Đề bài

Chứng minh các biểu thức sau không phụ thuộc vào x

a) \(A = \sin \left( {\frac{\pi }{4} + x} \right) - \cos \left( {\frac{\pi }{4} - x} \right)\);            b) \(B = \cos \left( {\frac{\pi }{6} - x} \right) - \sin \left( {\frac{\pi }{3} + x} \right)\); c) \(C = {\sin ^2}x + \cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right)\);                d) \(D = \frac{{1 - \cos 2x + \sin 2x}}{{1 + \cos 2x + \sin 2x}}.\cot x\).

Lời giải chi tiết

a) Ta có\(\begin{array}{l}A = \sin \left( {\frac{\pi }{4} + x} \right) - \cos \left( {\frac{\pi }{4} - x} \right) = \cos \left( {\frac{\pi }{2} - \left( {\frac{\pi }{4} + x} \right)} \right) - \cos \left( {\frac{\pi }{4} - x} \right)\\\,\,\,\,\,\, = \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{\pi }{4} - x} \right) = 0\end{array}\)b) Ta có\(\begin{array}{l}B = \cos \left( {\frac{\pi }{6} - x} \right) - \sin \left( {\frac{\pi }{3} + x} \right) = \cos \left( {\frac{\pi }{6} - x} \right) - \cos \left( {\frac{\pi }{2} - \left( {\frac{\pi }{3} + x} \right)} \right)\\\,\,\,\,\, = \cos \left( {\frac{\pi }{6} - x} \right) - \cos \left( {\frac{\pi }{6} - x} \right) = 0\end{array}\)c) Ta có\(\begin{array}{l}C = {\sin ^2}x + \cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right)\\\,\,\,\,\, = {\sin ^2}x + \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{3} - x + \frac{\pi }{3} + x} \right) + \cos \left( {\frac{\pi }{3} - x - \left( {\frac{\pi }{3} + x} \right)} \right)} \right]\\\,\,\,\,\, = {\sin ^2}x + \frac{1}{2}\left[ {\cos \frac{{2\pi }}{3} + \cos ( - 2x)} \right] = {\sin ^2}x + \frac{1}{2}\left( { - \frac{1}{2} + \cos 2x} \right)\\\,\,\,\,\, = {\sin ^2}x - \frac{1}{4} + \frac{1}{2}\left( {1 - 2{{\sin }^2}x} \right) = \frac{1}{4}\end{array}\) d) Ta có\(\begin{array}{l}D = \frac{{1 - \cos 2x + \sin 2x}}{{1 + \cos 2x + \sin 2x}}.\cot x\\\,\,\,\,\,\, = \frac{{1 - (1 - 2{{\sin }^2}x) + 2\sin x\cos x}}{{1 + 2{{\cos }^2}x - 1 + 2\sin x\cos x}}.\cot x\\\,\,\,\,\,\, = \frac{{2{{\sin }^2}x + 2\sin x\cos x}}{{2{{\cos }^2}x + 2\sin x\cos x}}.\cot x\\\,\,\,\,\,\, = \frac{{2\sin x(\sin x + \cos x)}}{{2\cos x(\cos x + \sin x)}}.\cot x = \frac{{\sin x}}{{\cos x}}.\cot x = \tan x.\cot x = 1\end{array}\)