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1) a) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=tan^2\alpha.cos^2=sin^2\alpha\left(đpcm\right)\)
b) ta có : \(cos^2\alpha+tan^2\alpha.cos^2\alpha=cos^2\alpha.\left(1+tan^2\alpha\right)\)
\(=cos^2\alpha\left(1+\dfrac{sin^2\alpha}{cos^2\alpha}\right)=cos^2\alpha\left(\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\right)=cos^2\alpha.\dfrac{1}{cos^2\alpha}=1\left(đpcm\right)\)
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=25\)
Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=9\)
b, \(\sin\alpha+\cos\alpha=1,4\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1,96\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha=1,96\\ \Leftrightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1,96-1}{2}=\dfrac{0,96}{2}=0,48\)
\(\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha\cdot\cos^2\alpha\\ =1^2+2\left(\sin\alpha\cdot\cos\alpha\right)^2=1+2\cdot\left(0,48\right)^2=1,4608\)
2/ \(\frac{sin^3a-cos^3a}{sin^3a+cos^3a}=\frac{tan^3a-1}{tan^3a+1}=\frac{3^3-1}{3^3+1}=\frac{13}{14}\) (chia tử mẫu cho cos3a)
1) a) ta có : \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)=tan^2\alpha\left(cos^2\alpha+cos^2\alpha+sin^2\alpha-1\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}\left(cos^2\alpha\right)=sin^2\alpha\)
b) \(\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha+cos^2\alpha-cos^2\alpha\)
\(=sin^2\alpha\)
2) a) ta có : \(6^2+8^2=10^2\Leftrightarrow AB^2+AC^2=BC^2\)
áp dụng Pytago \(\Rightarrow\) tam giác \(ABC\) là tam giác vuông tại \(A\)
b) ta có : \(sinB=\dfrac{AC}{BC}=\dfrac{8}{10}=\dfrac{4}{5}\Rightarrow\widehat{B}\simeq53^o\)
\(\Rightarrow\widehat{C}\simeq180-90-53=37^o\)
