sinπ/6-4cosπ/6+tgπ/6-5ctgπ/3. Created by jeremiahvaleska. algebra-ru

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jeremiahvaleska @jeremiahvaleska

August 2021 1 27 Report

sinπ/6-4cosπ/6+tgπ/6-5ctgπ/3

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Miroslava227

Ответ:

$\sin( \frac{\pi}{6} ) - 4\cos( \frac{\pi}{6} ) + tg( \frac{\pi}{6} ) - 5ctg( \frac{\pi}{3} ) = \\ = \frac{1}{2} - 4 \times \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{3} - \frac{5 \sqrt{3} }{3} = \\ = \frac{1}{2} - 2 \sqrt{3} - \frac{4 \sqrt{3} }{3} = \frac{3 - 12 \sqrt{3} - 8 \sqrt{3} }{6} = \frac{3 - 20 \sqrt{3} }{6}$

$\cos( \frac{\pi}{2} ) + 9 \sin( \frac{\pi}{2} ) - ctg( \frac{\pi}{4} ) - 7tg(0^{\circ} ) = \\ = 0 + 9 - 1 - 0 = 8$

$\sqrt{3} \cos( \frac{\pi}{6} ) + 2ctg( \frac{\pi}{3} ) - 11ctg( \frac{\pi}{4} ) = \\ = \sqrt{3} \times \frac{ \sqrt{3} }{2} + 2 \times \frac{ \sqrt{3} }{3} - 11 = 1.5 + \frac{2 \sqrt{3} }{3} - 11 = \frac{2 \sqrt{3} }{3} - 9.5$

$ctg( \frac{\pi}{2} ) - 5 \sin( \frac{\pi}{3} ) + 6 \cos( \frac{\pi}{3} ) - tg( \frac{\pi}{6} ) = \\ = 0 - 5 \times \frac{ \sqrt{3} }{2} + 6 \times \frac{1}{2} - \sqrt{3} = \\ = \frac{ - 5 \sqrt{3} + 6 - 2 \sqrt{3} }{2} = \frac{6 - 7 \sqrt{3} }{2}$

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jeremiahvaleska August 2022 | 0 Ответы

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at work) I come home at... (about six o'clock) I go to bed at... I go to English classes on... (Tuesday) I study a lot...

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