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)<span>/2+</span>cos(α+γ-β)/2 +cos(γ+β-α)<span>/2 =
</span>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.
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б) 4sin(α/2)*sin(β/2)*cos(γ/2) = sin α + sin <span>β</span> - sin γ
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sin α + sin β - sinγ =2sin((α+c)/2)*cos((α-β)/2) -sin(π-(α+<span>β))=
</span>2sin((α+β)/2)*cos((α-β)/2) -sin(α+<span>β)=
</span>2sin((α+β)/2)*cos((α-β)/2) -sin2*((α+<span>β)/2)=
</span>2sin((α+β)/2)*cos((α-β)/2) -2sin((α+β)/2)*cos((α<span>+β)/2) =
</span>2sin((α+β)/2)*(cos((α-β)/2) -cos((α<span>+β)/2) )=
</span>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) .
0,4(х-5) =0, 5(6+х) -2, 5
0,4х-2=3+0, 6х-2, 5
0,6х-0, 4=3+2-2, 5
0,2х=2, 5
х=2, 5: 0,2=25: 2
х=12, 5
Ответ: 12,5
Во второй системе там после неравенства 16? или 1/6?
Sin5П*cosx-sinx*cos5П=cos2x*cos7П-sin2x*sin7П
sinx=-cos2x
sinx=sin^2(x)-cos^2(x)
sinx=sin^2(x)-1+sin^2(x)
2sin^2(x)-sinx-1=0
sinx=y,-1<=y<=1
2y^2-y-1=0
y1=1,y2=-1/2
sinx=1
x=π/2+πn
или
sinx=-1/2
x=(-1)^n*arcsin(-1/2)+πn=(-1)^n*(-π/6)+πn