Bài tự luận · Bài 34
Dạng 2. Áp dụng tính chất của giá trị lượng giác Phương pháp: Giả sử $M$ có toạ độ $(x;y)$. Ta có: $\cos\alpha =x;\sin\alpha =y$ $\tan\alpha =\frac{\sin\alpha}{\cos\alpha}=\frac{y}{x}(\cos\alpha =x\neq 0);\cot\alpha =\frac{\cos\alpha}{\sin\alpha}=\frac{x}{y}(\sin\alpha =y\neq 0).$ - $\cos^{2}\alpha +\sin^{2}\alpha =1$ với mọi $\alpha$; - $\tan\alpha \cdot \cot\alpha =1$ với $\cos\alpha \neq 0,\sin\alpha \neq 0$; - $1+\tan^{2}\alpha =\frac{1}{\cos^{2}\alpha}$ với $\cos\alpha \neq 0$ - $1+\cot^{2}\alpha =\frac{1}{\sin^{2}\alpha}$ với $\sin\alpha \neq 0$.
Bài tự luận Giá trị lượng giác của góc lượng giác · Bài 34
(CTST 11) Chứng minh các đẳng thức lượng giác sau:
a) $\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x$;
b) $\frac{1+\cot x}{1-\cot x}=\frac{\tan x+1}{\tan x-1}$;
c) $\frac{\sin\alpha+\cos\alpha}{\sin^{3}\alpha}=\frac{1-\cot^{4}\alpha}{1-\cot\alpha}$;
d) $\frac{\tan^{2}\alpha+\cos^{2}\alpha-1}{\cot^{2}\alpha+\sin^{2}\alpha-1}=\tan^{6}\alpha$.
a) $\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x$;
b) $\frac{1+\cot x}{1-\cot x}=\frac{\tan x+1}{\tan x-1}$;
c) $\frac{\sin\alpha+\cos\alpha}{\sin^{3}\alpha}=\frac{1-\cot^{4}\alpha}{1-\cot\alpha}$;
d) $\frac{\tan^{2}\alpha+\cos^{2}\alpha-1}{\cot^{2}\alpha+\sin^{2}\alpha-1}=\tan^{6}\alpha$.
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Lời giải
a) $\sin^{4}x+\cos^{4}x=\left( \sin^{2}x+\cos^{2}x \right)^{2}-2\sin^{2}x\cos^{2}x=1-2\sin^{2}x\cos^{2}x$;
b) $\frac{1+\cot x}{1-\cot x}=\frac{1+\frac{1}{\tan x}}{1-\frac{1}{\tan x}}=\frac{\tan x+1}{\tan x-1}$;
c) $\frac{\sin\alpha+\cos\alpha}{\sin^{3}\alpha}=\frac{1}{\sin^{2}\alpha}+\frac{\cos\alpha}{\sin\alpha}\cdot \frac{1}{\sin^{2}\alpha}=\left( 1+\cot^{2}\alpha \right)+\cot\alpha\left( 1+\cot^{2}\alpha \right)=(1+\cot\alpha)\left( 1+\cot^{2}\alpha \right)$
$=\frac{\left( 1-\cot^{2}\alpha \right)\left( 1+\cot^{2}\alpha \right)}{1-\cot\alpha}=\frac{1-\cot^{4}\alpha}{1-\cot\alpha}$;
d)
$\begin{array}{ll} \frac{\tan^{2}\alpha+\cos^{2}\alpha-1}{\cot^{2}\alpha+\sin^{2}\alpha-1} & =\frac{\tan^{2}\alpha-\sin^{2}\alpha}{\cot^{2}\alpha-\cos^{2}\alpha}=\frac{\frac{\sin^{2}\alpha}{\cos^{2}\alpha}-\sin^{2}\alpha}{\frac{\cos^{2}\alpha}{\sin^{2}\alpha}-\cos^{2}\alpha} \\ & =\frac{\sin^{2}\alpha\left( \frac{1}{\cos^{2}\alpha}-1 \right)}{\cos^{2}\alpha\left( \frac{1}{\sin^{2}\alpha}-1 \right)}=\tan^{2}\alpha\cdot \frac{\tan^{2}\alpha}{\cot^{2}\alpha}=\tan^{6}\alpha. \end{array}$
b) $\frac{1+\cot x}{1-\cot x}=\frac{1+\frac{1}{\tan x}}{1-\frac{1}{\tan x}}=\frac{\tan x+1}{\tan x-1}$;
c) $\frac{\sin\alpha+\cos\alpha}{\sin^{3}\alpha}=\frac{1}{\sin^{2}\alpha}+\frac{\cos\alpha}{\sin\alpha}\cdot \frac{1}{\sin^{2}\alpha}=\left( 1+\cot^{2}\alpha \right)+\cot\alpha\left( 1+\cot^{2}\alpha \right)=(1+\cot\alpha)\left( 1+\cot^{2}\alpha \right)$
$=\frac{\left( 1-\cot^{2}\alpha \right)\left( 1+\cot^{2}\alpha \right)}{1-\cot\alpha}=\frac{1-\cot^{4}\alpha}{1-\cot\alpha}$;
d)
$\begin{array}{ll} \frac{\tan^{2}\alpha+\cos^{2}\alpha-1}{\cot^{2}\alpha+\sin^{2}\alpha-1} & =\frac{\tan^{2}\alpha-\sin^{2}\alpha}{\cot^{2}\alpha-\cos^{2}\alpha}=\frac{\frac{\sin^{2}\alpha}{\cos^{2}\alpha}-\sin^{2}\alpha}{\frac{\cos^{2}\alpha}{\sin^{2}\alpha}-\cos^{2}\alpha} \\ & =\frac{\sin^{2}\alpha\left( \frac{1}{\cos^{2}\alpha}-1 \right)}{\cos^{2}\alpha\left( \frac{1}{\sin^{2}\alpha}-1 \right)}=\tan^{2}\alpha\cdot \frac{\tan^{2}\alpha}{\cot^{2}\alpha}=\tan^{6}\alpha. \end{array}$