Prove that

Solution:

(i) $\sin 80^{\circ} \cos 20^{\circ}-\cos 80^{\circ} \sin 20^{\circ}=\sin \left(80^{\circ}-20^{\circ}\right)$

(using $\sin (A-B)=\sin A \cos B-\cos A \sin B$ )

$=\sin 60^{\circ}$

$=\frac{\sqrt{3}}{2}$

(ii) $\cos 45^{\circ} \cos 15^{\circ}-\sin 45^{\circ} \sin 15^{\circ}=\cos \left(45^{\circ}+15^{\circ}\right)$

$(U \operatorname{sing} \cos (A+B)=\cos A \cos B-\sin A \sin B)$

$=\cos 60^{\circ}$

$=\frac{1}{2}$

(iii) $\cos 75^{\circ} \cos 15^{\circ}+\sin 75^{\circ} \sin 15^{\circ}=\cos \left(75^{\circ}-15^{\circ}\right)$

(using $\cos (A-B)=\cos A \cos B+\sin A \sin B$ )

$=\cos 60^{\circ}$

$=\frac{1}{2}$

(iv) $\sin 40^{\circ} \cos 20^{\circ}+\cos 40^{\circ} \sin 20^{\circ}=\sin \left(40^{\circ}+20^{\circ}\right)$

(using $\sin (A+B)=\sin A \cos B+\cos A \sin B$ )

$=\sin 60^{\circ}$

$=\frac{\sqrt{3}}{2}$

(v) $\cos 130^{\circ} \cos 40^{\circ}+\sin 130^{\circ} \sin 40^{\circ}=\cos \left(130^{\circ}-40^{\circ}\right)$

(using $\cos (A-B)=\cos A \cos B+\sin A \sin B)$

$=\cos 90^{\circ}$

$=0$