Câu hỏi của khanh hoa

Question

cho tanα = $\dfrac{3}{4}$tính sinα, cosα, cotα

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}$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP