Bài tập nâng cao · Bài 17
Dạng 4. Biến đổi tổng thành tích
Bài tập nâng cao Công thức lượng giác · Bài 17
Chứng minh các đẳng thức sau:
a) $\cot x-\tan x-2\tan 2x=4\cot 4x$.
b) $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{1+\tan 2x}{1-\tan 2x}$.
c) $\frac{1}{\cos^{6}x}-\tan^{6}x=\frac{3\tan^{2}x}{\cos^{2}x}+1$.
d) $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$.
e) $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$.
f) $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$.
g) $\cos 5x.\cos 3x+\sin 7x.\sin x=\cos 2x.\cos 4x$.
h) Cho $\sin\left( 2a+b \right)=5\sin b$. Chứng minh: $\frac{2\tan\left( a+b \right)}{\tan a}=3$.
i) Cho $\tan\left( a+b \right)=3\tan a$. Chứng minh: $\sin\left( 2a+2b \right)+\sin 2a=2\sin 2b$.
a) $\cot x-\tan x-2\tan 2x=4\cot 4x$.
b) $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{1+\tan 2x}{1-\tan 2x}$.
c) $\frac{1}{\cos^{6}x}-\tan^{6}x=\frac{3\tan^{2}x}{\cos^{2}x}+1$.
d) $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$.
e) $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$.
f) $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$.
g) $\cos 5x.\cos 3x+\sin 7x.\sin x=\cos 2x.\cos 4x$.
h) Cho $\sin\left( 2a+b \right)=5\sin b$. Chứng minh: $\frac{2\tan\left( a+b \right)}{\tan a}=3$.
i) Cho $\tan\left( a+b \right)=3\tan a$. Chứng minh: $\sin\left( 2a+2b \right)+\sin 2a=2\sin 2b$.
Xem lời giải
Lời giải
a) $\cot x-\tan x-2\tan 2x=4\cot 4x$
Ta có: $\cot x-\tan x-2\tan 2x$ $=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}-2\frac{\sin 2x}{\cos 2x}$ $=\frac{2\cos^{2}x-2\sin^{2}x}{\sin 2x}-\frac{2\sin 2x}{\cos 2x}$
$=\frac{2\cos 2x}{\sin 2x}-\frac{2\sin 2x}{\cos 2x}$ $=\frac{4\left( \cos^{2}2x-\sin^{2}2x \right)}{\sin 4x}$ $=\frac{4\cos 4x}{\sin 4x}$ $=4\cot 4x$.
b) $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{1+\tan 2x}{1-\tan 2x}$
Ta có: $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{\cos 4x}{\left( \cos 2x-\sin 2x \right)^{2}}$ $=\frac{\cos^{2}2x-\sin^{2}2x}{\left( \cos 2x-\sin 2x \right)^{2}}$ $=\frac{\cos 2x+\sin 2x}{\cos 2x-\sin 2x}$ $=\frac{1+\tan 2x}{1-\tan 2x}$ (do $\cos 2x\neq 0$).
c) $\frac{1}{\cos^{6}x}-\tan^{6}x=\frac{3\tan^{2}x}{\cos^{2}x}+1$
Ta có: $\frac{1}{\cos^{6}x}-\tan^{6}x=\left( \frac{1}{\cos^{2}x} \right)^{3}-\tan^{6}x$ $=\left( 1+\tan^{2}x \right)^{3}-\tan^{6}x$ $=1+3\tan^{2}x+3\tan^{4}x+\tan^{6}x-\tan^{6}x$ $=1+3\tan^{2}x\left( 1+\tan^{2}x \right)$ $=\frac{3\tan^{2}x}{\cos^{2}x}+1$.
d) $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$
Ta có: $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 4x}{\cos 4x}-\frac{1}{\cos 4x}$ $=\frac{\sin 4x-1}{\cos 4x}$ $=\frac{-\left( \cos 2x-\sin 2x \right)^{2}}{\cos^{2}2x-\sin^{2}2x}$ $=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$.
e) $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$
Ta có: $\tan\left( 2x+4x \right)=\frac{\tan 2x+\tan 4x}{1-\tan 2x\tan 4x}$.
