.jkilfo,o7m5ijk

 Ta có $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alpha$

$=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)$

$=\sin \alpha .$

Vậy $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin \alpha$

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)

a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=25\)

Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=9\)

b, \(\sin\alpha+\cos\alpha=1,4\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1,96\)

\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha=1,96\\ \Leftrightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1,96-1}{2}=\dfrac{0,96}{2}=0,48\)

\(\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha\cdot\cos^2\alpha\\ =1^2+2\left(\sin\alpha\cdot\cos\alpha\right)^2=1+2\cdot\left(0,48\right)^2=1,4608\)

a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1​tanα+3tanα1​​=tan2α+1tan2α+3​=cos2α1​cos2α1​+2​=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169​=817​.

b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α​+cos3α3cos3α​+cos3α2sinα​cos3αsinα​−cos3αcosα​​=tan3α+3+2tanα(tan2α+1)tanα(

Bài 1:

Áp dụng định lí pytago trong tam giác vuông ABC ta có:

BC²=AC²+AB²

BC²=4²+3²

BC=\(\sqrt{25}\)=5(cm)

Ta có:

Sin B=\(\dfrac{AC}{BC}=\dfrac{4}{5}=0.8\)

Cos B=\(\dfrac{AB}{BC}=\dfrac{3}{5}=0.6\)

Tag B=\(\dfrac{AC}{AB}=\dfrac{4}{3}\)

Cotg B=\(\dfrac{AB}{AC}=\dfrac{3}{4}=0.75\)

bài 2:

\(\sin\alpha^2+\cos\alpha^2=1\)

=>0,6²+\(\cos\alpha^2=1\)

=>\(\cos\alpha=0,8\)

\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=>\tan\alpha=\dfrac{0,6}{0,8}=0,75\)

\(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{0,8}{0,6}\)\(\approx1,33\)