+25
1 год назад
Алгебра
10 - 11 классы
Відповідь:
Пояснення:
3) cosx/( 1 + sinx ) + ( 1 + sinx )/cosx = [ cos²x + ( 1 + sinx )²]/cosx( 1 + sinx ) =
= ( cos²x + 1 + 2sinx + sin²x )/[ cosx( 1 + sinx ) ] = [ ( sin²x + cos²x ) + 1 +
+ 2sinx ]/[ cosx( 1 + sinx ) ] = ( 1 + 1 + 2sinx )/[ cosx( 1 + sinx ) ] =
= ( 2 + 2sinx )/[ cosx( 1 + sinx ) ] = 2( 1 + sinx )/[ cosx( 1 + sinx ) ] = 2/cosx .
4) cos(- 8π/3) -1/2 sin(23π/6 ) - tg( 13π/4 ) = cos(8π/3 ) - 1/2 sin(4π - π/6 ) -
- tg( 3π + π/4 ) = cos( 3π - π/3 ) - 1/2 sin(- π/6 ) - tgπ/4 = cos( π - π/3 ) +
+ 1/2 sinπ/6 - tgπ/4 = - cosπ/3 + 1/2 sinπ/6 - tgπ/4 = - 1/2 + 1/2 * 1/2 - 1 =
= - 1 1/4 .
Відповідь:
Пояснення:
3) cosx/( 1 + sinx ) + ( 1 + sinx )/cosx = [ cos²x + ( 1 + sinx )²]/cosx( 1 + sinx ) =
= ( cos²x + 1 + 2sinx + sin²x )/[ cosx( 1 + sinx ) ] = [ ( sin²x + cos²x ) + 1 +
+ 2sinx ]/[ cosx( 1 + sinx ) ] = ( 1 + 1 + 2sinx )/[ cosx( 1 + sinx ) ] =
= ( 2 + 2sinx )/[ cosx( 1 + sinx ) ] = 2( 1 + sinx )/[ cosx( 1 + sinx ) ] = 2/cosx .
4) cos(- 8π/3) -1/2 sin(23π/6 ) - tg( 13π/4 ) = cos(8π/3 ) - 1/2 sin(4π - π/6 ) -
- tg( 3π + π/4 ) = cos( 3π - π/3 ) - 1/2 sin(- π/6 ) - tgπ/4 = cos( π - π/3 ) +
+ 1/2 sinπ/6 - tgπ/4 = - cosπ/3 + 1/2 sinπ/6 - tgπ/4 = - 1/2 + 1/2 * 1/2 - 1 =
= - 1 1/4 .