Chứng minh các biểu thức sau không phụ thuộc x
\(\begin{array}{l}
a)A = \sin \left( {\frac{\pi }{4} + x} \right) - \cos \left( {\frac{\pi }{4} - x} \right)\\
b)B = \cos \left( {\frac{\pi }{6} - x} \right) - \sin \left( {\frac{\pi }{3} + x} \right)\\
c)C = {\sin ^2}x + \cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right)\\
d)D = \frac{{1 - \cos 2x + \sin 2x}}{{1 + \cos 2x + \sin 2x}}.\cot x
\end{array}\)
a) Vì \(\left( {\frac{\pi }{4} + x} \right) + \left( {\frac{\pi }{4} - x} \right) = \frac{\pi }{2},\forall x\) nên \(\sin \left( {\frac{\pi }{4} + x} \right) = \cos \left( {\frac{\pi }{4} - x} \right)\)
Do đó \(A = 0,\forall x\)
b) Vì \(\left( {\frac{\pi }{6} - x} \right) + \left( {\frac{\pi }{3} + x} \right) = \frac{\pi }{2},\forall x\) nên \({\rm{cos}}\left( {\frac{\pi }{6} - x} \right) = \sin \left( {\frac{\pi }{3} + x} \right)\)
Do đó \(B = 0,\forall x\)
c) \(\begin{array}{l}
\cos \left( {\frac{\pi }{3} - x} \right).\cos \left( {\frac{\pi }{3} + x} \right) = \frac{1}{2}\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right) = \frac{1}{2}\left( { - \frac{1}{2} + 2{{\cos }^2}x - 1} \right)\\
\Rightarrow C = {\sin ^2}x + {\cos ^2}x - \frac{3}{4} = \frac{1}{4},\forall x
\end{array}\)
d) \(\begin{array}{l}
1 - \cos 2x + \sin 2x = 2{\sin ^2}x + 2\sin x\cos x = 2\sin x\left( {\cos x + \sin x} \right)\\
1 + \cos 2x + \sin 2x = 2{\cos ^2}x + 2\sin x\cos x = 2\cos x\left( {\cos x + \sin x} \right)\\
\Rightarrow D = \frac{{1 - \cos 2x + \sin 2x}}{{1 + \cos 2x + \sin 2x}}.\cot x = \frac{{\sin x}}{{\cos x}}.\cot x = 1,\forall x
\end{array}\)
-- Mod Toán 10