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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) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)
\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)
\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)
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
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
a) \(\frac{1}{cos^2x}=1+tan^2x=1+\frac{9}{16}=\frac{25}{16}\)
\(\Leftrightarrow cos^2x=\frac{16}{25}\Leftrightarrow\orbr{\begin{cases}cosx=\frac{4}{5}\\cosx=\frac{-4}{5}\end{cases}}\)
- \(cosx=\frac{4}{5}\):
\(sinx=cosxtanx=\frac{4}{5}.\frac{3}{4}=\frac{3}{5}\)
\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).
- \(cosx=\frac{-4}{5}\):
\(sinx=cosxtanx=\frac{-4}{5}.\frac{3}{4}=\frac{-3}{5}\)
\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).
b) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{49}{625}=\frac{576}{625}\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{24}{25}\\cosx=-\frac{24}{25}\end{cases}}\)
- \(cosx=\frac{24}{25}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)
- \(cosx=\frac{-24}{25}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{-24}{25}}=-\frac{7}{24}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\frac{7}{24}}=\frac{-24}{7}\)
a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)
- \(cosx=\frac{1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
- \(cosx=\frac{-1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)
b) Bạn làm tương tự câu a) nha.
1+\(cot^2\)x=\(\dfrac{1}{sin^2x}\)\(\Leftrightarrow\)1+\(\dfrac{3}{4}^2\)=\(\dfrac{1}{sin^2x}\)➩\(sin^2x\)=\(\dfrac{16}{25}\)\(\Rightarrow\)sinx=\(\dfrac{4}{5}\)
\(\sin^2x+\cos^2x=1\)\(\Rightarrow\)cosx=\(\sqrt{1-sin^2x}\)=\(\dfrac{3}{5}\)
tanx=\(\dfrac{\sin x}{\cos x}\)=\(\dfrac{4}{3}\)
\(\tan a=\dfrac{1}{\cot a}=\dfrac{15}{8}=\dfrac{\sin a}{\cos a}\\ \Rightarrow\sin a=\dfrac{15}{8}\cos a\\ \sin^2a+\cos^2a=1\\ \Rightarrow\dfrac{225}{64}\cos^2a+\cos^2a=1\\ \Rightarrow\dfrac{289}{64}\cos^2a=1\Rightarrow\cos^2a=\dfrac{64}{289}\\ \Rightarrow\cos a=\dfrac{8}{17}\Rightarrow\sin a=\dfrac{15}{17}\)
a) cos = 15/7
tan = 8/15
cot = 15/8
b) cos = 4/5
tan = 3/5
cot = 4/5