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

B=cos^2x-sin^2x+cosx-sinx

=(cosx-sinx)(cosx+sinx)+(cosx-sinx)

=(cosx-sinx)(cosx+sinx+1)

`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

\(a)\)

\(1-sin\left(x\right)\)

\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}-2.sin\frac{x}{2}.cos\frac{x}{2}\)

\(=\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2\)

\(b)\)

\(1+sin\left(x\right)\)

\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2.sin\frac{x}{2}.cos\frac{x}{2}\)

\(=\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2\)

\(d)\)

\(1+2cos\left(x\right)\)

\(=2\left(\frac{1}{2}+cosx\right)\)

\(=2\left(cos60^o+cosx\right)\)

\(=4\left(cos\frac{60^o+x}{2}cos\frac{60^o-x}{2}\right)\)

\(=4cos\left(30^o+\frac{x}{2}\right)cos\left(30^o-\frac{x}{2}\right)\)

a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)

Chọn A.

Áp dụng công thức nhân đôi và chú ý:  (đây là 2 góc phụ nhau)