3 \sin x = 2 \cos^{2} x
$$

Using the identity $$ \cos^{2} x = 1 - \sin^{2} x $$:
$$
3 \sin x = 2(1 - \sin^{2} x)
$$
Rearranging gives:
$$
2 \sin^{2} x + 3 \sin x - 2 = 0
$$

Let $$ y = \sin x $$:
$$
2y^{2} + 3y - 2 = 0
$$

Now we can solve this quadratic equation using the quadratic formula:
$$
y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^{2} - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}
$$
Calculating the discriminant:
$$
b^{2} - 4ac = 9 + 16 = 25
$$
Thus,
$$
y = \frac{-3 \pm 5}{4}
$$
Calculating the two possible values:
1. $$ y = \frac{2}{4} = \frac{1}{2} $$
2. $$ y = \frac{-8}{4} = -2 $$ (not valid since $$ \sin x $$ must be between -1 and 1)

Now, solving $$ \sin x = \frac{1}{2} $$:
$$
x = 30^{\circ} + 360^{\circ}k \quad \text{or} \quad x = 150^{\circ} + 360^{\circ}k \quad (k \in \mathbb{Z})
$$

**Problem 5.2 Solutions:** 1. $$ \sin 168^{\circ} = \sqrt{1 - m^2} $$ 2. $$ \cos 192^{\circ} = -m $$ 3. $$ \tan 372^{\circ} = \frac{\sqrt{1 - m^2}}{m} $$ **Problem 6.1 Simplification:** $$ \sin^{2}(180^{\circ} + x) - \cos(-x) \cdot \tan(90^{\circ} - x) \cdot \sin(540^{\circ} + x) = 1 $$ **Problem 6.2 Proof:** $$ \frac{\sin(x - 90^{\circ}) \cdot \tan(-x)}{\cos(450^{\circ} + x)} = -1 $$ **Problem 6.3 General Solution:** $$ x = 30^{\circ} + 360^{\circ}k \quad \text{or} \quad x = 150^{\circ} + 360^{\circ}k \quad (k \in \mathbb{Z}) $$

5.2 If , write down the following in terms of m :

5.2.2
5.2.3
Please to
5
Gr. 11/ Mathematics First Push
March

QUESTION 6
6.1 Simplify:

6.2 Prove the identity:

6.3 Determine the general solution for:

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Solución

Problem 5.2

5.2.1

5.2.2

5.2.3

Summary of 5.2

Problem 6.1

Problem 6.2

Problem 6.3

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5.2 If , write down the following in terms of m : 5.2.2 5.2.3 Please to 5 Gr. 11/ Mathematics First Push March QUESTION 6 6.1 Simplify: 6.2 Prove the identity: 6.3 Determine the general solution for: