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![](https://rs.olm.vn/images/avt/0.png?1311)
ta co \(sin^2a+cos^2a=1\Rightarrow cosa=0.36\)
\(\frac{sina}{cosa}=tana\Rightarrow tana=\frac{20}{9}\)
\(tana\cdot cotga=1\Rightarrow cotga=\frac{9}{20}\)
câu b tương tự nha cau c \(\frac{sina+cosa}{sina-cosa}=\) bn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sin\alpha=\frac{2}{5}\)
\(\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\)
\(=\sqrt{1-\frac{4}{25}}\)
\(=\sqrt{\frac{21}{25}}=\)\(\frac{\sqrt{21}}{5}\)
\(\Rightarrow\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{5}:\frac{\sqrt{21}}{5}=\frac{2}{\sqrt{21}}\)và \(\cot\alpha=\frac{\sqrt{21}}{2}\)
2. Tương tự a)
\(\cos B=\sqrt{1-\sin^2B}\)
\(=\sqrt{1-\frac{1}{4}}\)
\(=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)
\(\tan B,\cot B\)bạn tự tính nốt.
\(sin\alpha=0,4\Rightarrow sin^2\alpha=0,16\Rightarrow cos^2\alpha=1-sin^2\alpha=1-0,16=0,84\Rightarrow cos\alpha=\frac{\sqrt{21}}{5}\)
\(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{0,4}{\frac{\sqrt{21}}{5}}=\frac{2\sqrt{21}}{21}\)
\(cot\alpha=1:sin\alpha=1:\frac{2\sqrt{21}}{21}=\frac{21}{2\sqrt{21}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\cos\alpha=\dfrac{4}{5}\)
\(\tan\alpha=\dfrac{3}{4}\)
\(\cot\alpha=\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)
hay \(\cos\alpha=\dfrac{4}{5}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)
\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)
\(=\dfrac{141}{25}\)
c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: cos 88 0 < sin 40 0 (= cos 50 0 ) < cos 28 0 < sin 65 0 (= cos 25 0 ) < cos 20 0
b, Ta có: cot 67 0 18 ' (= tan 22 0 42 ' ) < tan 32 0 48 ' < tan 56 0 32 ' < cot 28 0 36 ' (= tan 61 0 24 ' )
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: cos 70 0 (= sin 20 0 ) < sin 24 0 < sin 54 0 < cos 35 0 (= sin 55 0 ) < sin 78 0
b, Ta có: tan 16 0 (= cot 74 0 ) < cot 57 0 67 ' < cot 30 0 < cot 24 0 < tan 80 0 (= cot 10 0 )
Làm tiêu biểu 1 bài thôi nhé. Các bài còn lại tương tự
a/ sin a = 0,8
Ta có: sin2 a + cos2 a = 1
=> cos2 a = 1 - sin2 a = 1 - 0,82 = 0,36
\(\Rightarrow\orbr{\begin{cases}cos\:a=0,6\\cos\:a=-0,6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}tan\:a=\frac{4}{3}\\tan\:a=-\frac{4}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}cot\:a=\frac{3}{4}\\cot\:a=-\frac{3}{4}\end{cases}}\)