A=\(\frac{\left(cos7x+cos10x\right)-\left(cos8x+cos9x\right)}{\left(sin7x+sin10x\right)-\left(sin8x+sin9x\right)}\) =\(\frac{2cos\frac{17x}{2}.cos\frac{3x}{2}-2cos\frac{17x}{2}.cos\frac{x}{2}}{2sin\frac{17x}{2}.cos\frac{3x}{2}-2sin\frac{17x}{2}.cos\frac{x}{2}}\)

=\(\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}\)=\(\frac{cos\frac{17x}{2}}{sin\frac{17x}{2}}\)=cotg\(\frac{17x}{2}\)

\(A=\frac{cos7x-cos8x-cos9x+cos10x}{sin7x-sin8x-sin9x+sin10x}=\frac{(cos10x+cos7x)-\left(cos9x+cos8x\right)}{\left(sin10x+sin7x\right)-\left(sin9x+sin8x\right)}.\) 

     \(=\frac{2cos\frac{17x}{2}cos\frac{3x}{2}-2cos\frac{17x}{2}cos\frac{x}{2}}{2sin\frac{17x}{2}cos\frac{3x}{2}-2sin\frac{17x}{2}cos\frac{x}{2}}=\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}=cotan\frac{17x}{2}.\)  

\(\begin{array}{l}1.\,\,\,\,\cos a.\cos b = \frac{1}{2}\left[ {\cos \left( {a + b} \right) + \cos \left( {a - b} \right)} \right] \Leftrightarrow 2\cos a.\cos b = \cos \left( {a + b} \right) + \cos \left( {a - b} \right)\\ \Leftrightarrow 2\cos \frac{{u + v}}{2}.\cos \frac{{u - v}}{2} = \cos u + \cos v\\2.\,\,\,\,\sin a.\sin b =  - \frac{1}{2}.\left[ {\cos \left( {a + b} \right) - \cos \left( {a - b} \right)} \right] \Leftrightarrow  - 2.\sin a.\sin b = \cos \left( {a + b} \right) - \cos \left( {a - b} \right)\\ \Leftrightarrow  - 2.\sin \frac{{u + v}}{2}.\sin \frac{{u - v}}{2} = \cos u - \cos v\\3.\,\,\,\,\sin a.\cos b = \frac{1}{2}\left[ {\sin \left( {a + b} \right) + \sin \left( {a - b} \right)} \right] \Leftrightarrow 2\sin a.\cos b = \sin \left( {a + b} \right) + \sin \left( {a - b} \right)\\ \Leftrightarrow 2\sin \frac{{u + v}}{2}.\cos \frac{{u - v}}{2} = \sin u + \sin v\\4.\,\,\,\,\sin \left( {a + b} \right) - \sin \left( {a - b} \right) = \sin a.\cos b + \cos a.\sin b - \sin a.\cos b + \cos a.\sin b = 2\cos a.\sin b\\ \Leftrightarrow \sin u - \sin v = 2.\cos \frac{{u + v}}{2}.\sin \frac{{u - v}}{2}\end{array}\)

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

`1) cos x + sin 2x - cos 3x`

`= -2sin 2x . (-sin x) + sin 2x`

`= sin 2x ( 2 sin x + 1 )`

Cấu 2 hình như sai đề bạn ạ phải là `sin 3x + sin x` chứ :v

\(\left(cos^4x-sin^4x\right)^2=\left(cos^2x-sin^2x\right)^2\left(cos^2x+sin^2x\right)^2=cos^22x\)

\(\Rightarrow cos^22x=\frac{1}{3}\Rightarrow\frac{1+cos4x}{2}=\frac{1}{3}\Rightarrow cos4x=-\frac{1}{3}\)

\(cos8x=2cos^24x-1=2.\left(-\frac{1}{3}\right)^2-1=-\frac{7}{9}\)

Mong m.n giúp mk vs