Câu b ) Ta có $sin^4x + cos^4x = ( sin^2 + cos^2 )^2 - 2sin^2x\times cos^2x$ ( 1)Mà $sinx + cosx = m => sin^2x + cos^2x + 2sinx\times cosx = m^2 $$<=> (m^2 - 1)/2 = sinx\times cox$$<=> sin^2x\times cox^2x = \frac{(m^2 - 1 )^2}{4}$Thay vào (1 ) ta có :$sin^4 + cos^4 = 1 - \frac{(m^2 - 1 )^2}{2}$
Câu b ) Ta có $sin^4x + cos^4x = ( sin^2 + cos^2 )^2 - 2sin^2x\times cos^2x$Mà $sinx + cosx = m => sin^2x + cos^2x + 2sinx\times cosx = m^2 $$<=> (m^2 - 1)/2 = sinx\times cox$
Câu b ) Ta có $sin^4x + cos^4x = ( sin^2 + cos^2 )^2 - 2sin^2x\times cos^2x$
( 1)Mà $sinx + cosx = m => sin^2x + cos^2x + 2sinx\times cosx = m^2 $$<=> (m^2 - 1)/2 = sinx\times cox$
$<=> sin^2x\times cox^2x = \frac{(m^2 - 1 )^2}{4}$Thay vào (1 ) ta có :$sin^4 + cos^4 = 1 - \frac{(m^2 - 1 )^2}{2}$