Question

Допоможіть будь ласка, без вас ніяк

Answer 1

$\displaystyle\bf\\1)\\\\Cos\Big(\frac{3\pi }{2} +\alpha \Big)=Sin\alpha\\\\2)\\\\Cos\Big(\pi -\alpha \Big)=-Cos\alpha \\\\3)\\\\Sin\Big(180^\circ+\alpha \Big)=-Sin\alpha \\\\4)\\\\Sin^{2} \Big(\frac{5\pi }{2} +\alpha \Big)=Sin^{2} \Big[ 2\pi +\Big(\frac{\pi }{2} +\alpha \Big)\Big]=Sin^{2} \Big(\frac{\pi }{2} +\alpha \Big)=Cos^{2} \alpha$