Verify each of the following:

Solution:

(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, $\sin 90^{\circ}=1$

$\therefore 2 \sin 45^{\circ} \cos 45^{\circ}=\sin 90^{\circ}$