(2sin^2Бета-1)/cosБета+sinБета
(дробь) Решение, желательно с объяснением

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.