Question

Chứng minh rằng:

*\(\tan3\alpha=\frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}\)

*\(\sin^6\alpha-\cos^6\alpha=-\cos2\alpha\left(1-\sin^2\alpha\cos^2\alpha\right)\)

Answer 1

a)\(tan3A=tan\left(A+2A\right)\)

\(=\frac{tanA+tan2A}{1-tanAtan2A}\)

\(=\frac{\frac{tanA+2tanA}{1-tan^2A}}{\frac{1-2tan^2A}{1-tan^2A}}\)

\(=\frac{\left(tanA-tan^3A+2tanA\right)}{1-tan^2A-2tan^2A}\)

\(=\frac{3tanA-tan^3A}{1-3tan^2A}\)

b)\(VT=cos^6A+sin^6A\)

\(=\left(cos^2A\right)^3+\left(sin^2A\right)^3\)

\(=\left(cos^2A+sin^2A\right)^3-3cos^2Asin^2A\left(cos^2A+sin^2A\right)^2\)

\(=1^3-3cos^2Asin^2A\left(1\right)^2\).Từ đó,\(sin^2A+cos^2A=1\)

\(=1-3cos^2Asin^2A=VP\)

Answer 2

phần b tui sai