1. Stating the problem: Factorize and simplify the expressions: (a) $$\frac{\sin x + \cos\left(x - \frac{\pi}{6}\right)}{\sin x + \sin\left(x - \frac{2\pi}{3}\right)}$$ (b) $$\frac{\sin(5x + 30^\circ) + \cos x}{\cos 4x + \cos(60^\circ - 2x)}$$ 2. Important formulas and rules: - Sum-to-product formulas: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ - Angle subtraction and addition identities for sine and cosine. 3. Solve (a): Numerator: $$\sin x + \cos\left(x - \frac{\pi}{6}\right) = \sin x + \sin\left(\frac{\pi}{2} - \left(x - \frac{\pi}{6}\right)\right) = \sin x + \sin\left(\frac{\pi}{2} - x + \frac{\pi}{6}\right) = \sin x + \sin\left(\frac{2\pi}{3} - x\right)$$ Using sum-to-product: $$= 2 \sin\left(\frac{x + (\frac{2\pi}{3} - x)}{2}\right) \cos\left(\frac{x - (\frac{2\pi}{3} - x)}{2}\right) = 2 \sin\left(\frac{2\pi}{3}}{2}\right) \cos\left(\frac{2x - \frac{2\pi}{3}}{2}\right) = 2 \sin\left(\frac{\pi}{3}\right) \cos\left(x - \frac{\pi}{3}\right)$$ Since $$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$, numerator simplifies to: $$2 \cdot \frac{\sqrt{3}}{2} \cos\left(x - \frac{\pi}{3}\right) = \sqrt{3} \cos\left(x - \frac{\pi}{3}\right)$$ Denominator: $$\sin x + \sin\left(x - \frac{2\pi}{3}\right) = 2 \sin\left(\frac{x + x - \frac{2\pi}{3}}{2}\right) \cos\left(\frac{x - (x - \frac{2\pi}{3})}{2}\right) = 2 \sin\left(x - \frac{\pi}{3}\right) \cos\left(\frac{2\pi}{3}}{2}\right) = 2 \sin\left(x - \frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)$$ Since $$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$, denominator simplifies to: $$2 \sin\left(x - \frac{\pi}{3}\right) \cdot \frac{1}{2} = \sin\left(x - \frac{\pi}{3}\right)$$ Therefore, the whole fraction is: $$\frac{\sqrt{3} \cos\left(x - \frac{\pi}{3}\right)}{\sin\left(x - \frac{\pi}{3}\right)} = \sqrt{3} \cot\left(x - \frac{\pi}{3}\right)$$ 4. Solve (b): Numerator: $$\sin(5x + 30^\circ) + \cos x$$ Rewrite cosine as sine: $$\cos x = \sin\left(90^\circ - x\right)$$ So numerator is: $$\sin(5x + 30^\circ) + \sin(90^\circ - x)$$ Using sum-to-product: $$= 2 \sin\left(\frac{5x + 30^\circ + 90^\circ - x}{2}\right) \cos\left(\frac{5x + 30^\circ - (90^\circ - x)}{2}\right) = 2 \sin\left(\frac{4x + 120^\circ}{2}\right) \cos\left(\frac{6x - 60^\circ}{2}\right) = 2 \sin(2x + 60^\circ) \cos(3x - 30^\circ)$$ Denominator: $$\cos 4x + \cos(60^\circ - 2x)$$ Using sum-to-product: $$= 2 \cos\left(\frac{4x + 60^\circ - 2x}{2}\right) \cos\left(\frac{4x - (60^\circ - 2x)}{2}\right) = 2 \cos\left(x + 30^\circ\right) \cos\left(3x - 30^\circ\right)$$ The fraction becomes: $$\frac{2 \sin(2x + 60^\circ) \cos(3x - 30^\circ)}{2 \cos(x + 30^\circ) \cos(3x - 30^\circ)} = \frac{\sin(2x + 60^\circ)}{\cos(x + 30^\circ)}$$ Rewrite $$\sin(2x + 60^\circ)$$ using angle addition: $$\sin(2x + 60^\circ) = \sin 2x \cos 60^\circ + \cos 2x \sin 60^\circ = \sin 2x \cdot \frac{1}{2} + \cos 2x \cdot \frac{\sqrt{3}}{2}$$ Rewrite $$\sin 2x = 2 \sin x \cos x$$ and $$\cos 2x = \cos^2 x - \sin^2 x$$, but better to use the identity: $$\sin(2x + 60^\circ) = 2 \sin(x + 30^\circ) \cos(x + 30^\circ)$$ (This is a known identity or can be verified by expansion.) So numerator is: $$2 \sin(x + 30^\circ) \cos(x + 30^\circ)$$ Therefore, the fraction is: $$\frac{2 \sin(x + 30^\circ) \cos(x + 30^\circ)}{\cos(x + 30^\circ)} = 2 \sin(x + 30^\circ)$$ Final answers: (a) $$\sqrt{3} \cot\left(x - \frac{\pi}{3}\right)$$ (b) $$2 \sin(x + 30^\circ)$$