Bài tập nâng cao · Bài 3
Dạng 1. Công thức cộng
Bài tập nâng cao Công thức lượng giác · Bài 3
Chứng minh các đẳng thức sau
a) $\sin\left( x+y \right).\sin\left( x-y \right)=\sin^{2}x-\sin^{2}y$;
b) $\tan x+\tan y=\frac{2\sin\left( x+y \right)}{\cos\left( x+y \right)+\cos\left( x-y \right)}$;
c) $\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan\left( x+\frac{2\pi}{3} \right).\tan x=-3$;
d) $\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{3\pi}{4} \right).\cos\left( x+\frac{\pi}{6} \right)=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right)$;
e)$\left( \cos 70^{\circ}+\cos 50^{\circ} \right)\left( \cos 230^{\circ}+\cos 290^{\circ} \right)-\left( \cos 40^{\circ}+\cos 160^{\circ} \right)\left( \cos 320^{\circ}+\cos 380^{\circ} \right)=0$;
f) $\tan x.\tan 3x=\frac{\tan^{2}2x-\tan^{2}x}{1-\tan^{2}x.\tan^{2}2x}$.
a) $\sin\left( x+y \right).\sin\left( x-y \right)=\sin^{2}x-\sin^{2}y$;
b) $\tan x+\tan y=\frac{2\sin\left( x+y \right)}{\cos\left( x+y \right)+\cos\left( x-y \right)}$;
c) $\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan\left( x+\frac{2\pi}{3} \right).\tan x=-3$;
d) $\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{3\pi}{4} \right).\cos\left( x+\frac{\pi}{6} \right)=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right)$;
e)$\left( \cos 70^{\circ}+\cos 50^{\circ} \right)\left( \cos 230^{\circ}+\cos 290^{\circ} \right)-\left( \cos 40^{\circ}+\cos 160^{\circ} \right)\left( \cos 320^{\circ}+\cos 380^{\circ} \right)=0$;
f) $\tan x.\tan 3x=\frac{\tan^{2}2x-\tan^{2}x}{1-\tan^{2}x.\tan^{2}2x}$.
Xem lời giải
Lời giải
a)
$\begin{array}{l}VT=\sin\left( x+y \right).\sin\left( x-y \right)=\left( \sin x.\cos y+\sin y.\cos x \right).\left( \sin x.\cos y-\sin y.\cos x \right)\\=\sin^{2}x.\cos^{2}y-\sin^{2}y.\cos^{2}x=\sin^{2}x.\left( 1-\sin^{2}y \right)-\sin^{2}y.\left( 1-\sin^{2}x \right)\\=\sin^{2}x-\sin^{2}y=VP\end{array}$
Điều phải chứng minh.
b)
$\begin{array}{l}VP=\frac{2\sin\left( x+y \right)}{\cos\left( x+y \right)+\cos\left( x-y \right)}=\frac{2\left( \sin x.\cos y+\sin y.\cos x \right)}{\cos x.\cos y-\sin x.\sin y+\cos x.\cos y+\sin x.\sin y}\\=\frac{2\left( \sin x.\cos y+\sin y.\cos x \right)}{2\cos x.\cos y}=\tan x+\tan y=VT\end{array}$
Điều phải chứng minh.
c)
Xét đẳng thức: $\tan\left( a-b \right)=\frac{\tan a-\tan b}{1+\tan a.\tan b}\Leftrightarrow \tan a.\tan b=\frac{\tan a-\tan b}{\tan\left( a-b \right)}-1$.
