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\(\frac{\cos a-\sin a}{cosa+sina}=\frac{\frac{cosa}{cosa}-\frac{sina}{cosa}}{\frac{cosa}{cosa}+\frac{sina}{cosa}}\)(chia ca tu va mau cho cosa)
\(=\frac{1-tana}{1+tana}=vt\left(dpcm\right)\)
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.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(0,8\right)^2=1-0,64=0,36\)
\(\Rightarrow\sin\alpha=0,6\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,6}{0,8}=\frac{3}{4}\)
\(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=25\)
Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=9\)
b, \(\sin\alpha+\cos\alpha=1,4\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1,96\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha=1,96\\ \Leftrightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1,96-1}{2}=\dfrac{0,96}{2}=0,48\)
\(\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha\cdot\cos^2\alpha\\ =1^2+2\left(\sin\alpha\cdot\cos\alpha\right)^2=1+2\cdot\left(0,48\right)^2=1,4608\)
tuỳ ý giá trị biểu thức
\(\tan\alpha=\frac{\sin\alpha}{\cos a}=\frac{3}{4}\)
\(\Rightarrow\sin\alpha=\frac{3}{4}.\cos\alpha\)
Ta có : \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\left(\frac{3}{4}.\cos\alpha\right)^2+\cos^2\alpha=1\)
\(\Rightarrow\frac{25}{16}.\cos^2\alpha=1\)
\(\Rightarrow\cos\alpha=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\Rightarrow\sin\alpha=\sqrt{1-\cos^2\alpha}=\frac{3}{5}\)