Bài 7 trang 21 SGK Toán 11 tập 1 – Cánh diều

Đề bài

Cho \(\cos 2x = \frac{1}{4}\).

Tính: \(A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x – \frac{\pi }{6}} \right)\); \(B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x – \frac{\pi }{3}} \right)\)

Lời giải chi tiết

\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x – \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x – \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} – x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)

\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x – \frac{\pi }{3}} \right) =  – \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x – \frac{\pi }{3}} \right) – \cos \left( {x + \frac{\pi }{3} – x + \frac{\pi }{3}} \right)} \right]\\B =  – \frac{1}{2}\left( {\cos 2x – \cos \frac{{2\pi }}{3}} \right) =  – \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) =  – \frac{3}{8}\end{array}\)