cos(α+β)+2sinαsinβ=cosαcosβ−sinαsinβ+2sinαsinβ=
cosαcosβ+sinαsinβ=cos(α−β)
если \alpha -\beta=\piα−β=π , то cos(\alpha -\beta ) =cos\pi =-1.cos(α−β)=cosπ=−1.
б)
\frac{sin^{2}\alpha +sin(\pi-\alpha)cos (\frac{\pi }{2} -\alpha) }{tq(\pi+\alpha)ctq( \frac{3\pi }{2} -\alpha ) } = \frac{sin^{2}\alpha +sin\alpha*sin\alpha }{tq\alpha*tq\alpha } =\frac{2sin^{2} \alpha }{tq^{2} \alpha } =\frac{2sin^{2}\alpha }{\frac{sin^{2} \alpha }{cos^{2} \alpha } } =2cos^{2} \alpha .
tq(π+α)ctq(
2
3π
−α)
sin
2
α+sin(π−α)cos(
2
π
−α)
=
tqα∗tqα
sin
2
α+sinα∗sinα
=
tq
2
α
2sin
2
α
=
cos
2
α
sin
2
α
2sin
2
α
=2cos
2
α.
в)
cos7xcos6x+sin7xsin6x=cos(7x-6x)=cosx.cos7xcos6x+sin7xsin6x=cos(7x−6x)=cosx.
ответ:2-A-B-C-D
5+A+B+C+D
4-A+B-C+D
-9A+9B-9C+9D
-6A-6B+6C+6D
Пошаговое объяснение: