Chứng minh các đẳng thức:
a) $\sin3a\cos^3a+\sin^3a\cos3a=\frac{3}{4}\sin4a                    (1)$
b) $\tan a\tan(\frac{\pi}{3}-a)\tan(\frac{\pi}{3}+a)=\tan3a                 (2)$
a) Ta có:   $\cos3a=4\cos^3a-3\cos a\Rightarrow \cos^3a=\frac{1}{4}(\cos3a+3\cos a)$
                   $\sin3a=3\sin a-4\sin^3a\Rightarrow \sin^3a=\frac{1}{4}(3\sin a-\sin3a)$
Thay vào vế trái (1) ta được:
     $\frac{1}{4}\sin3a(\cos3a+3\cos a)+\frac{1}{4}(3\sin a-\sin3 a)\cos3a$

$=\frac{3}{4}(\sin3a\cos a+\sin a\cos 3a)=\frac{3}{4}\sin(3a+a)=\frac{3}{4}\sin4a$.

b) $VT=\tan a.\frac{\tan\frac{\pi}{3}-\tan a}{1+\tan\frac{\pi}{3}\tan a}.\frac{\tan\frac{\pi}{3}+\tan a}{1-\tan\frac{\pi}{3}\tan a}$

             $=\tan a.\frac{\sqrt{3}-\tan a}{1+\sqrt{3}\tan a}.\frac{\sqrt{3}+\tan a}{1-\sqrt{3}\tan a}=\tan a.\frac{3-\tan^2a}{1-3\tan^2a}=\frac{3\tan 1-\tan^3a}{1-3\tan^2a}$

             $=\tan3a$.(đpcm)

Thẻ

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