Question

1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1

2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x

3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx

4. cos4x + 4cos2x + 3 = 8cos^4(x)

5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx

Answer 1

\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)

b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)

c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)

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

d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)

\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)

e/