Giao lưu

Đang tiếp cận đến sin cos

Lời giải

\(A=sin\left(2x\right).\left[2sin\left(3x\right).sin\left(x\right)\right]\)

\(A=sin\left(2x\right).\left\{-\left[cos\left(\dfrac{3x+x}{2}\right)-cos\left(\dfrac{3x-x}{2}\right)\right]\right\}\)

\(A=sin\left(2x\right).\left[cos\left(x\right)-cos\left(2x\right)\right]\)

\(A=sin\left(2x\right)cos\left(x\right)-sin\left(2x\right)cos\left(2x\right)\)

\(2A=\left[sin\left(\dfrac{2x+x}{2}\right)+sin\left(\dfrac{2x-x}{2}\right)\right]-\left[sin\left(\dfrac{2x+2x}{2}\right)+sin\left(\dfrac{2x-2x}{2}\right)\right]\)\(2A=sin\left(\dfrac{3x}{2}\right)+sin\left(\dfrac{x}{2}\right)-sin\left(2x\right)-0\)

\(A=\dfrac{sin\left(\dfrac{3x}{2}\right)+sin\left(\dfrac{x}{2}\right)-sin\left(2x\right)}{2}\)

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

Lời giải:
$A=\sin ^2a-\sin ^2b=(\sin a-\sin b)(\sin a+\sin b)$

$B$ không biến đổi được. Bạn coi lại đề.

\(A=2sin\dfrac{a+b}{2}cos\dfrac{a-b}{2}+2sin\dfrac{a+b}{2}cos\dfrac{a+b}{2}\)

\(=2sin\dfrac{a+b}{2}\left(cos\dfrac{a+b}{2}+cos\dfrac{a-b}{2}\right)\)

\(=2sin\dfrac{a+b}{2}.2cos\dfrac{a}{2}cos\dfrac{b}{2}\)

\(=4sin\dfrac{a+b}{2}cos\dfrac{a}{2}cos\dfrac{b}{2}\)

Sao Sin(a+b)=2Sin(a+b)/2Cos(a+b)/2 ạ

`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: \(A=cosa\cdot cosb\cdot cosc\)

\(=\frac12\cdot cosc\cdot\left\lbrack cos\left(a-b\right)+cos\left(a+b\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack cosc\cdot cos\left(a-b\right)+cosc\cdot cos\left(a+b\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\frac12\cdot cos\left(c-a+b\right)+\frac12\cdot\left(c+a-b\right)+\frac12\cdot cos\left(c-a-b\right)+\frac12\cdot cos\left(c+a+b\right)\right\rbrack\)

\(=\frac14\cdot cos\left(c-a+b\right)+\frac14\cdot cos\left(c+a-b\right)+\frac14\cdot cos\left(c-a-b\right)+\frac14\cdot cos\left(c+a+b\right)\)

b: \(B=4\cdot\sin2a\cdot\sin4a\cdot\sin6a\)

\(=4\cdot\frac12\cdot\left\lbrack cos\left(2a-4a\right)-cos\left(2a+4a\right)\right\rbrack\cdot\sin6a\)

\(=2\cdot\left\lbrack cos2a-cos6a\right\rbrack\cdot\sin6a=2\cdot\sin6a\cdot cos2a-2\cdot\sin6a\cdot cos2a\)

=sin(6a-2a)+sin(6a+2a)-sin12a

=sin4a+sin8a-sin12a

c: \(C=\sin\left(x+\frac{\pi}{6}\right)\cdot\sin\left(x-\frac{\pi}{6}\right)\cdot cos2x\)

\(=\frac12\cdot\left\lbrack cos\left(x+\frac{\pi}{6}-x+\frac{\pi}{6}\right)-cos\left(x+\frac{\pi}{6}+x-\frac{\pi}{6}\right)\right\rbrack\cdot cos2x\)

\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{3}\right)-cos2x\right\rbrack\cdot cos2x=\frac12\cdot\left\lbrack\frac12-cos2x\right\rbrack\cdot cos2x=\frac14\cdot cos2x-\frac12\cdot cos^22x\)