Suy ra: $\tan 6x-\tan 2x\tan 4x\tan 6x=\tan 2x+\tan 4x$.
Do đó: $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$.
f) $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$
Ta có: $\sin x\left( 1+2\cos 2x+2\cos 4x+2\cos 6x \right)$
$=\sin x+2.\frac{1}{2}\left( \sin 3x-\sin x \right)+2.\frac{1}{2}\left( \sin 5x-\sin 3x \right)+2.\frac{1}{2}\left( \sin 7x-\sin 5x \right)$
$=\sin x+\sin 3x-\sin x+\sin 5x-\sin 3x+\sin 7x-\sin 5x$
$=\sin 7x$.
Suy ra: $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$.
g) $\cos 5x.\cos 3x+\sin 7x.\sin x=\cos 2x.\cos 4x$
Ta có: $\cos 5x.\cos 3x+\sin 7x.\sin x=\frac{1}{2}\left( \cos 8x+\cos 2x \right)+\frac{1}{2}\left( \cos 6x-\cos 8x \right)$
$=\frac{1}{2}\left( \cos 2x+\cos 6x \right)$ $=\frac{1}{2}.2\cos 4x.\cos 2x$ $=\cos 2x\cos 4x$.
h) Cho $\sin\left( 2a+b \right)=5\sin b$. Chứng minh: $\frac{2\tan\left( a+b \right)}{\tan a}=3$.
Ta có: $\frac{2\tan\left( a+b \right)}{\tan a}=2.\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}.\frac{\cos a}{\sin a}$ $=2.\frac{\sin\left( 2a+b \right)+\sin b}{\sin\left( 2a+b \right)-\sin b}$ $=2.\frac{6\sin b}{4\sin b}$ $=3$.
i) Cho $\tan\left( a+b \right)=3\tan a$. Chứng minh: $\sin\left( 2a+2b \right)+\sin 2a=2\sin 2b$.
Ta có: $\tan\left( a+b \right)=3\tan a$
$\Leftrightarrow \frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}=3\frac{\sin a}{\cos a}$
$\Leftrightarrow \sin\left( a+b \right)\cos a=3\sin a\cos\left( a+b \right)$
$\Leftrightarrow \frac{1}{2}\left[ \sin\left( 2a+b \right)+\sin b \right]=\frac{3}{2}\left[ \sin\left( 2a+b \right)-\sin b \right]$ $\Leftrightarrow \sin\left( 2a+b \right)+\sin b=3\sin\left( 2a+b \right)-3\sin b$ $\Leftrightarrow \sin\left( 2a+b \right)=2\sin b$.
Khi đó: $\sin\left( 2a+2b \right)+\sin 2a=2\sin\left( 2a+b \right)\cos b$ $=2.2\sin b.\cos b$ $=2\sin 2b$.
Ta có: $\cot x-\tan x-2\tan 2x$ $=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}-2\frac{\sin 2x}{\cos 2x}$ $=\frac{2\cos^{2}x-2\sin^{2}x}{\sin 2x}-\frac{2\sin 2x}{\cos 2x}$
$=\frac{2\cos 2x}{\sin 2x}-\frac{2\sin 2x}{\cos 2x}$ $=\frac{4\left( \cos^{2}2x-\sin^{2}2x \right)}{\sin 4x}$ $=\frac{4\cos 4x}{\sin 4x}$ $=4\cot 4x$.
b) $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{1+\tan 2x}{1-\tan 2x}$
Ta có: $\frac{1-2\sin^{2}2x}{1-\sin 4x}=\frac{\cos 4x}{\left( \cos 2x-\sin 2x \right)^{2}}$ $=\frac{\cos^{2}2x-\sin^{2}2x}{\left( \cos 2x-\sin 2x \right)^{2}}$ $=\frac{\cos 2x+\sin 2x}{\cos 2x-\sin 2x}$ $=\frac{1+\tan 2x}{1-\tan 2x}$ (do $\cos 2x\neq 0$).
