См. вложение
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<span>Sin^2(3pi/2-y)+sin^2(3pi+y)+2tg(5pi/2-y)×tg (3pi+y)
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Решение
Вычислить Sin²(3π/2-y)+sin²(3π+y)+2tg(5π/2-y)*tg (3π+y) .
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Применяя формулы приведения ,получаем :
Sin²(3π/2-y)+sin²(3π+y)+2tg(5π/2-y)*tg (3π+y)= (cos²y +sin²y)+ 2ctqy*tqy =1 +2*`1 =3.
ответ : 3.
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sin (3π<span>/2-y)) = -cosy
</span>sin(3π +y ) =sin( (2π+(π +y ) )=sin(π<span> +y </span><span>) = </span> -siny
tg(5π/2-y) =tg((2π +(π<span>/2-y) )</span> =tq(π/2-y) =ctqy
tq(3π +y ) =tq( (2π+(π +y ) ) =tq(π<span> +y ) =tqy</span>
1)3х-5=2х+2-3
3х-2х=2-3+5
х=4
Что смогла=========================
4x-5 ≥ 2x-4
4x - 2x ≥ 5 - 4
2x ≥ 1
x ≥ 1/2
x є [ 1/2 ; +∞)