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

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\(\dfrac{\pi}{2}\)

`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

`cos α+sinα=5/4`

`<=>(cosα+sinα)^2=25/16`

`<=>(sin^2α+cos^2α)+2sinαcosα=25/16`

`<=>1+2sinαcosα=25/16`

`<=>sinα.cosα=9/32`

cos_a+sin_a=\(\dfrac{5}{4}\)

( cos_a+sin_a)²=\(\dfrac{5}{4}\)

1+2.sin_a.cos_a=\(\dfrac{25}{16}\)

sin_a.cos_a=\(\dfrac{9}{32}\)

$cosα+sinα=\dfrac{5}{4}\text{ (là cái đề bài)}\\ \Leftrightarrow \left(cosα+sinα\right)^2=\left(\dfrac{5}{4}\right)^2\text{ (bình phương 2 vế)}\\ \Leftrightarrow cos^2α+2cosα.sinα+sin^2α=\dfrac{25}{16} \\ \text{ (Vế trái khai triển hằng đẳng thức, vế phải bình phương)}\\ \Leftrightarrow \left(sin^2α+cos^2α\right)+2sinα.cosα=\dfrac{25}{16}\text{ (nhóm }sin^2α\text{ và }cos^2α\text{)}\\ \Leftrightarrow 1+2sinαcosα=\dfrac{25}{16}\left(sin^2α+cos^2α=1\right)\\ \Leftrightarrow sinαcosα=\left(\dfrac{25}{16}-1\right):2=\dfrac{9}{32}$

Ta có :

\(\cos \left( {a + b} \right)\cos \left( {a - b} \right) = \frac{1}{2}\left( {\cos 2a + \cos 2b} \right) = \frac{1}{2}\left( {2{{\cos }^2}a - 1 + 2{{\cos }^2}b - 1} \right) = 0\)

Chọn A

MN K BT?

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

\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)

\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)