c) $\frac{1}{\cos^{6}x}-\tan^{6}x=\frac{3\tan^{2}x}{\cos^{2}x}+1$
Ta có: $\frac{1}{\cos^{6}x}-\tan^{6}x=\left( \frac{1}{\cos^{2}x} \right)^{3}-\tan^{6}x$ $=\left( 1+\tan^{2}x \right)^{3}-\tan^{6}x$ $=1+3\tan^{2}x+3\tan^{4}x+\tan^{6}x-\tan^{6}x$ $=1+3\tan^{2}x\left( 1+\tan^{2}x \right)$ $=\frac{3\tan^{2}x}{\cos^{2}x}+1$.
d) $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$
Ta có: $\tan 4x-\frac{1}{\cos 4x}=\frac{\sin 4x}{\cos 4x}-\frac{1}{\cos 4x}$ $=\frac{\sin 4x-1}{\cos 4x}$ $=\frac{-\left( \cos 2x-\sin 2x \right)^{2}}{\cos^{2}2x-\sin^{2}2x}$ $=\frac{\sin 2x-\cos 2x}{\sin 2x+\cos 2x}$.
e) $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$
Ta có: $\tan\left( 2x+4x \right)=\frac{\tan 2x+\tan 4x}{1-\tan 2x\tan 4x}$.
Suy ra: $\tan 6x-\tan 2x\tan 4x\tan 6x=\tan 2x+\tan 4x$.
Do đó: $\tan 6x-\tan 4x-\tan 2x=\tan 2x.\tan 4x.\tan 6x$.
f) $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$
Ta có: $\sin x\left( 1+2\cos 2x+2\cos 4x+2\cos 6x \right)$
$=\sin x+2.\frac{1}{2}\left( \sin 3x-\sin x \right)+2.\frac{1}{2}\left( \sin 5x-\sin 3x \right)+2.\frac{1}{2}\left( \sin 7x-\sin 5x \right)$
$=\sin x+\sin 3x-\sin x+\sin 5x-\sin 3x+\sin 7x-\sin 5x$
$=\sin 7x$.
Suy ra: $\frac{\sin 7x}{\sin x}=1+2\cos 2x+2\cos 4x+2\cos 6x$.
g) $\cos 5x.\cos 3x+\sin 7x.\sin x=\cos 2x.\cos 4x$
Ta có: $\cos 5x.\cos 3x+\sin 7x.\sin x=\frac{1}{2}\left( \cos 8x+\cos 2x \right)+\frac{1}{2}\left( \cos 6x-\cos 8x \right)$
$=\frac{1}{2}\left( \cos 2x+\cos 6x \right)$ $=\frac{1}{2}.2\cos 4x.\cos 2x$ $=\cos 2x\cos 4x$.
h) Cho $\sin\left( 2a+b \right)=5\sin b$. Chứng minh: $\frac{2\tan\left( a+b \right)}{\tan a}=3$.
Ta có: $\frac{2\tan\left( a+b \right)}{\tan a}=2.\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}.\frac{\cos a}{\sin a}$ $=2.\frac{\sin\left( 2a+b \right)+\sin b}{\sin\left( 2a+b \right)-\sin b}$ $=2.\frac{6\sin b}{4\sin b}$ $=3$.
i) Cho $\tan\left( a+b \right)=3\tan a$. Chứng minh: $\sin\left( 2a+2b \right)+\sin 2a=2\sin 2b$.
Ta có: $\tan\left( a+b \right)=3\tan a$
$\Leftrightarrow \frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}=3\frac{\sin a}{\cos a}$
$\Leftrightarrow \sin\left( a+b \right)\cos a=3\sin a\cos\left( a+b \right)$
$\Leftrightarrow \frac{1}{2}\left[ \sin\left( 2a+b \right)+\sin b \right]=\frac{3}{2}\left[ \sin\left( 2a+b \right)-\sin b \right]$ $\Leftrightarrow \sin\left( 2a+b \right)+\sin b=3\sin\left( 2a+b \right)-3\sin b$ $\Leftrightarrow \sin\left( 2a+b \right)=2\sin b$.
Khi đó: $\sin\left( 2a+2b \right)+\sin 2a=2\sin\left( 2a+b \right)\cos b$ $=2.2\sin b.\cos b$ $=2\sin 2b$.