Question

Chứng minh các đẳng thức sau:

1/ \(sin^6\alpha+cos^6\alpha=\frac{5}{8}+\frac{3}{8}cos4\alpha\)

2/\(\frac{1+sin2\alpha-cos2\alpha}{1+cos2\alpha}=tan\alpha+tan^2\alpha\)

Answer 1

\(sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)\)

\(=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2\)

\(=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x\)

2/

\(\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}\)

\(=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a\)