Câu hỏi của khanh hoa

Question

cho tanα = $\dfrac{3}{4}$

tính sin

HT.Phong (9A5) · Accepted Answer

Ta có:

$cot\alpha\cdot tan\alpha=1$

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

$\Rightarrow cota=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$

Mà:

$cot^2\alpha+1=\dfrac{1}{sin^2\alpha}$

$\Rightarrow sin\alpha=\sqrt{\dfrac{1}{cot^2\alpha+1}}$

$\Rightarrow sin\alpha=\sqrt{\dfrac{1}{\left(\dfrac{4}{3}\right)^2+1}}=\dfrac{3}{5}$

Lại có:

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

$\Rightarrow cos\alpha=\sqrt{1-sin^2a}$

$\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

Câu trả lời:

Ta có:

$cot\alpha\cdot tan\alpha=1$

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

$\Rightarrow cota=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$

Mà:

$cot^2\alpha+1=\dfrac{1}{sin^2\alpha}$

$\Rightarrow sin\alpha=\sqrt{\dfrac{1}{cot^2\alpha+1}}$

$\Rightarrow sin\alpha=\sqrt{\dfrac{1}{\left(\dfrac{4}{3}\right)^2+1}}=\dfrac{3}{5}$

Lại có:

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

$\Rightarrow cos\alpha=\sqrt{1-sin^2a}$

$\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

Gia Huy · Answer

$tan\alpha=\dfrac{3}{4}\ \Rightarrow cot\alpha=1:\dfrac{3}{4}=\dfrac{4}{3}$

Có:

$1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\ \Rightarrow sin\alpha=\sqrt{1:\left(1+\left(\dfrac{4}{3}\right)^2\right)}=\dfrac{3}{5}$

$\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

Câu trả lời:

$tan\alpha=\dfrac{3}{4}\ \Rightarrow cot\alpha=1:\dfrac{3}{4}=\dfrac{4}{3}$

Có:

$1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\ \Rightarrow sin\alpha=\sqrt{1:\left(1+\left(\dfrac{4}{3}\right)^2\right)}=\dfrac{3}{5}$

$\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$