Добрый день, помогите с решением тригонометрии! Кто сколько может. с 34 по 37 но...

July 2022 1 15 Report

Добрый день, помогите с решением тригонометрии! Кто сколько может. с 34 по 37 номер. Заранее спасибо. (решение фоткой, пожалуйста)

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iamversus

34.
$\frac{2cos40^\circ-sin70^\circ}{sin340^\circ}=\frac{2cos(60^\circ-20^\circ)-sin(90^\circ-20^\circ)}{sin(360^\circ-20^\circ)}=\\\\
=\frac{2(cos60^\circ\cdot cos20^\circ-sin60^\circ\cdot
sin20^\circ)-cos20^\circ}{-sin20^\circ}=\\\\ =\frac{cos20^\circ-2(\frac{1}{2}\cdot
cos20^\circ-\frac{\sqrt{3}}{2}\cdot sin20^\circ)}{sin20^\circ}=\\\\
=\frac{cos20^\circ-cos20^\circ+\sqrt{3}\cdot sin20^\circ)}{sin20^\circ}=\\\\
=\sqrt{3}$

Ответ: 1) $\sqrt3$ .

35.
$\frac{2sin170^\circ+cos40^\circ}{cos130^\circ}=\frac{2sin(180^\circ-10^\circ)+cos(30^\circ+10^\circ)}{cos(90^\circ+40^\circ)}=\\\\
=\frac{2sin10^\circ+cos(30^\circ+10^\circ)}{-sin40^\circ}=\\\\
=-\frac{2sin10^\circ+cos30^\circ\cdot cos10^\circ-sin30^\circ\cdot
sin10^\circ}{sin30^\circ\cdot cos10^\circ+sin10^\circ\cdot cos30^\circ}=\\\\
=-\frac{2sin10^\circ+\frac{\sqrt3}{2}\cdot cos10^\circ-\frac{1}{2}\cdot
sin10^\circ}{sin30^\circ\cdot cos10^\circ+sin10^\circ\cdot cos30^\circ}=$
$=-\frac{\frac{3}{2}sin10^\circ+\frac{\sqrt3}{2}\cdot
cos10^\circ}{\frac{1}{2}\cdot cos10^\circ+\frac{\sqrt3}{2}\cdot
sin10^\circ}=\\\\ =-\frac{3sin10^\circ+\sqrt3\cdot cos10^\circ}{\sqrt3\cdot
sin10^\circ+cos10^\circ}=\\\\ =-\sqrt3\cdot \frac{\sqrt3\cdot
sin10^\circ+cos10^\circ}{\sqrt3\cdot sin10^\circ+cos10^\circ}=\\\\ =-\sqrt3
$

Ответ: 1) $-\sqrt3$ .

36.
$\frac{1}{cos105^\circ}+\frac{\sqrt3}{sin285^\circ}=\frac{1}{cos(90^\circ+15^\circ)}+\frac{\sqrt3}{sin(360^\circ-75^\circ)}=\\\\
=-\frac{1}{sin15^\circ}-\frac{\sqrt3}{sin75^\circ}=\\\\
=-\frac{1}{sin15^\circ}-\frac{\sqrt3}{sin(90^\circ-15^\circ)}=\\\\
=-\frac{1}{sin15^\circ}-\frac{\sqrt3}{cos15^\circ}=\\\\
=-\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{sin15^\circ\cdot cos15^\circ}=\\\\
=-2\cdot\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{2\cdot sin15^\circ\cdot
cos15^\circ}$
$=-2\cdot\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{sin30^\circ}=\\\\
=-4\cdot(cos15^\circ+\sqrt3\cdot sin15^\circ)=\\\\ =-8\cdot(\frac{1}{2}\cdot
cos15^\circ+\frac{\sqrt3}{2}\cdot sin15^\circ)=\\\\ =-8\cdot(sin30^\circ\cdot
cos15^\circ+cos30^\circ\cdot sin15^\circ)=\\\\
=-8\cdot(sin(30^\circ+15^\circ)=-8\cdot sin45^\circ=\\\\ =-4\sqrt2$

Ответ: 2) $-4\sqrt2$ .

37.
$\frac{\sqrt2\cdot sin\frac{\pi}{8}-2\cdot cos\frac{5\pi}{8}}{sin\frac{5\pi}{8}}=\frac{\sqrt2\cdot sin\frac{\pi}{8}-2\cdot cos(\frac{\pi}{2}+\frac{\pi}{8})}{sin(\frac{\pi}{2}+\frac{\pi}{8})}= \\\\
=\frac{\sqrt2\cdot sin\frac{\pi}{8}+2\cdot sin\frac{\pi}{8}}{cos\frac{\pi}{8}}= \\\\
=(2+\sqrt2)\cdot \frac{sin\frac{\pi}{8}}{cos\frac{\pi}{8}}=(2+\sqrt2)\cdot tg\frac{\pi}{8} \\\\\\
tg\frac{\pi}{8}=\sqrt{\frac{1-cos\frac{\pi}{4}}{1+cos\frac{\pi}{4}}}=
\sqrt{\frac{1-\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}}=$
$=\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}=\sqrt2-1 \\\\\\
(2+\sqrt2)\cdot(\sqrt2-1)=\sqrt2$

Ответ: 5) $\sqrt2$ .

Если тех не читается, то во вложениях фото.

2 votes Thanks 1

bimc дауш)

kak muhi pitayutsya kristalikami saharapomagite plizochnuzhno

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bimc March 2022 | 0 Ответы

blagodarya chemu muhi hodyat po potolku i okonnomu steklu

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kak muhi pitayutsya kristalikami saharapomagite plizochnuzhno

blagodarya chemu muhi hodyat po potolku i okonnomu steklu

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