SHORT TEST 1.4 Calculate the general Question 1 If \( \theta \) is a reztex angle and \( \tan \theta=-\frac{3}{4} \) with towing with \( \quad 2 \sin (3 x+20)=2 \cos x \) (6) determine the agram ald of a diagram (2) \( 1.11 \sin \theta \) (3) \( 11.2 \cos 2 \theta \) (3) \( 13 \cos \left(Q+30^{\circ}\right) \) \[ \text { Total }=27 \text { marks }+ \] Simplizy the following to a single trigonometric ratio. \( =\sin \left(900^{\circ}-x\right) \cdot \cos (-x)-\sin \left(x-180^{\circ}\right) \sin \left(90^{\circ}+x\right) \) (6) 3 given the identity \[ \frac{\sin 7 x+\sin x}{2 \cos 3 x}=\sin 4 x \] 1.3.1 Prove the identity (4) 1.3.1 For which values is the identity above undezined. (3)

Let's break down the problem step by step. ### Question 1 1. **Given Conditions:** - \( \theta \) is a reference angle. - \( \tan \theta = -\frac{3}{4} \). - The equation \( 2 \sin(3x + 20) = 2 \cos x \). 2. **Determine \( \sin \theta \) and \( \cos 2\theta \):** - Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can use the Pythagorean identity to find \( \sin \theta \) and \( \cos \theta \). - Let \( \sin \theta = -3k \) and \( \cos \theta = 4k \) for some \( k \). - Then, \( \sin^2 \theta + \cos^2 \theta = 1 \) gives us: \[ (-3k)^2 + (4k)^2 = 1 \implies 9k^2 + 16k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5} \] - Thus, \( \sin \theta = -\frac{3}{5} \) and \( \cos \theta = \frac{4}{5} \). 3. **Calculate \( \sin \theta \):** \[ \sin \theta = -\frac{3}{5} \] 4. **Calculate \( \cos 2\theta \):** - Using the double angle formula \( \cos 2\theta = 2\cos^2 \theta - 1 \): \[ \cos 2\theta = 2\left(\frac{4}{5}\right)^2 - 1 = 2 \cdot \frac{16}{25} - 1 = \frac{32}{25} - 1 = \frac{32}{25} - \frac{25}{25} = \frac{7}{25} \] 5. **Calculate \( 13 \cos(Q + 30^\circ) \):** - This requires more information about \( Q \). Assuming \( Q = \theta \): \[ \cos(Q + 30^\circ) = \cos(\theta + 30^\circ) = \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ \] - Using \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \): \[ \cos(\theta + 30^\circ) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} - \left(-\frac{3}{5}\right) \cdot \frac{1}{2} = \frac{4\sqrt{3}}{10} + \frac{3}{10} = \frac{4\sqrt{3} + 3}{10} \] - Therefore, \[ 13 \cos(Q + 30^\circ) = 13 \cdot \frac{4\sqrt{3} + 3}{10} = \frac{52\sqrt{3} + 39}{10} \] ### Simplify the Trigonometric Expression Now, simplify: \[ \sin(900^\circ - x) \cdot \cos(-x) - \sin(x - 180^\circ) \sin(90^\circ + x) \] 1. **Using Trigonometric Identities:** - \( \sin(900^\circ - x) = \sin(90^\circ + 810^\circ - x) = \cos(x) \) - \( \cos(-x) = \cos(x) \) - \( \sin(x - 180^\circ) = -\sin(x) \) - \( \sin(90^\circ + x) = \cos(x) \) 2. **Substituting:** \[ \cos(x) \cdot \cos(x) - (-\sin(x) \cdot \cos(x)) = \cos^2(x) + \sin(x) \cos(x) \] 3. **Final Simplification:** \[ = \cos^2(x) + \sin(x) \cos(x) = \cos(x)(\cos(x) + \sin(x)) \] ### Prove the Identity Given: \[ \frac{\sin 7x + \sin x}{2 \cos 3x} = \sin 4x \] 1. **Using the Sum-to-Product Identities:** \[ \sin 7x + \sin x = 2 \sin\left(\frac{7x + x}{2}\right) \cos\left(\frac{7x - x}{2}\right) = 2 \sin(4x) \cos(3x) \] - Therefore, \[ \frac{2 \sin(4x) \cos(3x)}{2 \cos(3x)} = \sin(4x) \] - This proves the identity. ### Values for Which the Identity is Undefined The identity is undefined when \( \cos(3x) = 0 \): \[ 3x = \frac{\pi}{2} + n\pi \implies x = \frac{\pi}{6} + \frac{n\pi}{3}, \quad n \in \mathbb{Z} \] ### Summary of Answers 1. \( \sin \theta = -\frac{3}{5} \) 2. \( \cos 2\theta = \frac{7}{25} \) 3. \( 13 \cos(Q + 30^\circ) = \frac{52\sqrt{3} + 39}{10} \) 4. Simplified expression: \( \cos(x)(\cos(x) + \sin(x)) \) 5. Identity proved. 6. Undefined for \( x = \frac{\pi}{6} + \frac{n\pi}{3} \).