Bài 9/

a/ $\tan\alpha=\dfrac{1}{2}\\↔\alpha≈26,6^\circ\\→\sin\alpha=\sin 26,6^\circ≈0,45$

Vậy $\sin\alpha≈0,45$

b/ $\cos\alpha=\cos26,6^\circ≈0,9\\\dfrac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\\=\dfrac{0,9+0,45}{0,9-0,45}\\=\dfrac{1,35}{0,45}\\=3$

Vậy $\dfrac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=3$

Bài 10/

a/ $A=(\sin\alpha+\cos\alpha)^2+(\sin\alpha-\cos\alpha)^2\\=\sin^2\alpha+2\sin\alpha.\cos\alpha+\sin^2\alpha+\sin^2\alpha-2\sin\alpha.\cos\alpha+\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)+(\sin^2\alpha+\cos^2\alpha)+(2\sin\alpha.\cos\alpha-2\sin\alpha.\cos\alpha)\\=1+1\\=2$

Vậy biểu thức không phụ thuộc vào $\alpha$

b/ $B=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)(\sin^4\alpha-\sin^2\alpha.\cos^2\alpha+\cos^4\alpha)+3\sin^2\alpha.\cos^2\alpha\\=1(\sin^4\alpha+2\sin^2\alpha.\cos^2\alpha+\cos^4\alpha-3\sin^2\alpha.\cos^2\alpha)+3\sin^2\alpha.\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)^2-3\sin^2\alpha.\cos^2\alpha\\=1^2-0\\=1$

Vậy biểu thức không phụ thuộc vào $\alpha$

Bài 11/

$\sin\alpha=\dfrac{5}{13}\\↔\sin^2\alpha=\dfrac{25}{169}$

mà $\sin^2\alpha+\cos^2\alpha=1$

$→\dfrac{25}{169}+\cos^2\alpha=1\\↔\cos^2\alpha=\dfrac{144}{169}\\↔\cos\alpha=±\dfrac{12}{13}\\\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{5}{13}}{±\dfrac{12}{13}}=±\dfrac{5}{12}\\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{±\dfrac{12}{13}}{\dfrac{5}{13}}=±\dfrac{52}{5}$

Vậy $\cos\alpha=\dfrac{12}{13},\tan\alpha=±\dfrac{5}{12},\cot\alpha=±\dfrac{52}{5}$

Bài 9:

a, Ta có: `tan\alpha=1/2`

`=>{sin\alpha}/{cos\alpha}=1/2<=>cos\alpha=2sin\alpha`

Lại có: `sin^2\alpha+cos^2\alpha=1`

`<=>sin^2\alpha+4sin^2\alpha=1`

`<=>5sin^2\alpha=1`

`<=>sin^2\alpha=1/5`

`<=>sin\alpha=\pm 1/{\sqrt5}`

b, `{cos\alpha+sin\alpha}/{cos\alpha-sin\alpha}={1+tan\alpha}/{1-tan\alpha}=`$\dfrac{1+\dfrac12}{1-\dfrac12}=3$

Bài 10:

a, `A=(sin\alpha+cos\alpha)^2+(sin\alpha-cos\alpha)^2`

`=sin^2\alpha+2sin\alphacos\alpha+cos^2\alpha+sin^2\alpha-2sin\alphacos\alpha+cos^2\alpha`

`=2sin^2\alpha+2cos\alpha=2(sin^2\alpha+cos^2\alpha)=2`

b, `B=sin^6\alpha+cos^6\alpha+3sin^2\alphacos^2\alpha`

`=(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alphacos^2\alpha+cos^4\alpha)+3sin^2\alphacos^2\alpha`

`=sin^4\alpha+2sin^2\alphacos^\alpha+cos^4\alpha` `=(sin^2\alpha+cos^2\alpha)^2=1`

Bài 11:

Ta có: `sin^2\alpha+cos^2\alpha=1`

`<=>25/169+cos^2\alpha=1`

`<=>cos^2\alpha=144/169`

`<=>cos\alpha=\pm 12/13`

+) Với `cos\alpha=-12/13` ta có:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac5{13}}{-\dfrac{12}{13}}=-\dfrac5{12}$

$\Rightarrow \cot\alpha=\dfrac{1}{\tan\alpha}=-\dfrac{12}5$

+) Với `cos\alpha=12/13` ta có:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac5{13}}{\dfrac{12}{13}}=\dfrac5{12}$

$\Rightarrow \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{12}5$