Question

[Nguyen Thi Mai]

CMR :

a) \(\frac{sinx+sin3x+sin4x}{1+cosx+cos3x+cos4x}=tan2x\)

b) \(\frac{sin^22x+2cos\left(3\pi+2x\right)-2}{-3+4cos2x+cos\left(4x-\pi\right)}=\frac{1}{2}cot^4x\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)

\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)

\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)

\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

Answer 2

[Nguyen Thi Mai]

Mình cảm ơn nhé :))