Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°
(i) sin 60o cos 30o − cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}\right) \times\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Also, $\sin 30^{\circ}=\frac{1}{2}$
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o
(ii) cos 60o cos 30o + sin 60o sin 30o
$=\left(\frac{1}{2}\right) \times\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right) \times\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
Also, $\cos 30^{\circ}=\frac{\sqrt{3}}{2}$
$\therefore \cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}=\cos 30^{\circ}$
(iii) $2 \sin 30^{\circ} \cos 30^{\circ}$
$=2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
Also, $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\therefore 2 \sin 30^{\circ} \cos 30^{\circ}=\sin 60^{\circ}$
(iv) $2 \sin 45^{\circ} \cos 45^{\circ}$
$=2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=1$
Also, $\sin 90^{\circ}=1$
$\therefore 2 \sin 45^{\circ} \cos 45^{\circ}=\sin 90^{\circ}$
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