Question

Найдите значение выражения 1 + co( x+ pi/2) + cos ( x - pi/2) при х= 5pi/12.


3-6 sin^2 x/6 при х= 3p/4.


1-2(sin 36 * cos 9 + sin 9 * cos 36)^24.


( sin 49 * cos 14 + sin 14 * cos 49)^2 + (cos 28 * cos 35 - sin 28 * sin 35)^2.

Answer 1

1+cos(x+π/2)+cos(x-π/2)=1-sinx+sinx=1

3-6sin²x/6=3(1-2sin²x/6)=3cosx/3=3cosπ/4=3√2/2

1-2(sin 36 * cos 9 + sin 9 * cos 36)^24.
=1-2(sin(36+9))^24=1-2(sin45)^24=1-2*(1/√2)^24=1-2/2^12=1-1/2^11=
=1-1/2048=2047/2048

( sin 49 * cos 14 + sin 14 * cos 49)^2 + (cos 28 * cos 35 - sin 28 * sin 35)^2=
=(sin(49+14))²+(cos(28+35))²=sin²63+cos²63=1