Представьте в виде многочлена степень: (1 пятая а в квадрате + 0, 36 в степени 4)

Объяснение:

1

\begin{gathered}1 - 8 \sin(2 \beta ) \times \cos( 2\beta ) = 1 - 4 \times 2 \sin( 2\beta ) \cos( 2\beta ) = \\ = 1 - 4 \sin( 4\beta ) \end{gathered}

1−8sin(2β)×cos(2β)=1−4×2sin(2β)cos(2β)=

=1−4sin(4β)

2

\begin{gathered}tg \beta (1 + \cos(2 \beta ) - \sin( 2\beta ) = \\ = tg \beta \times (1 + { \cos }^{2} (\beta) - { \sin}^{2}( \beta )) - \sin( 2\beta ) = \\ = tg \beta \times 2 { \cos }^{2} (\beta ) - \sin( 2\beta ) = \\ = 2 \sin( \beta ) \cos( \beta ) - 2 \sin( \beta ) \cos( \beta ) = 0\end{gathered}

tgβ(1+cos(2β)−sin(2β)=

=tgβ×(1+cos

2

(β)−sin

2

(β))−sin(2β)=

=tgβ×2cos

2

(β)−sin(2β)=

=2sin(β)cos(β)−2sin(β)cos(β)=0

3

\begin{gathered} \frac{2 \sin( \beta ) - \sin( 2\beta ) }{ 2\sin( \beta ) + \sin( 2\beta ) } = \\ = \frac{2 \sin( \beta ) - 2 \sin( \beta ) \cos( \beta ) }{ 2\sin( \beta ) + 2 \sin( \beta ) \cos( \beta ) } = \\ = \frac{2 \sin( \beta )(1 - \cos( \beta )) }{ 2\sin( \beta ) (1 + \cos( \beta )) } = \frac{1 - \cos( \beta ) }{1 + \cos( \beta ) } \end{gathered}

2sin(β)+sin(2β)

2sin(β)−sin(2β)

=

=

2sin(β)+2sin(β)cos(β)

2sin(β)−2sin(β)cos(β)

=

=

2sin(β)(1+cos(β))

2sin(β)(1−cos(β))

=

1+cos(β)

1−cos(β)

4

\begin{gathered} \frac{ctg(45 - \beta )}{1 - {ctg}^{2}(45 - \beta ) } = - \frac{ctg(45 - \beta )}{ {ctg}^{2} (45 - \beta ) - 1} = \\ = - \frac{2ctg(45 - \ \beta )}{2( {ctg}^{2}(45 - \beta ) - 1) } = - \frac{1}{2ctg(45 - \beta )} \end{gathered}

1−ctg

2

(45−β)

ctg(45−β)

=−

ctg

2

(45−β)−1

ctg(45−β)

=

=−

2(ctg

2

(45−β)−1)

2ctg(45− β)

=−

2ctg(45−β)

1