Question

Хотя бы часть. поставить знак (< 0 или > 0) sin (-23п/9) cos1 tg(x-1) 2.вычислить значение tg 315* (* - градусы) tg 5п/6 tg 5п/4 sin 10п/3 sin105*sin75* cos135*cos105* (sin75*-cos75*)/(cos15*+sin15*)

Answer 1

1)Sin (-23п/9)>0
Cos1>0
Tg(x-1) зависит от х

2)а) $tg(315`) = tg(360`-45`) = -tg45` = -1$

б) $tg(\frac{5\pi}{6})=tg(\pi-\frac{\pi}{6}) = -tg\frac{\pi}{6}=-\frac{1}{\sqrt{3}}$

в) $tg\frac{5\pi}{4} = tg(\pi+\frac{\pi}{4}) = tg\frac{\pi}{4} = 1$

г) $sin\frac{10\pi}{3} = sin(3\pi+\frac{\pi}{3})} = -sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}$

д) $sin105`sin75`=0.5(cos(105`-75`)-cos(105`+75`)) = 0.5(cos30`-cos180`) = 0.5(\frac{\sqrt{3}}{2}+1) = \frac{2+\sqrt{3}}{4}$

е) $cos135`cos105` = 0.5(cos240`+cos30`) = 0.5(cos(180`+60`)+cos30`) = 0.5(cos30`-cos60`) = \frac{1}{2}(\frac{\sqrt{3}}{2}-\frac{1}{2})=\frac{\sqrt{3}-1}{4}$

ж) $\frac{sin75`-cos75`}{cos15`+sin15`}=\frac{\sqrt{2}sin(75`-45`)}{\sqrt{2}sin(15`+45`)}=\frac{sin30`}{sin60`}=\frac{1}{2}:\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}$