ta có : \(sinx-cosx=\dfrac{1}{3}\Leftrightarrow sin^2x+cos^2x-2sinx.cosx=\dfrac{1}{9}\)

\(\Leftrightarrow-sinx.cosx=\dfrac{-4}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}sinx-cosx=\dfrac{1}{3}\\sinx\left(-cos\right)=\dfrac{-4}{9}\end{matrix}\right.\)

sử dụng vi ét đảo \(\Rightarrow\) \(sinx\) và \(-cosx\) là nghiệm của phương trình :

\(X^2-\dfrac{1}{3}X-\dfrac{4}{9}\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=\dfrac{1+\sqrt{17}}{6}\\-cosx=\dfrac{1-\sqrt{17}}{6}\end{matrix}\right.\\\left\{{}\begin{matrix}sinx=\dfrac{1-\sqrt{17}}{6}\\-cosx=\dfrac{1+\sqrt{17}}{6}\end{matrix}\right.\end{matrix}\right.\)

từ đó \(\Rightarrow\) \(sinx\overset{.}{,}cosx\) rồi thế vào các bài toán bấm máy là ra .

ta có : \(\left(sin\alpha+cos\alpha\right)^2=1+2sin\alpha.cos\alpha=\dfrac{49}{25}\)

\(\Rightarrow sin\alpha+cos\alpha=\pm\dfrac{7}{5}\)

ta có : \(A=sin^3\alpha+cos^3\alpha=\left(sin\alpha+cos\alpha\right)^3-3sin\alpha.cos\alpha\left(sin\alpha+cos\alpha\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}A=\left(\dfrac{7}{5}\right)^3-3\left(0,48\right)\left(\dfrac{7}{5}\right)=\dfrac{91}{125}\\A=\left(\dfrac{-7}{5}\right)^3-3\left(0,48\right)\left(\dfrac{-7}{5}\right)=\dfrac{-91}{125}\end{matrix}\right.\)

vậy \(sin^3\alpha+cos^3\alpha=\pm\dfrac{91}{125}\)

ta có : \(sin^2x+cos^2x=1\Leftrightarrow\left(sinx+cosx\right)^2-2sinx.cosx=1\)

\(\Leftrightarrow\left(sinx+cosx\right)^2-0,96=1\) \(\Leftrightarrow sinx+cosx=\pm\sqrt{1,96}=\pm1,4\)

ta có : \(sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)

th1: \(sinx+cosx=1,4\Rightarrow sin^3x+cos^3x=0,728\)

th2: \(sinx+cosx=-1,4\Rightarrow sin^3x+cos^3x=-0,728\)

vậy ............................................................................................................

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

Có : ΔABC vuông tại A => sinB = cosC = \(\frac{3}{4}\)

Mà lại có : sin² B + cos²B = 1

=> cos²B = 1 - sin²B

=> cosB = 1 - \(\frac{3}{4}\)= \(\frac{1}{4}\)

ta có : \(A=\dfrac{sin^3\alpha+cos^3\alpha}{2sin\alpha.cos^2\alpha+cos^2\alpha.sin^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\dfrac{sin^3\alpha}{cos^3\alpha}+\dfrac{cos^3\alpha}{cos^3\alpha}}{\dfrac{2sin\alpha.cos^2\alpha}{cos^3\alpha}+\dfrac{cos\alpha.sin^2\alpha}{cos^3\alpha}}=\dfrac{tan^3\alpha+1}{2tan\alpha+tan^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\left(\dfrac{3}{4}\right)^3+1}{2\left(\dfrac{3}{4}\right)+\left(\dfrac{3}{4}\right)^2}=\dfrac{91}{132}\)

VT = sin3a.cos^3a + sin^3a.cos3a 
= sin3a.cosa.cos^2a + sin^2a.sina.cos3a 
= 1/2.(sin2a + sin4a).cos^2a + 1/2.sin^2a.(sin(-2a) + sin4a) 
= 1/2.(sin2a + sin4a).cos^2a + 1/2.sin^2a.(sin4a - sin2a) 
= 1/2.sin2a.cos^2a + 1/2.sin4a.cos^2a + 1/2.sin^2a.sin4a - 1/2.sin^2a.sin2a 
= 1/2.sin2a.(cos^2a - sin^2a) + 1/2.sin4a.(cos^2a + sin^2a) 
= 1/2.sin2a.cos2a + 1/2.sin4a 
= 1/4.sin4a + 1/2.sin4a 
= 3/4.sin4a = VP 
=> đpcm

P/s: Chỉ sợ you ko hiểu