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

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

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