QUESTION 4: FREE STATE 4.1. If \( \tan 58^{\circ}=m \), deternine the following in terms of \( m \) without using a calculator. 4.1.1 \( \sin 58^{\circ} \) \( 4.1 .2 \sin 296^{\circ} \) (2) \( 4.1 .3 \cos 2^{\circ} \) (3) 4.2 If \( 5 \tan \theta+2 \sqrt{6}=0 \) and \( 0^{\circ}<\theta<270^{\circ} \), determine with the aid of a sketch and withour using a calculator, the value of: 4.2.1 \( \sin \theta \) \( 4.2 .2 \cos \theta \) (2) \( 4.2 .3 \frac{14 \cos \theta+7 \sqrt{6} \sin \theta}{\cos \left(-240^{\circ}\right) \cdot \tan 225^{\circ}} \) (I) 4.3 Determine the value of: 4.3 Determine the value of: (6) \[ \frac{\cos \left(180^{\circ}+x\right) \cdot \tan \left(360^{\circ}-x\right) \cdot \sin ^{2}\left(90^{\circ}-x\right)}{\sin \left(180^{\circ}-x\right)}+\sin ^{2} x \] 2.5 Prove the identity; \( \cos (A-B)-\cos (A+B)=2 \sin A \sin B \) . 5 Hence calcalate, without using a calculator, the value of : \( \cos 15^{\circ}-\cos 75^{\circ} \) .6 Find the value of \( \tan \theta \), if the distance between \( A(\cos \theta ; \sin \theta) \) and \( B(6 ; 7) \) is \( \sqrt{86} \) units.

**4.1.** We are given that \[ \tan 58^\circ = m. \] Since \[ \tan 58^\circ = \frac{\sin 58^\circ}{\cos 58^\circ}, \] and using the Pythagorean identity, we have \[ 1+\tan^2 58^\circ=\frac{1}{\cos^2 58^\circ}\quad\Longrightarrow\quad \cos 58^\circ=\frac{1}{\sqrt{1+m^2}}. \] Then, \[ \sin 58^\circ=\tan 58^\circ\cdot\cos 58^\circ=\frac{m}{\sqrt{1+m^2}}. \] **4.1.1.** \[ \sin 58^\circ=\frac{m}{\sqrt{1+m^2}}. \] --- **4.1.2.** Write \[ 296^\circ=360^\circ-64^\circ, \] so that \[ \sin 296^\circ = -\sin 64^\circ. \] Notice that \[ 64^\circ=2\cdot 32^\circ. \] Also, observe that \[ 58^\circ+32^\circ=90^\circ, \] which implies \[ \sin 58^\circ=\cos 32^\circ\quad \text{and}\quad \cos 58^\circ=\sin 32^\circ. \] From **4.1.1** we already found \[ \cos 58^\circ=\frac{1}{\sqrt{1+m^2}}\quad\Longrightarrow\quad \sin 32^\circ=\frac{1}{\sqrt{1+m^2}}. \] Also, \[ \tan 58^\circ = m=\frac{\sin 58^\circ}{\cos 58^\circ}=\frac{\sin 58^\circ}{\sin 32^\circ}\quad\Longrightarrow\quad \sin 58^\circ=m\sin 32^\circ=\frac{m}{\sqrt{1+m^2}}. \] Now, use the double-angle formula for sine: \[ \sin 64^\circ=2\sin 32^\circ\cos 32^\circ. \] But \[ \cos 32^\circ = m\sin 32^\circ, \] so \[ \sin 64^\circ=2\sin 32^\circ\cdot(m\sin 32^\circ)=2m\sin^2 32^\circ. \] And since \[ \sin 32^\circ=\frac{1}{\sqrt{1+m^2}}, \] we have \[ \sin 64^\circ=2m\left(\frac{1}{\sqrt{1+m^2}}\right)^2=\frac{2m}{1+m^2}. \] Therefore, \[ \sin 296^\circ=-\sin 64^\circ=-\frac{2m}{1+m^2}. \