Выражение 2cos^2(пи+а)/4-2sin^2(пи+а)/4 - вопрос №2995860224224 от SanyaZv 27.10.2022 16:51

Question

Алгебра: Выражение 2cos^2(пи+а)/4-2sin^2(пи+а)/4

SanyaZv · Answer

= 2 сos^2 (pi/4 + a/4) - 2 sin^2 (pi/4 + a/4) =
=2(cospi/4*cos a/4-sin pi/4 *sin a/4)^2 - 2(sin pi/4 *cos a/4+cos pi/4*sin a/4)^ =
= 2(sgrt 2/2 *cos a/4-sgrt2/2*sin a/4)^2 -2( sgrt2/2 *cos a/4 + sgrt2/2*sin a/4)^2=
=2*(sgrt2/2)^2 (cos a/4 - sin a/4)^2 - 2*(sgrt2/2)^2 *(cos a/4 +sin a/4)^2 =
= 2*2/4 (cos^2(a/4) - 2sin a/4*cos a/4 + sin^2(a/4) -
- 2*2/4(cos^2(a/4) + 2 sin a/4 cos a/4 + cos^2(a/4) =
= (1 - 2 sin(a/4)cos(a/4)) - (1 + 2 sin (a/4)cos(a/4)=
=1 - sin(a/2) - 1 - sin(a/2) = - 2 sin(a/2)