a) cos (α - β) - cos (α + β);
b) sin(-α) + cos(π + α)
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1 + 2 cos (π/2 - α) cos(-α)
b) cos 3α + cos 2α + 2sin² α
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2 cos α
первый способ по формуле
[m]cosx- cosy=-2\cdot sin\frac{x+y }{2}\cdot sin\frac{x-u }{2}[/m]
[m]cos( α - β )-cos( α + β )=-2\cdot sin\frac{( α- β)+( α+ β ) }{2}\cdot sin\frac{( α- β)-( α+ β ) }{2}=-2\cdot sin α \cdot sin(- β )=-2\cdot sin α \cdot(-sin β)=2sin α\cdot sin β [/m]
второй способ по формулам:
[m]cos( α - β )=cos α \cdot cos β +sin α sin β[/m]
[m]cos( α + β )=cos α \cdot cos β -sin α sin β[/m]
[m]cos( α - β )-cos( α + β )=cos α \cdot cos β +sin α sin β -(cos α \cdot cos β -sin α sin β)=2\cdot sin α \cdot sin β [/m]
б)
[m]sin(- α )=-sin α [/m]
[m]cos(π + α )=-cos α [/m]
[m]cos(\frac{π}{2}- α )=sin α [/m]
[m]cos(- α )=cos α [/m]
Тогда
[m]\frac{sin(- α )+cos(π+ α) }{1+2cos(\frac{π}{2}- α )\cdot cos(- α) }=\frac{-sin α-cos α }{sin^2 α +cos^2 α +2sin α\cdot cos α }=\frac{-(sin α+cos α ) }{(sin α+cos α)^2 }=-\frac{1}{sin α+cos α} [/m]
в)
[m]cos3 α +cos α =2cos\frac{3 α + α }{2}\cdot cos\frac{3 α- α }{2}=2cos(2 α) \cdot cos α [/m]
Тогда
[m]\frac{cos3 α +cos α}{2cos α }+2sin^2 α =\frac{2cos(2 α)\cdot cos α}{2cos α }+2sin^2 α=cos2 α +2sin^2 α=cos^2 α-sin^2 α +2sin^2 α=cos^2 α+sin^2 α =1 [/m]