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\(\cos^4\alpha-\sin^4\alpha+1\\ =\left(\sin^2\alpha+\cos^2\alpha\right)\left(-\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\cos^2\alpha\right)\\ =-\sin^2\alpha+\cos^2\alpha+\sin^2\alpha+\cos^2\alpha=2\cos^2\alpha\)
\(cos^4a-sin^4a+1=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1\)
\(=cos^2a-sin^2a+1=cos^2a-sin^2a+sin^2a+cos^2a=2cos^2a\)
Vậy ta có đpcm
b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm
bài 1 : ta có : \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(0,6\right)^2=\dfrac{16}{25}\)
\(\Rightarrow cosa=\pm\dfrac{4}{5}\)
\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\pm\dfrac{3}{4}\) \(\Rightarrow cotx=\dfrac{1}{tanx}=\pm\dfrac{4}{3}\)
bài 2)
ý 1 : a) ta có : \(\dfrac{1}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=tan^2a+1\left(đpcm\right)\)
b) ta có : \(\dfrac{1}{sin^2a}=\dfrac{sin^2a+cos^2a}{sin^2a}=1+cot^2a\left(đpcm\right)\)
c) \(cos^4a-sin^4a=\left(sin^2a+cos^2a\right)\left(cos^2a-sin^2a\right)\)
\(=cos^2a-sin^2a=2cos^2a-cos^2a-sin^2a=2cos^2a-1\left(đpcm\right)\)
ý 2 :
ta có : \(tana=2\Rightarrow cota=\dfrac{1}{2}\)
ta có : \(tan^2a+1=\dfrac{1}{cos^2a}\Leftrightarrow cos^2a=\dfrac{1}{tan^2a+1}=\dfrac{1}{5}\)
\(\Rightarrow cosa=\pm\dfrac{1}{\sqrt{5}}\Rightarrow sin^2a=1-cos^2a=\dfrac{4}{5}\) \(\Rightarrow sina=\pm\dfrac{2}{\sqrt{5}}\)
vậy ............................................................................
bài 3 bạn tự luyện tập như bài 2 cho quen nha :)
a) \(\tan^2\alpha+1=\frac{\sin^2\alpha}{\cos^2\alpha}+1=\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha}\)
b) \(\cot^2\alpha+1=\frac{\cos^2\alpha}{\sin^2\alpha}+1=\frac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}=\frac{1}{\sin^2\alpha}\)
c) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)
\(=2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2-1\)