Cho cos alpha = 1/căn 3 và α ∈ ( -pi /2;0).Khi đó:

Đáp án: a) Sai. b) Đúng. c) Sai. d) Đúng.

a) Sai. sin²α + cos²α = 1 nên \({\sin ^2}\alpha = 1 - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} = \frac{2}{3}\).

Do α ∈ \(\left( {\frac{{ - \pi }}{2};0} \right)\) nên \(\sin \alpha = \frac{{ - \sqrt 6 }}{3}\).

b) Đúng. \(\sin \left( {\alpha + \frac{\pi }{4}} \right) = \sin \alpha \cdot \cos \frac{\pi }{4} + \cos \alpha \cdot \sin \frac{\pi }{4} = \frac{{ - \sqrt 6 }}{3} \cdot \frac{{\sqrt 2 }}{2} + \frac{1}{{\sqrt 3 }} \cdot \frac{{\sqrt 2 }}{2}\)

= \(\frac{{\sqrt 6 \cdot \left( {1 - \sqrt 2 } \right)}}{6}\).

c) Sai. \(A = \sin 3\alpha \cdot \cos \alpha - \frac{1}{4}\sin 4\alpha = \left( {\sin 3\alpha \cdot \cos \alpha } \right) - \frac{1}{4}\sin 4\alpha \)

= \(\frac{1}{2}\left[ {\sin \left( {3\alpha + \alpha } \right) + \sin \left( {3\alpha - \alpha } \right)} \right] - \frac{1}{4}\sin 4\alpha = \frac{1}{2}\left( {\sin 4\alpha + \sin 2\alpha } \right) - \frac{1}{4}\sin 4\alpha \)

= \(\frac{1}{4}\sin 4\alpha + \frac{1}{2}\sin 2\alpha = \frac{1}{4} \cdot \left( {2\sin 2\alpha \cdot \cos 2\alpha } \right) + \frac{1}{2}\sin 2\alpha = \frac{1}{2}\sin 2\alpha \cdot \left( {\cos 2\alpha + 1} \right)\)

= \(\frac{1}{2} \cdot 2 \cdot \cos \alpha \cdot \sin \alpha \cdot \left( {2{{\cos }^2}\alpha - 1 + 1} \right) = \cos \alpha \cdot \sin \alpha \cdot 2{\cos ^2}\alpha = 2{\cos ^3}\alpha \cdot \sin \alpha \)

= \(2 \cdot {\left( {\frac{1}{{\sqrt 3 }}} \right)^3} \cdot \left( {\frac{{ - \sqrt 6 }}{3}} \right) = \frac{{ - 2\sqrt 2 }}{9}\).

d) Đúng. \(B = \sin \left( {\alpha + \frac{\pi }{6}} \right) \cdot \cos \left( {\alpha - \frac{\pi }{6}} \right) - \frac{1}{2} \cdot \sin 2\alpha \)

= \(\frac{1}{2} \cdot \sin 2\alpha + \frac{1}{2} \cdot \sin \frac{\pi }{3} - \frac{1}{2} \cdot \sin 2\alpha = \frac{1}{2} \cdot \sin \frac{\pi }{3} = \frac{1}{2} \cdot \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{4}\).