Найдите значение функции у=2х+5 при х=-3

Question

Алгебра: Найдите значение функции у=2х+5 при х=-3

Ftgbjjjm · Accepted Answer

А)
4cos a/2*cos b/2*cos y/2 = sin a + sin b + sin y
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4cos α/2*cos β/2*cosγ/2 =
2(cos(α+β)/2 +cos(α-β)/2)*cosγ/2 =
2cos(α+β)/2*cosγ/2 +2cosγ/2 *cos(α-β)/2=
cos(α+β+γ)/2 +cos(α+β-γ)/2+cos(α+γ-β)/2 +cos(γ+β-α)/2 =
cosπ/2 +cos(α+β+γ -2γ)/2+cos(α+β+γ-2β)/2 +cos(β+γ+α-2α)/2=
cos(π -2γ)/2+cos(π-2β)/2 +cos(π-α)/2=
cos(π/2 -γ)+cos(π/2-β) +cos(π/2-α) = sinα +sinβ+sinγ.

б) 4sin(α/2)*sin(β/2)*cos(γ/2) = sin α + sin β - sin γ
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sin α + sin β - sinγ =2sin((α+c)/2)*cos((α-β)/2) -sin(π-(α+β))=
2sin((α+β)/2)*cos((α-β)/2) -sin(α+β)=
2sin((α+β)/2)*cos((α-β)/2) -sin2*((α+β)/2)=
2sin((α+β)/2)*cos((α-β)/2) -2sin((α+β)/2)*cos((α+β)/2) =
2sin((α+β)/2)*(cos((α-β)/2) -cos((α+β)/2) )=
2sin((π-γ)/2) *(-2sin(α/2)*sin(-β/2) =2sin(π/2-γ/2) *2sin(α/2)*sin(β/2)=
2cos(γ/2) *2sin(α/2)*sin(β/2) =4sin(α/2)*sin(β/2)*cos(γ/2) .

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