Question

Prove that:
(i) $\sin 65°+\cos 65°=\sqrt{2}\cos 20°$
(ii) sin 47° + cos 77° = cos 17°

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 65°+\cos 65° \\ =\sin 65°+\cos \left(90°-25°\right) \\ =\sin 65°+\sin 25° \\ =2\sin \left(\frac{65°+25°}{2}\right)\cos \left(\frac{65°-25°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 45°\cos 20° \\ =2\times \frac{1}{\sqrt{2}}\cos 20° \\ =\sqrt{2}\cos 20° \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin 47°+\cos 77° \\ =\sin 47°+\cos \left(90°-13°\right) \\ =\sin 47°+\sin 13° \\ =2\sin \left(\frac{47°+13°}{2}\right)\cos \left(\frac{47°-13°}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 30°\cos 17° \\ =2\times \frac{1}{2}\cos 17° \\ =\cos 17° \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$