Verify the equality without using tables: $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ = 1 + \sqrt{2} \sin 65^\circ$
To Prove That (TPT) $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ = 1 + \sqrt{2} \sin 65^\circ$
i.e. $\;$ TPT $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ - \sqrt{2} \sin 65^\circ = 1$
$\begin{aligned}
LHS & = \cos 20^\circ + 1 - \cos 110^\circ - \sqrt{2} \sin 65^\circ \;\;\; \left(\because \; 2 \sin^2 \theta = 1 - \cos 2 \theta\right) \\\\
& = \cos 20^\circ + 1 - \cos \left(90^\circ + 20^\circ\right) - \sqrt{2} \sin 65^\circ \\\\
& = \cos 20^\circ + 1 - \left(- \sin 20^\circ\right) - \sqrt{2} \sin 65^\circ \\\\
& = \left(\cos 20^\circ + \sin 20^\circ\right) + 1 - \sqrt{2} \sin 65^\circ \\\\
& = \sqrt{2} \left(\dfrac{1}{\sqrt{2}} \sin 20^\circ + \dfrac{1}{\sqrt{2}} \cos 20^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\
& = \sqrt{2} \left(\sin 20^\circ \cos 45^\circ + \cos 20^\circ \sin 45^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\
& = \sqrt{2} \sin \left(20^\circ + 45^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\
& = \sqrt{2} \sin 65^\circ - \sqrt{2} \sin 65^\circ + 1 \\\\
& = 1 = RHS
\end{aligned}$
Hence verified.