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Ta có \(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{8}{17}\right)^2=\frac{225}{289}\)
Vậy B=\(4.\left(\frac{8}{17}\right)^2+3.\frac{225}{289}=\)\(\frac{931}{289}\)
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#
\(\sin\alpha=\frac{8}{17}\Rightarrow sin^2\alpha=\frac{64}{289}\Rightarrow cos^2\alpha=1-sin^2\alpha=1-\frac{64}{289}=\frac{225}{289}\)
\(\Rightarrow cos\alpha=\frac{15}{17}\)
từ đó tính ra \(tan\alpha;cot\alpha\)
Ta có: \(\sin^2\alpha+\tan^2\alpha=1\)
\(\Leftrightarrow\frac{64}{289}+\tan^2\alpha=1\)
\(\Leftrightarrow\tan^2\alpha=\frac{225}{289}\)
\(\Rightarrow\tan\alpha=\frac{15}{17}\)
Đến đây thì dễ rồi:
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{15}{8}\) ; \(\cot\alpha=\frac{8}{15}\)
\(M=\frac{\sin^3a+3\cos^3a}{27\sin^3a-25\cos^3a}\)
\(M=\frac{\frac{\sin^3a+3\cos^3a}{\cos^3a}}{\frac{27\sin^3a-25\cos^3a}{\cos^3a}}\)
\(M=\frac{\tan^3a+3}{27\tan^3a-25}\)
\(M=\frac{\frac{8}{27}+3}{27.\frac{8}{27}-25}\)
\(M=\frac{\frac{89}{27}}{-17}\)
\(M=-\frac{89}{459}\)
P/s haphuong