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a, ta có \(\cos^2\alpha\)+ \(\sin^2\alpha\)= 1
1/5 + \(\cos^2\alpha\)= 1
\(\cos^2\alpha\)= 4/5
\(4\cos^2\alpha\)+6 \(\sin^2\alpha\)= 4 . 4/5 + 6.1/5=22/5
b, \(\sin\alpha\)= 2/3
\(\sin^2\alpha\)= 4/9
\(\cos^2\alpha=\frac{5}{9}\)
\(5\cos^2\alpha+2\sin^2=\frac{5.5}{9}+\frac{2.4}{9}=\frac{33}{9}\)
#mã mã#
a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha\)= 3 \(\sin\alpha\)
ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)
#mã mã#
a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)
\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)
\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
\(\frac{\sin^4\alpha-\cos^2\alpha+2\cos^4\alpha-\cos^6\alpha}{\cos^4\alpha-\sin^2\alpha+2\sin^4\alpha-\sin^6\alpha}=\frac{\sin^4\alpha-\cos^2\alpha\left(1-\cos^2\alpha\right)^2}{\cos^4\alpha-\sin^2\alpha\left(1-\sin^2\alpha\right)^2}\)
\(=\tan^4\alpha.\frac{1-\cos^2\alpha}{1-\sin^2\alpha}=\tan^6\alpha\)
ap dung sin2a+cos2a=1 =>4cos2a -6sin2a=4 -4sin2a-6sin2a=4-10sin2a=4-10.1/25=3,6