Cmr: (sin4a/1+cos4a)×(cos2a/1+cos2a)=tan a 18/11/2021 Bởi Reagan Cmr: (sin4a/1+cos4a)×(cos2a/1+cos2a)=tan a
Đáp án: $\begin{array}{l}\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}}\\ = \frac{{2.\sin 2a.\cos 2a}}{{2.{{\cos }^2}2a}}.\frac{{\cos 2a}}{{2{{\cos }^2}a}}\\ = \frac{{\sin 2a}}{{2.{{\cos }^2}a}}\\ = \frac{{2.\sin a.\cos a}}{{2.{{\cos }^2}a}}\\ = \frac{{\sin a}}{{\cos a}}\\ = \tan a\end{array}$ Vậy $\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}} = \tan a$ Bình luận
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Đáp án:
$\begin{array}{l}
\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}}\\
= \frac{{2.\sin 2a.\cos 2a}}{{2.{{\cos }^2}2a}}.\frac{{\cos 2a}}{{2{{\cos }^2}a}}\\
= \frac{{\sin 2a}}{{2.{{\cos }^2}a}}\\
= \frac{{2.\sin a.\cos a}}{{2.{{\cos }^2}a}}\\
= \frac{{\sin a}}{{\cos a}}\\
= \tan a
\end{array}$
Vậy $\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}} = \tan a$