K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a: sin a=2/3

=>cos^2a=1-(2/3)^2=5/9

=>\(cosa=\dfrac{\sqrt{5}}{3}\)

\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)

\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

b: cos a=1/5

=>sin^2a=1-(1/5)^2=24/25

=>\(sina=\dfrac{2\sqrt{6}}{5}\)

\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

c: cot a=1/tana=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>1/cos^2a=1+4=5

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

NV
26 tháng 7 2021

Lớp 9 nên coi như các góc này đều nhọn

a.

\(cosa=\sqrt{1-sin^2a}=\dfrac{15}{17}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{8}{15}\)

\(cota=\dfrac{1}{tana}=\dfrac{15}{8}\)

b.

\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)

\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)

\(tana=\dfrac{1}{cota}=\dfrac{4}{3}\)

a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)

\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)

\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)

5 tháng 8 2023

\(tanx=\dfrac{4}{3}\)

\(\Rightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

\(\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}\)

\(=\dfrac{1}{1+\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{1+\dfrac{16}{9}}=\dfrac{1}{\dfrac{25}{9}}=\dfrac{9}{25}\)

\(\Rightarrow cosx=\dfrac{3}{5}\)

\(sin^2x+cos^2x=1\)

\(\Rightarrow sin^2x=1-cos^2x=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)

\(\Rightarrow sinx=\dfrac{4}{5}\)

5 tháng 8 2023

Có \(tan.\alpha=\dfrac{4}{3}\)

Mà \(tan.\alpha.cot.\alpha=1\)

\(\Rightarrow cot.\alpha=1:\dfrac{4}{3}=\dfrac{3}{4}\)

Lại có \(sin^2\alpha+cos^2\alpha=1\\ \Leftrightarrow sin^2\alpha=1-cos^2\alpha\\ \Leftrightarrow sin\alpha=\sqrt{1-cos^2\alpha}\)

Vì \(tan.\alpha=\dfrac{sin.\alpha}{cos.\alpha}\)

\(\Leftrightarrow\dfrac{4}{3}=\dfrac{\sqrt{1-cos^2\alpha}}{cos.\alpha}\)

\(\Leftrightarrow\dfrac{4}{3}=\dfrac{1-cos^2\alpha}{cos^2\alpha}\\ \Leftrightarrow4.cos^2\alpha=3.\left(1-cos^2\alpha\right)\\ \Leftrightarrow4.cos^2\alpha=3-3cos^2\alpha\\ \Leftrightarrow cos.\alpha=\dfrac{\sqrt{21}}{7}\)

\(\Rightarrow sin.\alpha=\sqrt{1-\left(\dfrac{\sqrt{21}}{7}\right)^2}=\dfrac{4}{7}\)

 

 

 

 

4 tháng 8 2023

\(sin\) \(a\) \(=\dfrac{\sqrt{3}}{2}\Rightarrow\widehat{a}=60^o\)

\(\Rightarrow cos\) \(a\) \(=cot\) \(60^o\) \(=\dfrac{1}{2}\)

\(tan\) \(a\) \(=tan\) \(60^o\) \(=\sqrt{3}\)

\(cot\) \(a\) \(\dfrac{1}{tana}\) \(=\dfrac{1}{\sqrt{3}}\) \(=\dfrac{\sqrt{3}}{\sqrt{3}}\)