Cho góc nhọn \(\alpha\) biết \(tan\alpha=\frac{3}{4}\) .Tính \(sin\alpha\) và \(cos\alpha\) (bằng 2 cách).
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3 tháng 7 2017
\(\frac{\cos a-\sin a}{cosa+sina}=\frac{\frac{cosa}{cosa}-\frac{sina}{cosa}}{\frac{cosa}{cosa}+\frac{sina}{cosa}}\)(chia ca tu va mau cho cosa)
\(=\frac{1-tana}{1+tana}=vt\left(dpcm\right)\)
Y
24 tháng 9 2023
\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 9 2023
Ta có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)
\( \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{10}} \Leftrightarrow \cos \alpha = \pm \frac{{\sqrt {10} }}{{10}}\)
Vì \({0^o} < \alpha < {180^o}\) nên \(\sin \alpha > 0\).
Mà \(\tan \alpha = 3 > 0 \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \frac{{\sqrt {10} }}{{10}}\)
Lại có: \(\sin \alpha = \cos \alpha .\tan \alpha = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)
\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2022
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \({\cos ^2}\alpha + {\sin ^2}\alpha = 1\)
b) \(\tan \alpha .\cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\)
c) \(\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha + 1\)
d) \(\frac{1}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha \)
AN
18 tháng 10 2019
1.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}\)
\(\Rightarrow\cos\alpha=\frac{4}{5}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\)
\(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\)
2.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(0,8\right)^2=1-0,64=0,36\)
\(\Rightarrow\sin\alpha=0,6\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,6}{0,8}=\frac{3}{4}\)
\(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
8 tháng 11 2021
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=25\)
Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=9\)
b, \(\sin\alpha+\cos\alpha=1,4\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1,96\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha=1,96\\ \Leftrightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1,96-1}{2}=\dfrac{0,96}{2}=0,48\)
\(\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha\cdot\cos^2\alpha\\ =1^2+2\left(\sin\alpha\cdot\cos\alpha\right)^2=1+2\cdot\left(0,48\right)^2=1,4608\)
\(\tan\alpha=\frac{\sin\alpha}{\cos a}=\frac{3}{4}\)
\(\Rightarrow\sin\alpha=\frac{3}{4}.\cos\alpha\)
Ta có : \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\left(\frac{3}{4}.\cos\alpha\right)^2+\cos^2\alpha=1\)
\(\Rightarrow\frac{25}{16}.\cos^2\alpha=1\)
\(\Rightarrow\cos\alpha=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\Rightarrow\sin\alpha=\sqrt{1-\cos^2\alpha}=\frac{3}{5}\)