Question

[Nyusu TegoKato]

Rút gọn các biểu thức :

A= \(\sqrt{3}\) sin ( x - \(\dfrac{\pi}{3}\) ) + sin ( x + \(\dfrac{\pi}{6}\) )

B= cos7x cos5x - \(\sqrt{3}\) sin2x + sin7x sin5x

C= 2sin( 2x -\(\dfrac{\pi}{6}\)) + 4sin + 1

D= \(\sqrt{3}\) cos2x + sin2x + 2sin(2x - \(\dfrac{\pi}{6}\))

E= sin2x + 2\(\sqrt{2}\) cosx + 2sin(x + \(\dfrac{\pi}{4}\)) +3

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)

\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)

\(=\sqrt{3}sinx-cosx\)

c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)

\(=\sqrt{3}sin2x-cos2x+4sinx+1\)

d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)

\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)

\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)