Áp dụng:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)=\frac{\tan x-\tan\left( x+\frac{\pi}{3} \right)}{\tan\left( -\frac{\pi}{3} \right)}-1=-\frac{1}{\sqrt{3}}\left[ \tan x-\tan\left( x+\frac{\pi}{3} \right) \right]-1$
$\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{\pi}{3} \right)-\tan\left( x+\frac{2\pi}{3} \right) \right]-1$
$\tan\left( x+\frac{2\pi}{3} \right).\tan x=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan x \right]-1$
Cộng theo vế ta được: $\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan\left( x+\frac{2\pi}{3} \right).\tan x=-3$
Điều phải chứng minh.
d)
$\begin{array}{l}VT=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{\pi}{6} \right).\cos\left( x+\frac{3\pi}{4} \right)\\=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\sin\left( \frac{\pi}{3}-x \right).\sin\left( -\frac{\pi}{4}-x \right)\\=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\sin\left( x-\frac{\pi}{3} \right).\sin\left( x+\frac{\pi}{4} \right)\end{array}$
$\begin{array}{l}=\cos\left[ \left( x-\frac{\pi}{3} \right)-\left( x+\frac{\pi}{4} \right) \right]=\cos\left( \frac{\pi}{3}+\frac{\pi}{4} \right)\\=\cos\frac{\pi}{3}.\cos\frac{\pi}{4}-\sin\frac{\pi}{3}.\sin\frac{\pi}{4}=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right).\end{array}$
Điều phải chứng minh.
Cách 2.
$\begin{array}{l}VT=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{\pi}{6} \right).\cos\left( x+\frac{3\pi}{4} \right)\\=\left( \cos x.\cos\frac{\pi}{3}+\sin x.\sin\frac{\pi}{3} \right).\left( \cos x.\cos\frac{\pi}{4}-\sin x.\sin\frac{\pi}{4} \right)\\+\left( \cos x.\cos\frac{\pi}{6}-\sin x.\sin\frac{\pi}{6} \right).\left( \cos x.\cos\frac{3\pi}{4}-\sin x.\sin\frac{3\pi}{4} \right)\\=\left( \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \right).\left( \frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x \right)+\left( \frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x \right).\left( -\frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x \right)\end{array}$
$\begin{array}{l}=\frac{\sqrt{2}}{4}\left( \cos x+\sqrt{3}\sin x \right).\left( \cos x-\sin x \right)-\frac{\sqrt{2}}{4}\left( \sqrt{3}\cos x-\sin x \right).\left( \cos x+\sin x \right)\\=\frac{\sqrt{2}}{4}.\left( \cos^{2}x+\sqrt{3}\sin x.\cos x-\sin x.\cos x-\sqrt{3}\sin^{2}x \right)-\frac{\sqrt{2}}{4}\left( \sqrt{3}\cos^{2}x+\sqrt{3}\sin x.\cos x-\sin^{2}x-\sin x.\cos x \right)\\=\frac{\sqrt{2}}{4}\left( \sin^{2}x+\cos^{2}x-\sqrt{3}\sin^{2}x-\sqrt{3}\cos^{2}x \right)=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right).\end{array}$ Điều phải chứng minh.
e)
Ta có
+ $\cos 230^{\circ}=\cos\left( 180^{\circ}+50^{\circ} \right)=-\cos 50^{\circ}$
+ $\cos 290^{\circ}=\cos\left( 360^{\circ}-70^{\circ} \right)=\cos 70^{\circ}$
+ $\cos 160^{\circ}=\cos\left( 90^{\circ}+70^{\circ} \right)=-\sin 70^{\circ}$
+ $\cos 320^{\circ}=\cos\left( 360^{\circ}-40^{\circ} \right)=\cos 40^{\circ}=\sin 50^{\circ}$
+ $\cos 380^{\circ}=\cos 20^{\circ}=\sin 70^{\circ}$
Khi đó
$\begin{array}{l}VT=\left( \cos 70^{\circ}+\cos 50^{\circ} \right).\left( \cos 230^{\circ}+\cos 290^{\circ} \right)-\left( \cos 40^{\circ}+\cos 160^{\circ} \right).\left( \cos 320^{\circ}+\cos 380^{\circ} \right)\\=\left( \cos 70^{\circ}+\cos 50^{\circ} \right).\left( -\cos 50^{\circ}+\cos 70^{\circ} \right)-\left( \sin 50^{\circ}-\sin 70^{\circ} \right).\left( \sin 50^{\circ}+\sin 70^{\circ} \right)\\=\cos^{2}70^{\circ}-\cos^{2}50^{\circ}-\sin^{2}50^{\circ}+\sin^{2}70^{\circ}\\=-\left( \cos^{2}50^{\circ}+\sin^{2}50^{\circ} \right)+\left( \cos^{2}70^{\circ}+\sin^{2}70^{\circ} \right)=-1+1=0.\end{array}$
Điều phải chứng minh.
