Chứng minh sin(a)^6+cos(a)^6=1-3sin(a)^2*cos(a)^2

$VT= sin^6a+cos^6a$

$= (sin^2a)^3 + (cos^2a)^3$

$= (sin^2a+cos^2a)(sin^4a-sin^2a.cos^2a+cos^4a)$

$= sin^4a-sin^2a.cos^2a+cos^4x$

$= (sin^2a+cos^2a)^2-2sin^2a.cos^2a-sin^2a.cos^2a$

$= 1-3sin^2a.cos^2a$

$= VP$ (đpcm)

Đáp án:

Giải thích các bước giải:

 Ta có: sin$^{6}$a+cos$^{6}$a=(sin$^{2}$a)$^{3}$+(cos$^{2}$a)$^{3}$

=(sin$^{2}$a+cos$^{2}$a)(sin$^{4}$a-sin$^{2}$a.cos$^{2}$a+cos$^{4}$a)

=(sin$^{2}$a+cos$^{2}$a)$^{2}$-3sin$^{2}$a.cos$^{2}$a (vì sin$^{2}$a+cos$^{2}$a)

=1-3sin$^{2}$a.cos$^{2}$a