\(sin\alpha=\dfrac{3}{4}\)

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

\(\Leftrightarrow cos^2\alpha=1-sin^2\alpha\)

\(\Leftrightarrow cos^2\alpha=1-\dfrac{9}{16}=\dfrac{7}{16}\)

\(\Leftrightarrow cos\alpha=-\dfrac{\sqrt[]{7}}{4}\left(\dfrac{\pi}{2}< \alpha< \pi\right)\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{4}}{-\dfrac{\sqrt[]{7}}{4}}=-\dfrac{3}{\sqrt[]{7}}=-\dfrac{3\sqrt[]{7}}{7}\)

\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=-\dfrac{\sqrt[]{7}}{3}\)

\(0< a< \frac{\pi}{2}\Rightarrow sina;cosa;tana>0\)

\(tana+\frac{1}{tana}=3\Leftrightarrow tan^2a-3tana+1=0\) \(\Rightarrow\left[{}\begin{matrix}tana=\frac{3-\sqrt{5}}{2}\\tana=\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)

- Với \(tana=\frac{3-\sqrt{5}}{2}\)

\(\Rightarrow cota=\frac{1}{tana}=\frac{3+\sqrt{5}}{2}\)

\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=\frac{2}{\sqrt{18-6\sqrt{5}}}\)

\(sina=\sqrt{1-cos^2a}=\frac{2}{\sqrt{18+6\sqrt{5}}}\)

\(cos\left(\frac{3\pi}{2}-a\right)=cos\left(2\pi-\frac{\pi}{2}-a\right)=-sina=...\)

\(sin\left(2\pi+a\right)=sina=...\)

\(tan\left(\pi-a\right)=-tana=...\)

\(cot\left(\pi+a\right)=cota=...\)

TH2: \(tana=\frac{3+\sqrt{5}}{2}\)

Tương tự như trên

a) sin anpha = 2/3 => góc anpha = 42^o 

cos 42^o = 0,743

tan 42^o =  0,9

cot  42^o = 1/tan 42^o = 1/0,9 = 1,111

b) tan anpha + cot anpha = 3

<=> tan anpha + 1/tan anpha = 3

<=> tan²  anpha = 2

<=> tan anpha = \(\sqrt{2}\)

=> góc anpha =  55^o 

Ta có: a = sin 55^o . cos 55^o

<=> a = 0,469

Bài 3: 

a: cos B=0,8 nên AC/BC=4/5

=>AC=8cm

=>AB=6cm

b: sin C=cos B=4/5

cos C=3/5

tan C=4/3

cot C=3/4

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\sin^2=1-\dfrac{16}{49}=\dfrac{33}{49}\)

Ta có: \(4\cdot\cos^2\alpha-3\cdot\sin^2\alpha\)

\(=4\cdot\dfrac{16}{49}-3\cdot\dfrac{33}{49}\)

\(=\dfrac{64-99}{49}=-\dfrac{5}{7}\)

`cos^2α=16/49`

`sin^2α+cos^2α=1`

`<=>sin^2α+(4/7)^2=1`

`<=>sin^2α=33/49`

`4cos^2α-3sin^2α=4. 16/49 - 3. 33/49 = -5/7`

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

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

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

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

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