Question

cos^6+sin^6 упростить

Answer 1

(sin^2a+cos^2a)(sin^4a-sin^2a*cos^2a+cos^4a)=(sin^2a+cos^2a)^2-3*sin^2a*cos^2a=

1-3*(4*sin^2a*cos^2a)/4=1-(3/4)*(sin 2a)^2

Answer 2

I think I was wrong

Answer 3

$1); ; cos^6x+sin^6x=(cos^2x)^3+(sin^2x)^3=\\=(underbrace {cos^2x+sin^2x}_{1})cdot (cos^4x-cos^2xcdot sin^2x+sin^4x)=\\=Big ((cos^2x)^2-2cos^2xcdot sin^2x+(sin^2x)^2Big )+cos^2xcdot sin^2x=\\=(underbrace {cos^2x-sin^2x}_{cos2x})^2+(underbrace {sinxcdot cosx}_{frac{1}{2}sin2x})^2=cos^22x+frac{1}{4}sin^22x\\\2)cos^6x+sin^6x=cos^4x-cos^2xcdot sin^2x+sin^4x=\\=Big ((cos^2x)^2+2cos^2xcdot sin^2x+(sin^2x)^2Big )-3, cos^2xcdot sin^2x=$

$=(underbrace {cos^2x+sin^2x}_{1})^2-3(underbrace {sinxcdot cosx}_{frac{1}{2}sin2x})^2=1-frac{3}{4}cdot sin^22x\\\star ; ; 2, sinxcdot cosx=sin2x; ; Rightarrow ; ; sinxcdot cosx=frac{1}{2}sin2x; ; star$