Biết tan = 4/3 tính sin, cos, cot
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9 tháng 8 2023
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
4 tháng 8 2023
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
QE
2
NV
Nguyễn Việt Lâm
Giáo viên
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.
\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)
\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)
\(tana=\dfrac{1}{cota}=\dfrac{4}{3}\)
26 tháng 7 2021
a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)
\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)
\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
NT
1
NT
1
14 tháng 8 2019
Ta có : \(\tan\alpha.\cot\alpha=1\); \(1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\); \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\cot\alpha=\frac{1}{\tan\alpha}=\frac{4}{3}\); \(\frac{1}{\cos^2\alpha}=\frac{25}{16}\Rightarrow\cos\alpha=\frac{4}{5}\); \(\sin\alpha=\tan\alpha.\cos\alpha=\frac{3}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2022
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
\(tanx=\dfrac{4}{3}\)
\(\Rightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
\(\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}\)
\(=\dfrac{1}{1+\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{1+\dfrac{16}{9}}=\dfrac{1}{\dfrac{25}{9}}=\dfrac{9}{25}\)
\(\Rightarrow cosx=\dfrac{3}{5}\)
\(sin^2x+cos^2x=1\)
\(\Rightarrow sin^2x=1-cos^2x=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow sinx=\dfrac{4}{5}\)
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 \(tan.\alpha=\dfrac{sin.\alpha}{cos.\alpha}\)
\(\Leftrightarrow\dfrac{4}{3}=\dfrac{\sqrt{1-cos^2\alpha}}{cos.\alpha}\)
\(\Leftrightarrow\dfrac{4}{3}=\dfrac{1-cos^2\alpha}{cos^2\alpha}\\ \Leftrightarrow4.cos^2\alpha=3.\left(1-cos^2\alpha\right)\\ \Leftrightarrow4.cos^2\alpha=3-3cos^2\alpha\\ \Leftrightarrow cos.\alpha=\dfrac{\sqrt{21}}{7}\)
\(\Rightarrow sin.\alpha=\sqrt{1-\left(\dfrac{\sqrt{21}}{7}\right)^2}=\dfrac{4}{7}\)