Cho cot a=5 ( a là góc nhọn ) Tính sin a, tan a, cos a

Đáp án + Giải thích các bước giải:

Ta có: `tan\alpha.cot\alpha=1`

`=>tan\alpha=1/(cot\alpha)`

`=>tan\alpha=1/5`

Ta có: `1/(cos^2\alpha)=1+tan^2\alpha`

`=>1/(cos^2\alpha)=1+(1/5)^2`

`=>1/(cos^2\alpha)=1+1/25`

`=>1/(cos^2\alpha)=26/25`

`=>cos^2\alpha=25/26`

`=>cos\alpha=(\sqrt{26})/5`

Ta có: `cos^2\alpha+sin^2\alpha=1`

`=>25/26+sin^2\alpha=1`

`=>sin^2\alpha=1/26`

`=>sin\alpha=1/(\sqrt{26})`

Đáp án:

$\sin\alpha=\dfrac{\sqrt{26}}{26}\\\cos\alpha=\dfrac{5\sqrt{26}}{26}\\\tan\alpha=\dfrac{1}{5}$

Giải thích các bước giải:

Do $\alpha$ là góc nhọn nên $0<\sin \alpha,\cos \alpha<1;\;\tan \alpha,\cot \alpha>0$

Ta có:

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

+) $1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}$

$\Leftrightarrow 1+\dfrac{1}{25}=\dfrac{1}{\cos^2\alpha}$

$\Rightarrow\cos^2\alpha=\dfrac{25}{26}$

$\Rightarrow \cos\alpha=\dfrac{5\sqrt{26}}{26}$

+) $\sin^2\alpha+\cos^2\alpha=1$

$\Leftrightarrow \sin^2\alpha+\dfrac{25}{26}=1$

$\Leftrightarrow \sin^2\alpha=\dfrac{1}{26}$

$\Rightarrow \sin\alpha=\dfrac{\sqrt{26}}{26}$