Question

Пожалуйста помогите

Answer 1

Ответ:

1.

$- \sin( \frac{\pi}{2} ) (2tg( \frac{\pi}{4} ) - \cos( \frac{\pi}{6} ) ) = \\ = - 1 \times (2 - \frac{ \sqrt{3} }{2} ) = \\ = \frac{ \sqrt{3} }{2} - 2 = \frac{ \sqrt{3} - 4}{2}$

2.

$ctg( \frac{\pi}{6} ) \times ( \sin( \frac{\pi}{3} ) - 3 \cos( \frac{\pi}{3} ) ) = \\ = \sqrt{3} \times ( \frac{ \sqrt{3} }{2} - \frac{3} {2} ) = \\ = \sqrt{3} \times \frac{ \sqrt{3} - 3}{2} = \frac{3 - 3 \sqrt{3} }{2}$

3.

$\cos( \frac{\pi}{4} ) (2ctg( \frac{\pi}{4} ) - 3 \sin( \frac{\pi}{6} ) ) = \\ = \frac{ \sqrt{2} } {2} \times (2 - \frac{3}{2} ) = \frac{ \sqrt{2} }{2} \times \frac{1}{2} = \frac{ \sqrt{2} } {4}$

Answer 2

Ответ:

Ответ:

1.

\begin{gathered} - \sin( \frac{\pi}{2} ) (2tg( \frac{\pi}{4} ) - \cos( \frac{\pi}{6} ) ) = \\ = - 1 \times (2 - \frac{ \sqrt{3} }{2} ) = \\ = \frac{ \sqrt{3} }{2} - 2 = \frac{ \sqrt{3} - 4}{2} \end{gathered}

−sin(

2

π

)(2tg(

4

π

)−cos(

6

π

))=

=−1×(2−

2

3

)=

=

2

3

−2=

2

3

−4

2.

\begin{gathered}ctg( \frac{\pi}{6} ) \times ( \sin( \frac{\pi}{3} ) - 3 \cos( \frac{\pi}{3} ) ) = \\ = \sqrt{3} \times ( \frac{ \sqrt{3} }{2} - \frac{3} {2} ) = \\ = \sqrt{3} \times \frac{ \sqrt{3} - 3}{2} = \frac{3 - 3 \sqrt{3} }{2} \end{gathered}

ctg(

6

π

)×(sin(

3

π

)−3cos(

3

π

))=

=

3

×(

2

3

−

2

3

)=

=

3

×

2

3

−3

=

2

3−3

3

3.

\begin{gathered} \cos( \frac{\pi}{4} ) (2ctg( \frac{\pi}{4} ) - 3 \sin( \frac{\pi}{6} ) ) = \\ = \frac{ \sqrt{2} } {2} \times (2 - \frac{3}{2} ) = \frac{ \sqrt{2} }{2} \times \frac{1}{2} = \frac{ \sqrt{2} } {4} \end{gathered}

cos(

4

π

)(2ctg(

4

π

)−3sin(

6

π

))=

=

2

2

×(2−

2

3

)=

2

2

×

2

1

=

4

2