Question:
sin 43° cos 47° + cos 43° sin 47° = ?
(a) 0
(b) 1
(c) sin 4°
(d) cos 4°
Solution:
$\sin 43^{\circ} \cos 47^{\circ}+\cos 43^{\circ} \sin 47^{\circ}$
$=\sin \left(90^{\circ}-47^{\circ}\right) \cos 47^{\circ}+\cos \left(90^{\circ}-47^{\circ}\right) \sin 47^{\circ}$
$=\cos 47^{\circ} \cos 47^{\circ}+\sin 47^{\circ} \sin 47^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$
$=\cos ^{2} 47^{\circ}+\sin ^{2} 47^{\circ}$
$=1$ (using the identity: $\sin ^{2} \theta+\cos ^{2} \theta=1$ )
Hence, the correct option is (b).