Rút gọn: (2cos2a – sin4a )/(2cos2a + sin4a)= tan^2( pi/4 – a ). Giúp e với ạ !!!! 16/10/2021 Bởi Julia Rút gọn: (2cos2a – sin4a )/(2cos2a + sin4a)= tan^2( pi/4 – a ). Giúp e với ạ !!!!
Đáp án: $0$ Giải thích các bước giải: $VT=\dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}\\=\dfrac{2\cos2x-2\sin2a\cos2x}{2\cos2a+2\sin2a\cos2a}\\=\dfrac{2\cos2x(1-\sin2a)}{2\cos2a(1+\sin2a)}\\=\dfrac{1-\sin2a}{1+\sin2a}\\VP=\tan^2\left ( \dfrac{\pi}{4}-a \right )\\=\dfrac{1}{\cos^2\left ( \dfrac{\pi}{4}-a \right )}-1\\=\dfrac{2}{\cos\left ( 2\dfrac{\pi}{4}-2a \right )+1}-1\\=\dfrac{2}{\cos\left ( \dfrac{\pi}{2}-2a \right )+1}-1\\=\dfrac{2}{\sin2a+1}-1\\=\dfrac{2-\sin2a-1}{\sin2a+1}\\=\dfrac{1-\sin2a}{\sin2a+1}\\\Rightarrow \dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}-\tan^2\left ( \dfrac{\pi}{4}-a \right )\\=\dfrac{1-\sin2a}{1+\sin2a}-\dfrac{1-\sin2a}{\sin2a+1}=0$ Bình luận
Đáp án:
$0$
Giải thích các bước giải:
$VT=\dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}\\
=\dfrac{2\cos2x-2\sin2a\cos2x}{2\cos2a+2\sin2a\cos2a}\\
=\dfrac{2\cos2x(1-\sin2a)}{2\cos2a(1+\sin2a)}\\
=\dfrac{1-\sin2a}{1+\sin2a}\\
VP=\tan^2\left ( \dfrac{\pi}{4}-a \right )\\
=\dfrac{1}{\cos^2\left ( \dfrac{\pi}{4}-a \right )}-1\\
=\dfrac{2}{\cos\left ( 2\dfrac{\pi}{4}-2a \right )+1}-1\\
=\dfrac{2}{\cos\left ( \dfrac{\pi}{2}-2a \right )+1}-1\\
=\dfrac{2}{\sin2a+1}-1\\
=\dfrac{2-\sin2a-1}{\sin2a+1}\\
=\dfrac{1-\sin2a}{\sin2a+1}\\
\Rightarrow \dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}-\tan^2\left ( \dfrac{\pi}{4}-a \right )\\
=\dfrac{1-\sin2a}{1+\sin2a}-\dfrac{1-\sin2a}{\sin2a+1}=0$
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