f) $VP=\frac{\tan^{2}2x-\tan^{2}x}{1-\tan^{2}x.\tan^{2}2x}=\frac{\tan 2x-\tan x}{1+\tan 2x.\tan x}.\frac{\tan 2x+\tan x}{1-\tan 2x.\tan x}=\tan x.\tan 3x=VT$.
Điều phải chứng minh.
$\begin{array}{l}VT=\sin\left( x+y \right).\sin\left( x-y \right)=\left( \sin x.\cos y+\sin y.\cos x \right).\left( \sin x.\cos y-\sin y.\cos x \right)\\=\sin^{2}x.\cos^{2}y-\sin^{2}y.\cos^{2}x=\sin^{2}x.\left( 1-\sin^{2}y \right)-\sin^{2}y.\left( 1-\sin^{2}x \right)\\=\sin^{2}x-\sin^{2}y=VP\end{array}$
Điều phải chứng minh.
b)
$\begin{array}{l}VP=\frac{2\sin\left( x+y \right)}{\cos\left( x+y \right)+\cos\left( x-y \right)}=\frac{2\left( \sin x.\cos y+\sin y.\cos x \right)}{\cos x.\cos y-\sin x.\sin y+\cos x.\cos y+\sin x.\sin y}\\=\frac{2\left( \sin x.\cos y+\sin y.\cos x \right)}{2\cos x.\cos y}=\tan x+\tan y=VT\end{array}$
Điều phải chứng minh.
c)
Xét đẳng thức: $\tan\left( a-b \right)=\frac{\tan a-\tan b}{1+\tan a.\tan b}\Leftrightarrow \tan a.\tan b=\frac{\tan a-\tan b}{\tan\left( a-b \right)}-1$.
Áp dụng:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)=\frac{\tan x-\tan\left( x+\frac{\pi}{3} \right)}{\tan\left( -\frac{\pi}{3} \right)}-1=-\frac{1}{\sqrt{3}}\left[ \tan x-\tan\left( x+\frac{\pi}{3} \right) \right]-1$
$\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{\pi}{3} \right)-\tan\left( x+\frac{2\pi}{3} \right) \right]-1$
$\tan\left( x+\frac{2\pi}{3} \right).\tan x=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan x \right]-1$
Cộng theo vế ta được: $\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan\left( x+\frac{2\pi}{3} \right).\tan x=-3$
Điều phải chứng minh.
d)
$\begin{array}{l}VT=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{\pi}{6} \right).\cos\left( x+\frac{3\pi}{4} \right)\\=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\sin\left( \frac{\pi}{3}-x \right).\sin\left( -\frac{\pi}{4}-x \right)\\=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\sin\left( x-\frac{\pi}{3} \right).\sin\left( x+\frac{\pi}{4} \right)\end{array}$
$\begin{array}{l}=\cos\left[ \left( x-\frac{\pi}{3} \right)-\left( x+\frac{\pi}{4} \right) \right]=\cos\left( \frac{\pi}{3}+\frac{\pi}{4} \right)\\=\cos\frac{\pi}{3}.\cos\frac{\pi}{4}-\sin\frac{\pi}{3}.\sin\frac{\pi}{4}=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right).\end{array}$
Điều phải chứng minh.
Cách 2.
