Đáp án:
\[B = \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \frac{{1 - \cos 2x}}{2}\\
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \frac{{1 + \cos 2x}}{2}\\
\cos x + \cos y = 2.\cos \frac{{x + y}}{2}.\cos \frac{{x - y}}{2}\\
B = {\cos ^2}10^\circ + {\cos ^2}110^\circ + {\cos ^2}130^\circ \\
= {\cos ^2}10^\circ + {\sin ^2}\left( {90^\circ - 110^\circ } \right) + {\sin ^2}\left( {90^\circ - 130^\circ } \right)\\
= {\cos ^2}10^\circ + {\sin ^2}\left( { - 20^\circ } \right) + {\sin ^2}\left( { - 40^\circ } \right)\\
= {\cos ^2}10^\circ + {\sin ^2}20^\circ + {\sin ^2}40^\circ \\
= \frac{{1 + \cos 20^\circ }}{2} + \frac{{1 - \cos 40^\circ }}{2} + \frac{{1 - \cos 80^\circ }}{2}\\
= \frac{3}{2} + \frac{1}{2}\left( {\cos 20^\circ - \cos 40^\circ - \cos 80^\circ } \right)\\
= \frac{3}{2} + \frac{1}{2}.\left( {\cos 20^\circ - 2.\cos 60^\circ .\cos 20^\circ } \right)\\
= \frac{3}{2} + \frac{1}{2}.\left( {\cos 20^\circ - \cos 20^\circ } \right)\\
= \frac{3}{2}
\end{array}\)
