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: \( 3 \sin x=2 \cos ^{2} x \)

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