Bài 1.33 trang 41 SGK Toán 11 tập 1 - Cùng khám phá

Đề bài

Chứng minh các đẳng thức sau:

a) \({\left( {\cos a - \sin a} \right)^2} = 1 - \sin 2a;\)

b) \(\cos \left( {a + b} \right)\cos \left( {a - b} \right) = {\cos ^2}a - {\sin ^2}b;\)

c) \(\frac{{\sin a + \sin 3a}}{{1 + \cos 2a}} = 2\sin a\,\,\,\,\left( {{\rm{khi }}\cos 2a \ne  - 1} \right)\)

d) \(\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9} + \cos \frac{{7\pi }}{9} = 0.\)

Lời giải chi tiết

a) \({\left( {\cos a - \sin a} \right)^2} = {\cos ^2}a - 2\cos a\sin a + {\sin ^2}a = 1 - \sin 2a\)

b)

\(\begin{array}{l}\cos \left( {a + b} \right)\cos \left( {a - b} \right) = \frac{1}{2}\left[ {\cos \left( {2a} \right) + \cos \left( {2b} \right)} \right] = \frac{1}{2}\left( {2{{\cos }^2}a - 1 + 2{{\cos }^2}b - 1} \right)\\ = \frac{1}{2}\left( {2{{\cos }^2}a - 2{{\cos }^2}b} \right) = {\cos ^2}a - {\sin ^2}b\end{array}\)

c) \(\frac{{\sin a + \sin 3a}}{{1 + \cos 2a}} = \frac{{2\sin \left( {2a} \right)\cos a}}{{2{{\cos }^2}a}} = \frac{{\sin \left( {2a} \right)}}{{\cos a}} = \frac{{2\sin a\cos a}}{{\cos a}} = 2\sin a\)

d)

\(\begin{array}{l}\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9} + \cos \frac{{7\pi }}{9} = \cos \frac{\pi }{9} + 2\cos \frac{{2\pi }}{3}\cos \frac{\pi }{9}\\ = \cos \frac{\pi }{9}\left( {1 + 2\cos \frac{{2\pi }}{3}} \right) = \cos \frac{\pi }{9}\left( {1 + 2.\left( { - \frac{1}{2}} \right)} \right) = \cos \frac{\pi }{9}.0 = 0\end{array}\)