Нужна помощь с тригонометрией!!!. Created by lenivs. algebra-ru

Нужна помощь с тригонометрией!!!...

WaizuKun

Ответ:

2; -3,5; -6

Объяснение:

$\sqrt{32}\;cos^2\dfrac{7\pi}{8}-\sqrt8\;\;=\;\;4\sqrt{2}\cdot\dfrac{1+cos\dfrac{7\pi}{4}}{2}-2\sqrt2\;\;=\\\\\\=\;\;2\sqrt{2}\cdot\Bigg(1+\dfrac{\sqrt2}{2}\Bigg)-2\sqrt2\;\;=\;\;2\sqrt2+2\sqrt2\cdot\dfrac{\sqrt2}{2}-2\sqrt2\;\;=\;\;2$

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$7\sqrt2\;sin\dfrac{15\pi}{8}\cdot cos\dfrac{15\pi}{8}=\sqrt2\cdot\dfrac{7}{2}\cdot sin\dfrac{15\pi}{4}=\dfrac{7\sqrt2}{2}\cdot\Big(-\dfrac{\sqrt2}{2}\Big)=\\\\\\=-\dfrac{14}{4}=-\dfrac{7}{2}=-3{,}5$

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$\dfrac{-6\cdot sin374^{\circ}}{sin14^{\circ}}=\dfrac{-6\cdot sin(360^{\circ}+14^{\circ})}{sin14^{\circ}}=\dfrac{-6\cdot sin14^{\circ}}{sin14^{\circ}}=-6$

1 votes Thanks 2

ZlataSergeeva2000

Решение:

$4.~~\sqrt{32} \cdot cos^2\dfrac{7\pi}{8} -\sqrt{8} = \sqrt{8}\cdot( 2cos^2\dfrac{7\pi}{8}-1)=\sqrt{8} \cdot cos\dfrac{7\pi}{4} =\sqrt{8} \cdot \dfrac{\sqrt{2} }{2} =2;$

$5.~~7\sqrt{2} \cdot sin\dfrac{15\pi}{8}\cdot cos\dfrac{15\pi}{8} = 3.5\sqrt{2} \cdot sin\dfrac{15\pi}{4} = 3.5\sqrt{2} \cdot \Big (-\dfrac{\sqrt{2} }{2}\Big ) =-3.5;$

$6.~~~ \dfrac{-6sin~374^\circ}{sin~14^\circ}= \dfrac{-6sin~(360^\circ + 14^\circ)}{sin~14^\circ}= \dfrac{-6sin~ 14^\circ}{sin~14^\circ}= -6;$

0 votes Thanks 1

Answers & Comments

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