Question

Упростите выражения 23,4-23,5

Answer 1

$234\\\\1)1-Sin^{2}(270^{0}+\alpha)=1-Cos^{2}\alpha=\boxed{Sin^{2}\alpha}\\\\2)1-Cos^{2}(270^{0} -\alpha)=1-Sin^{2}\alpha=\boxed{Cos^{2}\alpha}\\\\3)1-Sin^{2}(360^{0}-\alpha)=1-Sin^{2}\alpha=\boxed{Cos^{2}\alpha}\\\\4)1-Cos^{2}(360^{0}+\alpha)=1-Cos^{2}\alpha=\boxed{Sin^{2}\alpha}$

$5)1+Ctg^{2}(270^{0}-\alpha)=1+tg^{2}\alpha=\boxed{\frac{1}{Cos^{2}\alpha}}\\\\6)1+tg^{2}(360^{0}-\alpha)=1+tg^{2}\alpha=\boxed{\frac{1}{Cos^{2}\alpha}}\\\\\\23.5\\\\1)Sin(90^{0}-\alpha)+Cos(180^{0}+\alpha)+Ctg(270^{0}-\alpha)+tg(360^{0}-\alpha)=\\\\=Cos\alpha -Cos\alpha+tg\alpha-tg\alpha=\boxed0\\\\2)Cos(90^{0}+\alpha)-Sin(180^{0}+\alpha)+Ctg(270^{0}+\alpha)+tg(360^{0}+\alpha)=\\\\=-Sin\alpha+Sin\alpha-tg\alpha+tg\alpha=\boxed{0}$

$3)Cos(\frac{3\pi }{2}+\alpha)-Sin(\pi+\alpha)+Ctg(\frac{\pi }{2}+\alpha)+tg(2\pi+\alpha)=\\\\=Sin\alpha+Sin\alpha-tg\alpha+tg\alpha=\boxed{2Sin\alpha} \\\\4)Sin(\frac{3\pi }{2}-\alpha)-Cos(\pi+\alpha)+tg(\frac{3\pi }{2}+\alpha)+Ctg(2\pi+\alpha)=\\\\=-Cos\alpha+Cos\alpha-Ctg\alpha+Ctg\alpha=\boxed0$