Question

преобразовать ;1) sin 9pi/2=        2) sin^2 x+cos^4x-sin^4x=

Answer 1

1)
sin (9pi/2) = sin ( 4pi + pi/2) = sin (pi/2) = 1 

 2)
Формула разности квадратов 
a^2 - b^2 = (a - b) ( a  + b) 

sin^2 x+ cos^4x - sin^4x = 
= sin^2x + (cos^2x - sin^2x)( cos^2x +  sin^2x) = 
= sin^2x + (cos^2x - sin^2x) *1 
= sin^2x + cos^2x - sin^2x =  cos^2x 

Answer 2

благодарю!!!

Answer 3

не за что

Answer 4

$1) sin frac{9 pi }{2} =sin( frac{ pi }{2} )=1 \ 2)sin^{2} x+cos^{4} x-sin^{4} x=sin^{2}x+(cos^{2} x+sin^{2} x)(cos^{2} x-sin^{2}x)= \ =sin^{2} x+1*(cos^{2} x-sin^{2}x)= sin^{2} x+cos^{2} x-sin^{2}x= cos^{2} x$