Toán chứng minh sin² 3a/ sin² a -cos² 3 a /cos ² a = 8× cos 2a 19/10/2021 By Aaliyah chứng minh sin² 3a/ sin² a -cos² 3 a /cos ² a = 8× cos 2a
$\dfrac{sin^23a}{sin^2a}-\dfrac{cos^23a}{cos^2a}=8cos2a$ $VT=(\dfrac{sin3a}{sina}+\dfrac{cos3a}{cosa})(\dfrac{sin3a}{sina}-\dfrac{cos3a}{cosa})$ $=\dfrac{(sin3acosa+sinacos3a)(sin3acosa-sinacos3a)}{sin^2a.cos^2a}$ $=\dfrac{sin4a.sin2a}{sin^2.cos^2a}$ $=\dfrac{4sina.cosa.cos2a.2sinacosa}{sin^2acos^2a}$ $=\dfrac{8sin^2acos^2a.cos2a}{sin^2acos^2a}$ $=8cos2a=VP(đúng)$ $\Rightarrow đpcm$ Trả lời
Giải thích các bước giải: $VT=\dfrac{\sin^23a}{\sin^2a}-\dfrac{\cos^23x}{\cos^2a}\\=\dfrac{\sin^23a\cos^2a-\sin^2a\cos^23x}{\sin^2a\cos^2a}\\=\dfrac{(\sin3a\cos a)^2-(\sin a\cos3x)^2}{\sin^2a\cos^2a}\\=\dfrac{(\sin3a\cos a+\sin a\cos3x).(\sin3a\cos a-\sin a\cos3x)}{\sin^2a\cos^2a}\\=\dfrac{\sin(3a+a).\sin(3a-a)}{\sin^2a\cos^2a}\\=\dfrac{\sin4a.\sin2a}{\dfrac{1}{4}\sin^22a}\\=\dfrac{4\sin4a}{\sin2a}\\=\dfrac{4.2\sin2a\cos2x}{\sin2a}\\=8\cos2x=VP\Rightarrow đpcm$ Trả lời
$\dfrac{sin^23a}{sin^2a}-\dfrac{cos^23a}{cos^2a}=8cos2a$
$VT=(\dfrac{sin3a}{sina}+\dfrac{cos3a}{cosa})(\dfrac{sin3a}{sina}-\dfrac{cos3a}{cosa})$
$=\dfrac{(sin3acosa+sinacos3a)(sin3acosa-sinacos3a)}{sin^2a.cos^2a}$
$=\dfrac{sin4a.sin2a}{sin^2.cos^2a}$
$=\dfrac{4sina.cosa.cos2a.2sinacosa}{sin^2acos^2a}$
$=\dfrac{8sin^2acos^2a.cos2a}{sin^2acos^2a}$
$=8cos2a=VP(đúng)$
$\Rightarrow đpcm$
Giải thích các bước giải:
$VT=\dfrac{\sin^23a}{\sin^2a}-\dfrac{\cos^23x}{\cos^2a}\\
=\dfrac{\sin^23a\cos^2a-\sin^2a\cos^23x}{\sin^2a\cos^2a}\\
=\dfrac{(\sin3a\cos a)^2-(\sin a\cos3x)^2}{\sin^2a\cos^2a}\\
=\dfrac{(\sin3a\cos a+\sin a\cos3x).(\sin3a\cos a-\sin a\cos3x)}{\sin^2a\cos^2a}\\
=\dfrac{\sin(3a+a).\sin(3a-a)}{\sin^2a\cos^2a}\\
=\dfrac{\sin4a.\sin2a}{\dfrac{1}{4}\sin^22a}\\
=\dfrac{4\sin4a}{\sin2a}\\
=\dfrac{4.2\sin2a\cos2x}{\sin2a}\\
=8\cos2x=VP\Rightarrow đpcm$