8. Pave the following identities (a) \( \cos \left(\theta+60^{\circ}\right)-\sqrt{3} \cos \left(\theta+30^{\circ}\right)=-\cos \beta \) (b) \( \cos (\alpha+\beta)+\cos (\alpha-\beta)=2 \cos \alpha \cos \beta \) (c) \( \sin (x+30)-\sin \left(x-30^{\circ}\right)=\cos x \) (d) \( \cos 7 A+\cos A=2 \cos 4 A \cos 3 A \)

**(a)** Expand both cosine terms using the cosine angle addition formula: \[ \cos(\theta + 60^\circ) = \cos\theta\cos60^\circ - \sin\theta\sin60^\circ = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta, \] \[ \cos(\theta + 30^\circ) = \cos\theta\cos30^\circ - \sin\theta\sin30^\circ = \frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta. \] Substitute these into the left-hand side: \[ \begin{aligned} \cos(\theta+60^\circ)-\sqrt{3}\cos(\theta+30^\circ) &=\left[\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\right]-\sqrt{3}\left[\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta\right]\\[1mm] &=\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta-\frac{3}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\\[1mm] &=\left(\frac{1}{2}-\frac{3}{2}\right)\cos\theta\\[1mm] &=-\cos\theta. \end{aligned} \] Thus, the expression simplifies to \[ -\cos\theta. \] Comparing with the given identity \[ \cos (\theta+60^\circ)-\sqrt{3}\cos(\theta+30^\circ)=-\cos\beta, \] we deduce that \[ \beta=\theta. \] --- **(b)** Recall the cosine sum-to-product formula: \[ \cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). \] Let \( x = \alpha+\beta \) and \( y = \alpha-\beta \), then \[ \frac{x+y}{2}=\frac{(\alpha+\beta)+(\alpha-\beta)}{2}=\alpha, \] \[ \frac{x-y}{2}=\frac{(\alpha+\beta)-(\alpha-\beta)}{2}=\beta. \] Thus, \[ \cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta. \] --- **(c)** Use the sine subtraction formula: \[ \sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right). \] Let \( A=x+30^\circ \) and \( B=x-30^\circ \). Then, \[ \frac{A+B}{2}=\frac{(x+30^\circ)+(x-30^\circ)}{2}=x, \] \[ \frac{A-B}{2}=\frac{(x+30^\circ)-(x-30^\circ)}{2}=30^\circ. \] Thus, \[ \sin(x+30^\circ)-\sin(x-30^\circ)=2\cos(x)\sin(30^\circ). \] Since \[ \sin(30^\circ)=\frac{1}{2}, \] we obtain \[ 2\cos(x)\cdot\frac{1}{2}=\cos(x). \] --- **(d)** Apply the cosine sum formula: \[ \cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). \] Here, let \( x=7A \) and \( y=A \). Then, \[ \frac{x+y}{2}=\frac{7A+A}{2}=4A, \] \[ \frac{x-y}{2}=\frac{7A-A}{2}=3A. \] Thus, \[ \cos7A+\cos A=2\cos4A\cos3A. \]