# Prove that

Question:

Prove that

(i) $\sin \left(50^{\circ}+\theta\right) \cos \left(20^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right) \sin \left(20^{\circ}+\theta\right)=\frac{1}{2}$

(ii) $\cos \left(70^{\circ}+\theta\right) \cos \left(10^{\circ}+\theta\right)+\sin \left(70^{\circ}+\theta\right) \sin \left(10^{\circ}+\theta\right)=\frac{1}{2}$

Solution:

(i) $\sin \left(50^{\circ}+\theta\right) \cos \left(20^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right) \sin \left(20^{\circ}+\theta\right)$

$=\sin \left(50^{\circ}+\theta-\left(20^{\circ}+\theta\right)\right)($ using $\sin (A-B)=\sin A \cos B-\cos A \sin B)$

$=\sin \left(50^{\circ}+\theta-20^{\circ}-\theta\right)$

$=\sin 30^{\circ}$

$=\frac{1}{2}$

(ii) $\cos \left(70^{\circ}+\theta\right) \cos \left(10^{\circ}+\theta\right)+\sin \left(70^{\circ}+\theta\right) \sin \left(10^{\circ}+\theta\right)$

$=\cos \left(70^{\circ}+\theta-\left(10^{\circ}+\theta\right)\right)($ using $\cos (A-B)=\cos A \cos B+\sin A \sin B)$

$=\cos \left(70^{\circ}+\theta-10^{\circ}-\theta\right)$

$=\cos 60^{\circ}$

$=\frac{1}{2}$