S2 Simplify the following without using a calculator: \[ \frac{\sin \left(90^{\circ}-\theta\right) \cdot \cos 480^{\circ}+\cos \left(180^{\circ}-\theta\right)}{\cos \theta \cdot \sin 150^{\circ}-\tan 180^{\circ}} \] 53 Prove that \( \frac{\cos x}{\sin 2 x}-\frac{\cos 2 x}{2 \sin x}=\sin x \) 5.4. Givex: \( \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 \) 5.4.1 Show that the equation \( \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 \) can be written as \[ \cos \left(x+60^{\circ}\right)=\cos \left(90^{\circ}-2 x\right) \] 5.4.2 Ienec, or otherwise, determine the general solution of \( \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 \) 55 Given that \( \cos 22,5^{\circ}=\frac{a}{c} \) and \( a^{2}+b^{2}=c^{2} \). With the aid of a diagram, or otherwise, show that \( \frac{2 a b}{c^{2}}=\frac{\sqrt{2}}{2} \).

**Problem S2** We want to simplify \[ \frac{\sin \left(90^\circ-\theta\right) \cdot \cos 480^\circ+\cos \left(180^\circ-\theta\right)}{\cos \theta \cdot \sin 150^\circ-\tan 180^\circ}. \] 1. Use the identity \(\sin(90^\circ-\theta)=\cos \theta\). 2. Note that \(\cos 480^\circ=\cos(480^\circ-360^\circ)=\cos 120^\circ=-\frac{1}{2}\). 3. Use \(\cos(180^\circ-\theta)=-\cos \theta\). Thus, the numerator becomes \[ \begin{aligned} \sin(90^\circ-\theta)\cdot \cos 480^\circ + \cos(180^\circ-\theta) &=\cos\theta \cdot \left(-\frac{1}{2}\right) - \cos\theta\\[1mm] &=-\frac{1}{2}\cos\theta - \cos\theta\\[1mm] &=-\frac{3}{2}\cos\theta. \end{aligned} \] 4. For the denominator, note that \(\sin150^\circ=\frac{1}{2}\) and \(\tan 180^\circ=0\). Hence, \[ \cos \theta \cdot \sin150^\circ - \tan 180^\circ=\cos \theta \cdot \frac{1}{2}-0=\frac{1}{2}\cos\theta. \] 5. Dividing numerator and denominator we have \[ \frac{-\frac{3}{2}\cos\theta}{\frac{1}{2}\cos\theta}=-3, \] provided \(\cos\theta\ne 0\). --- **Problem 53** We need to prove that \[ \frac{\cos x}{\sin 2x}-\frac{\cos 2x}{2\sin x}=\sin x. \] 1. Use the double-angle identity \(\sin 2x=2\sin x \cos x\). Thus, \[ \frac{\cos x}{\sin2x}=\frac{\cos x}{2\sin x\cos x}=\frac{1}{2\sin x}. \] 2. Substitute back into the expression: \[ \frac{1}{2\sin x}-\frac{\cos2x}{2\sin x}=\frac{1-\cos2x}{2\sin x}. \] 3. Use the identity \(1-\cos2x=2\sin^2 x\). Hence, \[ \frac{2\sin^2 x}{2\sin x}=\sin x. \] Thus, the given equation is proved. --- **Problem 5.4** We are given \[ \frac{\cos 60^\circ}{\sin x}-\frac{\sin 60^\circ}{\cos x}=2. \] *Step 5.4.1* 1. Note that \(\cos60^\circ=\frac{1}{2}\) and \(\sin60^\circ=\frac{\sqrt{3}}{2}\). Then the equation becomes \[ \frac{\frac{1}{2}}{\sin x}-\frac{\frac{\sqrt{3}}{2}}{\cos x}=2. \] 2. Multiply both sides by 2: \[ \frac{1}{\sin x}-\frac{\sqrt{3}}{\cos x}=4. \] 3. Multiply through by \(\sin x \cos x\) to obtain \[ \cos x-\sqrt{3}\sin x=4\sin x\cos x. \] 4. Recognize that \(4\sin x\cos x=2\sin2x\), so \[ \cos x-\sqrt{3}\sin x=2\sin2x. \] 5. Express the left-hand side in the form \(R\cos(x+\phi)\). Since \[ \cos(x+60^\circ)=\cos x\cos60^\circ-\sin x\sin60^\circ=\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x, \] multiplying by 2 gives \[ 2\cos(x+60^\circ)=\cos x-\sqrt{3}\sin x. \] 6. Therefore, the equation becomes \[ 2\cos(x+60^\circ)=2\sin2x. \] Dividing both sides by 2: \[ \cos(x+60^\circ)=\sin2x. \] 7. Recall that \(\sin2x=\cos(90^\circ-2x)\), so finally we have \[ \cos(x+60^\circ)=\cos(90^\circ-2x). \] *Step 5.4.2* To find the general solution of \[ \cos(x+60^\circ)=\cos(90^\circ-2x), \] we use the cosine identity: If \(\cos A=\cos B\), then either \[ A=B+360^\circ k\quad \text{or}\quad A=-B+360^\circ k,\quad k\in\mathbb Z. \] **Case 1:** \[ x+60^\circ=90^\circ-2x+360^\circ k. \] Solve for \(x\): \[ \begin{aligned} x+60^\circ+2x&=90^\circ+360^\circ k,\\[1mm] 3x&=30^\circ+360^\circ k,\\[1mm] x&=10^\circ+120^\circ k. \end{aligned} \] **Case 2:** \[ x+60^\circ=-(90^\circ-2x)+360^\circ k. \] Simplify the right-hand side: \[ x+60^\circ=-90^\circ+2x+360^\circ k. \] Solve for \(x\): \[ \begin{aligned} 60^\circ+90^\circ &=2x-x+360^\circ k,\\[1mm] 150^\circ&=x+360^\circ k,\\[1mm] x&=150^\circ-360^\circ k. \end{aligned} \] Since \(k\) is any integer, writing \(k'=-k\) gives the general solution: \[ x=150^\circ+360^\circ k,\quad k\in\mathbb Z. \] Thus, the general solutions are \[ x=10^\circ+120^\circ k\quad \text{or}\quad x=150^\circ+360^\circ k,\quad k\in\mathbb Z. \] --- **Problem 55** We are given that \[ \cos22.5^\circ=\frac{a}{c} \] and that \(a\), \(b\), and \(c\) are the sides of a right triangle (with \(a^2+b^2=c^2\)). In such a triangle, if we assume that the angle corresponding to \(a\) is \(22.5^\circ\), then \[ \sin22.5^\circ=\frac{b}{c}. \] To show that \[ \frac{2ab}{c^2}=\frac{\sqrt{2}}{2}, \] follow these steps: 1. Write the left-hand side in terms of ratios: \[ \frac{2ab}{c^2}=2\left(\frac{a}{c}\right)\left(\frac{b}{c}\right)=2\cos22.5^\circ\,\sin22.5^\circ. \] 2. Use the double-angle identity for sine: \[ 2\cos22.5^\circ\,\sin22.5^\circ=\sin(2\times22.5^\circ)=\sin45^\circ. \] 3. Recall that \[ \sin45^\circ=\frac{\sqrt{2}}{2}. \] Thus, \[ \frac{2ab}{c^2}=\frac{\sqrt{2}}{2}. \]