Question

прошу помочь)Пожалуйста
б,г,е,з плиз

Answer 1

b)(sinα+sinβ)\(sinα-sinβ)=(2sin(α+β)\2·cos(α-β)\2) \2sin(α-β)\2·cos(α+β)/2=
=tg(α+β\2·ctg(α-β)\2
г)(  cosα-cosβ)\(sinα+sinβ)=(2sin(α+β)\2·sin(β-α)\2))|2sin(α+β)\2·cos(α-β)\2=
=-2tg(α-β)\2
e) (cos2x -cos3x) \(sin2x+sin3x)=(2sin(2x+3x)\2sin(3x-2x)\2))\2sin(2x+3x)\2cos(2x-3x)\2=sinx\2:cosx|2=tgx\2
з(sin4α -sin6α)\(cos3α+cos7α)=(2sin(4α-6α)\2·cos(4α+6α)\2))\2cos(3α+7α)\2·
·cos(3α-7α)\2=(-2sinα·cos5α)\(2cos5α·cos2α=-sinα\cos2α