$\begin{array}{l}VT=\cos\left( x-\frac{\pi}{3} \right).\cos\left( x+\frac{\pi}{4} \right)+\cos\left( x+\frac{\pi}{6} \right).\cos\left( x+\frac{3\pi}{4} \right)\\=\left( \cos x.\cos\frac{\pi}{3}+\sin x.\sin\frac{\pi}{3} \right).\left( \cos x.\cos\frac{\pi}{4}-\sin x.\sin\frac{\pi}{4} \right)\\+\left( \cos x.\cos\frac{\pi}{6}-\sin x.\sin\frac{\pi}{6} \right).\left( \cos x.\cos\frac{3\pi}{4}-\sin x.\sin\frac{3\pi}{4} \right)\\=\left( \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \right).\left( \frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x \right)+\left( \frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x \right).\left( -\frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x \right)\end{array}$
$\begin{array}{l}=\frac{\sqrt{2}}{4}\left( \cos x+\sqrt{3}\sin x \right).\left( \cos x-\sin x \right)-\frac{\sqrt{2}}{4}\left( \sqrt{3}\cos x-\sin x \right).\left( \cos x+\sin x \right)\\=\frac{\sqrt{2}}{4}.\left( \cos^{2}x+\sqrt{3}\sin x.\cos x-\sin x.\cos x-\sqrt{3}\sin^{2}x \right)-\frac{\sqrt{2}}{4}\left( \sqrt{3}\cos^{2}x+\sqrt{3}\sin x.\cos x-\sin^{2}x-\sin x.\cos x \right)\\=\frac{\sqrt{2}}{4}\left( \sin^{2}x+\cos^{2}x-\sqrt{3}\sin^{2}x-\sqrt{3}\cos^{2}x \right)=\frac{\sqrt{2}}{4}\left( 1-\sqrt{3} \right).\end{array}$ Điều phải chứng minh.
e)
Ta có
+ $\cos 230^{\circ}=\cos\left( 180^{\circ}+50^{\circ} \right)=-\cos 50^{\circ}$
+ $\cos 290^{\circ}=\cos\left( 360^{\circ}-70^{\circ} \right)=\cos 70^{\circ}$
+ $\cos 160^{\circ}=\cos\left( 90^{\circ}+70^{\circ} \right)=-\sin 70^{\circ}$
+ $\cos 320^{\circ}=\cos\left( 360^{\circ}-40^{\circ} \right)=\cos 40^{\circ}=\sin 50^{\circ}$
+ $\cos 380^{\circ}=\cos 20^{\circ}=\sin 70^{\circ}$
Khi đó
$\begin{array}{l}VT=\left( \cos 70^{\circ}+\cos 50^{\circ} \right).\left( \cos 230^{\circ}+\cos 290^{\circ} \right)-\left( \cos 40^{\circ}+\cos 160^{\circ} \right).\left( \cos 320^{\circ}+\cos 380^{\circ} \right)\\=\left( \cos 70^{\circ}+\cos 50^{\circ} \right).\left( -\cos 50^{\circ}+\cos 70^{\circ} \right)-\left( \sin 50^{\circ}-\sin 70^{\circ} \right).\left( \sin 50^{\circ}+\sin 70^{\circ} \right)\\=\cos^{2}70^{\circ}-\cos^{2}50^{\circ}-\sin^{2}50^{\circ}+\sin^{2}70^{\circ}\\=-\left( \cos^{2}50^{\circ}+\sin^{2}50^{\circ} \right)+\left( \cos^{2}70^{\circ}+\sin^{2}70^{\circ} \right)=-1+1=0.\end{array}$
Điều phải chứng minh.
f) $VP=\frac{\tan^{2}2x-\tan^{2}x}{1-\tan^{2}x.\tan^{2}2x}=\frac{\tan 2x-\tan x}{1+\tan 2x.\tan x}.\frac{\tan 2x+\tan x}{1-\tan 2x.\tan x}=\tan x.\tan 3x=VT$.
Điều phải chứng minh.