ПОМОГИТЕ РЕШИТЬ!!!!
1)
sin(3/2п-п/6) ·cos(п/2+п/6)·tg(п+п/4)
2)
cos(п+п/6) ·tg(3/2п+п/6)·ctg(п+п/3)
3)
sin(п+п/6)·tg(п/2+п/4)·sin(п/2+п/3)
4)
ctg(3/2п+п/6)·sin(2п-п/4)·cos(п/2+п/4)
1)
sin(3/2п-п/6) ·cos(п/2+п/6)·tg(п+п/4)
2)
cos(п+п/6) ·tg(3/2п+п/6)·ctg(п+п/3)
3)
sin(п+п/6)·tg(п/2+п/4)·sin(п/2+п/3)
4)
ctg(3/2п+п/6)·sin(2п-п/4)·cos(п/2+п/4)
Answers & Comments
Verified answer
[tex]\displaystyle\bf\\1)\\\\Sin\Big(\frac{3\pi }{2} -\frac{\pi }{6} \Big)\cdot Cos\Big(\frac{\pi }{2} +\frac{\pi }{6} \Big)\cdot tg\Big(\pi +\frac{\pi }{4} \Big)=\\\\\\=-Cos\frac{\pi }{6} \cdot \Big(-Sin\frac{\pi }{6} \Big)\cdot tg\frac{\pi }{4} =\frac{\sqrt{3} }{2} \cdot \frac{1}{2} \cdot 1=\frac{\sqrt{3} }{4} \\\\\\2)\\\\Cos\Big(\pi+\frac{\pi }{6} \Big)\cdot tg\Big(\frac{3\pi }{2} +\frac{\pi }{6} \Big)\cdot Ctg\Big(\pi +\frac{\pi }{3} \Big)=[/tex]
[tex]\displaystyle\bf\\=-Cos\frac{\pi }{6}\cdot\Big(-Ctg\frac{\pi }{6} \Big)\cdot Ctg\frac{\pi }{3} = \frac{\sqrt{3} }{2} \cdot \sqrt{3} \cdot \frac{1}{\sqrt{3} } =\frac{\sqrt{3} }{2} \\\\\\3)\\\\Sin\Big(\pi +\frac{\pi }{6} \Big)\cdot tg\Big(\frac{\pi }{2} +\frac{\pi }{4}\Big)\cdot Sin\Big(\frac{\pi }{2} +\frac{\pi }{3} \Big)=\\\\\\=-Sin\frac{\pi }{6} \cdot\Big(-Ctg\frac{\pi }{4}\Big)\cdot Cos\frac{\pi }{3} =\frac{1}{2} \cdot 1\cdot\frac{1}{2} =\frac{1}{4} \\\\\\4)[/tex]
[tex]\displaystyle\bf\\Ctg\Big(\frac{3\pi }{2} +\dfrac{\pi }{6} \Big)\cdot Sin\Big(2\pi -\dfrac{\pi }{4} \Big)\cdot Cos\Big(\frac{\pi }{2} +\frac{\pi }{4} \Big)=\\\\\\=-tg\frac{\pi }{6} \cdot \Big(-Sin\frac{\pi }{4} \Big)\cdot \Big(-Sin\frac{\pi }{4}\Big) =-\frac{1}{\sqrt{3} } \cdot \frac{\sqrt{2} }{2}\cdot \frac{\sqrt{2} }{2} =-\frac{1}{2\sqrt{3} }[/tex]