Question

Используя тождество sin²a+cos²a=1 ,упростите выражения:1)sin²a-1. 2) sin⁴a+2sin²a cos⁴a.3)(sina+cosa)²+(sina-cosa)².4)cos²a–cos⁴a+sin⁴a.​

Answer 1

Ответ:

                      $\boxed{\ \boxed {\ sin^2a+cos^2a=1\ }\ }$

                   

$1)\ \ sin^2a-1=-(1-sin^2a)=-cos^2a\\\\2)\ \ sin^4a+2\, sin^2a\cdot cos^2a+cos^4a=(sin^2a+cos^2a)^2=1^2=1\\\\3)\ \ (sina+cosa)^2+(sina-cosa)^2=2sin^2a+2cos^2a=2(sin^2a+cos^2a)=2\\\\4)\ \cos^2a-cos^4a+sin^4a=cos^2a(1-cos^2a)+sin^4a=cos^2a\cdot sin^2a+sin^4a=\\\\=sin^2a(cos^2a+sin^2a)=sin^2a$

Answer 2

$1)sin^2a-1=-cos^2x\\\\2)sin^4a+2sin^acos^2a+cos^4a=(sin^2a+cos^2a)^2=1^2=1\\\\3)(sina+cosa)^2+(sina-cosa)^2=sin^2a+2sinacosa+cos^2a+sin^2a-\\\\-2sinacosa+cos^2a=2sin^2a+2cos^2a=2(sina^2a+cos^2a)=2*1=2\\\\4)cos^2a-cos^4a+sin^4a=(cos^2a-cos^4a)+sin^4a=cos^2a(1-cos^2a)+\\\\=sin^4a=cos^2a*sin^2a+sin^4a=sin^2a(cos^2a+sin^2a)=sin^2a*1=sin^2a$