Question

CM các đẳng thức:

a) \(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=tan\left(2x+\frac{15}{4}\right)\)

b) \(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2=-2sin8x.cos2x\)

Answer 1

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

\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)

\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)

\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)

\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)

\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)

\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)

\(=-sin10x-sin6x=-2sin8x.cos2x\)