Question

Помогите пожалуйста! просто я на уроках не был и не понял задание 89,90

Answer 1

$\boxed {sin(arcsinx)=x\; \; (-1\leq x\leq 1)\; \; ,\; \; cos(arccosx)=x\; \; (-1\leq x\leq 1)}\\\\\boxed {tg(arctgx)=x\; \; (-\infty <x<+\infty )\; \; ,\; \; ctg(arctgx)=x\; \; (-\infty <x<+\infty )}\\\\89.\; \; sin(arcsin\frac{2}{3})=\frac{2}{3}\; \; ,\; \; cos(arccos\frac{3}{4})= \frac{3}{4}\\\\tg(arctg\frac{3}{2})=\frac{3}{2}\; \; ,\; \; ctg(arcctg\frac{5}{3})=\frac{5}{3}\; .$

$90.\; \; 1)\; \; 2arcsin(-\frac{1}{2})+arctg(-1)+\underbrace {arccos\frac{\sqrt2}{2}}_{cos\frac{\pi}{4}=\frac{\sqrt2}{2}}=\\\\=-2\underbrace {arcsin\frac{1}{2}}_{sin\frac{\pi}{6}=\frac{1}{2}}-\underbrace {arctg1}_{tg1=\frac{\pi}{4}}+\frac{\pi}{4}=-2\cdot \frac{\pi}{6}-\frac{\pi }{4}+\frac{\pi }{4}=-\frac{\pi }{3}\; ;\\\\\\2)\; \; arctg(-\frac{\sqrt3}{3})+arccos(-\frac{1}{2})+\underbrace {arcsin1}_{sin\frac{\pi}{2}=1}=\\\\=-\underbrace {arctg\frac{\sqrt3}{3}}_{tg\frac{\pi}{6}=\frac{\sqrt3}{3}}+(\pi -\underbrace {arccos\frac{1}{2}}_{cos\frac{\pi}{3}+\frac{1}{2}})+\frac{\pi }{2}=-\frac{\pi}{6}+(\pi -\frac{\pi }{3})+\frac{\pi }{2}=\pi \; ;$

$3)\; \; 3\, arcsin(-1)-\frac{3}{2}\, arccos(-\frac{\sqrt3}{2})-7,5\, arctg(-\frac{1}{\sqrt3})=\Big [\, \frac{1}{\sqrt3}=\frac{\sqrt3}{3}\m \Big ]=\\\\=3\, \cdot (-\underbrace {arcsin1}_{sin\frac{\pi}{2}=1})-\frac{3}{2}\, (\pi -\underbrace {arccos\frac{\sqrt3}{2}}_{cos\frac{\pi}{6}=\frac{\sqrt3}{2}})-7,5\, (-\underbrace {arctg\frac{\sqrt3}{3}}_{tg\frac{\sqrt3}{3}=\frac{\pi}{6}})=\\\\\\=-\frac{3\pi}{2}-\frac{3}{2}\cdot (\pi -\frac{\pi }{6})+\frac{15}{2}\cdot \frac{\pi }{6}=-\frac{3\pi }{2}-\frac{3}{2}\cdot \frac{5\pi }{6}+\frac{15\pi }{2\cdot 6}=-\frac{3\pi }{2}\; ;$

$4)\; \; 5\cdot arcsin(-\frac{\sqrt3}{2})+8\cdot arccos(-1)-6\cdot \underbrace {arcctg\frac{\sqrt3}{3}}_{ctg\frac{\pi}{3}=\frac{\pi}{3}}=\\\\=5\cdot (-\underbrace {arcsin\frac{\sqrt3}{2}}_{sin\frac{\pi}{3}=\frac{\sqrt3}{2}})-8\cdot \underbrace {arccos1}_{cos\, 0 =1}-6\cdot \frac{\pi}{3}=-\frac{5\pi }{3}-8\cdot 0-2\pi =-\frac{11\pi }{3}\; .$