Question

35 б. Помогите!!!Очень нужно и прошу!!!!Знайдіть значення виразу:а)arccos(-$\frac{\sqrt{3} }{2}$)+arcsin(-$\frac{\sqrt{3} }{2}$)
б)arccos$\frac{\sqrt{2} }{2}$-arcsin$\frac{\sqrt{3} }{2}$
в)arctg1-arctg(-1)
г)arcctg(-1)+arctg$\sqrt{3}$

Answer 1

$a)arccos(-\frac{\sqrt{3} }{2})+arcsin(-\frac{\sqrt{3} }{2})=\frac{5\pi}{6}-\frac{\pi}{3}=\frac{\pi}{2} \\ \\ b)arccos\frac{\sqrt{2} }{2}-arcsin\frac{\sqrt{3} }{2}=\frac{\pi}{4}-\frac{\pi}{3}=-\frac{\pi}{12} \\ \\ c)arctg1-arctg(-1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2} \\ \\ d)arcctg(-1)+arctg\sqrt{3}=-\frac{\pi}{4}+\frac{\pi}{3}=\frac{\pi}{12}$

Answer 2

$1)arcCos(-\frac{\sqrt{3} }{2})+arcSin(-\frac{\sqrt{3} }{2})=\frac{5\pi }{6}+(-\frac{\pi }{3})=\frac{5\pi }{6}-\frac{2\pi }{6} =\frac{3\pi }{6}=\frac{\pi }{2}\\\\2)arcCos\frac{\sqrt{2} }{2}-arcSin\frac{\sqrt{3} }{2}=\frac{\pi }{4}-\frac{\pi }{3}=\frac{3\pi-4\pi }{12}=-\frac{\pi }{12}$

$3)arctg1-arctg(-1)=\frac{\pi }{4}-(-\frac{\pi }{4})=\frac{\pi }{4}+\frac{\pi }{4}=\frac{2\pi }{4}=\frac{\pi }{2}\\\\4)arcCtg(-1)+arctg\sqrt{3} =\frac{3\pi }{4}+\frac{\pi }{3} = \frac{9\pi+4\pi}{12}=\frac{13\pi }{12}$