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July 2022 1 4 Report

Найдите значение выражения:
1) [tex]27 \sqrt{6} tg \frac{ \pi }{3} sin \frac{ \pi }{4} [/tex]
2)[tex]6 \sqrt{2} cos \frac{ \pi }{4} cos \frac{7 \pi }{3} [/tex]
3)[tex] \frac{32}{sin(- \frac{3 \pi }{4})cos \frac{25 \pi }{4} } [/tex]
4)[tex]32 \sqrt{3} cos (-\frac{ \pi }{6} )sin(- \frac{ \pi }{6)} [/tex]

Answers & Comments

artalex74 $1)\ 27 \sqrt{6} tg \frac{ \pi }{3} sin \frac{ \pi }{4} = 27 \sqrt{6} * \sqrt{3} * \frac{ \sqrt{2} }{2} =27*3=81$
$2)\ 6 \sqrt{2} cos \frac{ \pi }{4} cos \frac{ 7\pi }{3} =6 \sqrt{2} cos \frac{ \pi }{4} cos(2 \pi + \frac{ \pi }{3} ) =6 \sqrt{2} cos \frac{ \pi }{4} cos \frac{ \pi }{3} = \\ =6 \sqrt{2} * \frac{ \sqrt{2} }{2}* \frac{1}{2} =6* \frac{1}{2} =3$
$3) \dfrac{32}{sin(- \frac{3 \pi }{4})cos \frac{25 \pi }{4} }=\dfrac{32}{-sin \frac{3 \pi }{4}cos(4 \pi + \frac{ \pi }{4} ) }=\dfrac{32}{-sin \frac{3 \pi }{4}cos \frac{ \pi }{4} }= \\ =\dfrac{32}{- \frac{ \sqrt{2} }{2}*\frac{ \sqrt{2} }{2} }=- \dfrac{32}{ \frac{1}{2} } =-64$
$4)\ 32 \sqrt{3} cos (-\frac{ \pi }{6}) sin(- \frac{ \pi }{6}) = -32 \sqrt{3} cos \frac{ \pi }{6} sin \frac{ \pi }{6} =-16 \sqrt{3} sin \frac{ \pi }{3} = \\ =-16 \sqrt{3} * \frac{ \sqrt{3} }{2}=-8*3=-24$

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Answers & Comments

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