Chứng mình:(1-2sin ^2x)/(2cot(pi/4+x).cos^2(pi/2-x)=1 09/08/2021 Bởi Natalia Chứng mình:(1-2sin ^2x)/(2cot(pi/4+x).cos^2(pi/2-x)=1
$\begin{array}{l} {\cos ^2}\left( {\dfrac{\pi }{4} – x} \right) = {\cos ^2}\left( {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{4} + x} \right)} \right) = {\sin ^2}\left( {\dfrac{\pi }{4} + x} \right)\\ \to A = \dfrac{{1 – 2{{\sin }^2}x}}{{2\cot \left( {\dfrac{\pi }{4} + x} \right){{\cos }^2}\left( {\dfrac{\pi }{4} – x} \right)}} = \dfrac{{\cos 2x}}{{2\cot \left( {\dfrac{\pi }{4} + x} \right){{\sin }^2}\left( {\dfrac{\pi }{4} + x} \right)}}\\ \to A = \dfrac{{\cos 2x}}{{2\cos \left( {\dfrac{\pi }{4} + x} \right)\sin \left( {\dfrac{\pi }{4} + x} \right)}} = \dfrac{{\cos 2x}}{{\sin \left( {\dfrac{\pi }{2} + 2x} \right)}} = \dfrac{{\cos 2x}}{{\cos 2x}} = 1 \end{array}$ Bình luận
$\begin{array}{l} {\cos ^2}\left( {\dfrac{\pi }{4} – x} \right) = {\cos ^2}\left( {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{4} + x} \right)} \right) = {\sin ^2}\left( {\dfrac{\pi }{4} + x} \right)\\ \to A = \dfrac{{1 – 2{{\sin }^2}x}}{{2\cot \left( {\dfrac{\pi }{4} + x} \right){{\cos }^2}\left( {\dfrac{\pi }{4} – x} \right)}} = \dfrac{{\cos 2x}}{{2\cot \left( {\dfrac{\pi }{4} + x} \right){{\sin }^2}\left( {\dfrac{\pi }{4} + x} \right)}}\\ \to A = \dfrac{{\cos 2x}}{{2\cos \left( {\dfrac{\pi }{4} + x} \right)\sin \left( {\dfrac{\pi }{4} + x} \right)}} = \dfrac{{\cos 2x}}{{\sin \left( {\dfrac{\pi }{2} + 2x} \right)}} = \dfrac{{\cos 2x}}{{\cos 2x}} = 1 \end{array}$