Value of $\cos 20^\circ + 2 \sin^2 55^\circ – \sqrt{2} \sin 65^\circ$.
Frankly speaking, I have no idea how to start in this question. I tried simplifying $\cos 20^\circ$ in terms of $\cos 55^\circ$ & $35^\circ$ and $\sin 65^\circ$ in terms of $\sin 30^\circ$ & $35^\circ$, however this didn't help.
Any hint would be extremely helpful. 🙂
Best Answer
$$\cos 20^{\circ} + 2\sin^2 55^{\circ} - \sqrt{2}\sin 65^{\circ} \\ = \cos 20^{\circ}+ 1 - \cos 110^{\circ} - \sqrt{2}\sin 65^{\circ} \\ = 2\sin(\frac{110^{\circ}+20^{\circ}}{2})\sin(\frac{110^{\circ} - 20^{\circ}}{2}) - \sqrt{2}\sin 65^{\circ} +1\\ = 2\sin 45^{\circ}\sin65^{\circ} - \sqrt{2}\sin 65^{\circ} +1 \\ = \sqrt{2}\sin 65^{\circ} - \sqrt{2}\sin 65^{\circ} +1\\ = 1$$