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Lời giải:
a)
\(\cos ^2a+\cos ^2b+\cos ^2a\sin ^2b+\sin ^2a\)
\(=(\cos ^2a+\sin ^2a)+\cos ^2b+\cos ^2a\sin ^2b\)
\(=1+1-\sin ^2b+\cos ^2a\sin ^2b\)
\(=2-\sin ^2b(1-\cos ^2a)=2-\sin ^2b\sin ^2a\)
b)
\(2(\sin a-\cos a)^2-[(\sin a+\cos a)^2+\sin a\cos a]\)
\(=2(\sin ^2a-2\sin a\cos a+\cos ^2a)-[\sin ^2+2\sin a\cos a+\cos ^2a+\sin a\cos a]\)
\(=2(1-2\sin a\cos a)-(1+3\sin a\cos a)\)
\(=1-7\sin a\cos a\)
c)
\((\tan a-\cot a)^2-(\tan a+\cot a)^2\)
\(=\tan ^2a+\cot ^2a-2\tan a\cot a-(\tan ^2a+\cot ^2a+2\tan a\cot a)\)
\(=-4\tan a\cot a=-4\)
1.Cậu bấm máy tính
Có cosα=0,6 →α=cos-1(0,6)≃53,130.Từ đó ta có tanα=tan53,130≃1,33. cotα=1/tan53,130≃0,75. Tương tự các câu còn lại. 2.Dùng các CT lượng giác
ta có \(\sin\alpha:\cos\alpha=\frac{đ}{h}:\frac{k}{h}=\frac{đ}{h}.\frac{h}{k}=\frac{đ}{k}=\tan\alpha \left(đpcm\right)\)
Góc B bằng alpha suy ra:
\(\frac{sin\alpha}{cos\alpha}=\frac{\frac{AC}{BC}}{\frac{AB}{BC}}=\frac{AC}{AB}=tan\alpha\)(đpcm)
a: cos a=0.8
tan a=0,6/0,8=3/4
b: \(sina=\sqrt{1-0.7^2}=\dfrac{\sqrt{51}}{10}\)
\(tana=\dfrac{\sqrt{51}}{7}\)
c: \(1+tan^2a=\dfrac{1}{cos^2a}=1.64\)
\(\Leftrightarrow cos^2a=\dfrac{25}{41}\)
=>\(cosa=\dfrac{5}{\sqrt{41}}\)
=>\(sina=\sqrt{1-\dfrac{25}{41}}=\sqrt{\dfrac{16}{41}}\)
a)\(\sin\alpha=\dfrac{9}{15}\Rightarrow\sin^2\alpha=\dfrac{81}{225}\)
Có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{81}{225}=\dfrac{144}{225}\)
\(\Rightarrow\cos\alpha=\sqrt{\dfrac{144}{225}}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{9}{15}:\dfrac{4}{5}=\dfrac{3}{4}\)
\(\cot\alpha=\dfrac{\cos\alpha}{\tan\alpha}=\dfrac{4}{5}:\dfrac{9}{15}=\dfrac{4}{3}\)
b)\(\cos\alpha=\dfrac{3}{5}\Rightarrow\cos^2\alpha=\dfrac{9}{25}\)
Có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow\sin\alpha=\dfrac{4}{5}\)
\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
\(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\)