426(1,3) 427(1,3) 428(1,3) 429(1,3,5)

Question 1

426(1,3)
427(1,3)
428(1,3)
429(1,3,5)

Question 2

Решите задачу:

$426.\; \; 1)\; ctgx+\frac{sinx}{1+cosx}= \frac{cosx}{sinx} + \frac{2sin\frac{x}{2}\cdot cos\frac{x}{2}}{2cos^2\frac{x}{2}} = \\\\=\frac{2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}\cdot cos\frac{x}{2}} +\frac{sin\frac{x}{2}}{cos\frac{x}{2}}= \frac{2cos^2\frac{x}{2}-1+2sin^2\frac{x}{2}}{2sin\frac{x}{2}\cdot cos\frac{x}{2}} =\frac{2-1}{sinx}=\frac{1}{sinx} \\\\3)\; \; \frac{1-sin^2x}{1-cos^2x}=\frac{cos^2x}{sin^2x} =ctg^2x=\frac{1}{tg^2x}$

$427.\; \; 1)\; y=1-(cos^2x-sin^2x)=1-cos2x\\\\-1 \leq cos2x \leq 1\; \; \; \; \to \; \; \; \; -1 \leq -cos2x \leq 1\\\\-1+1 \leq 1-cos2x \leq 1+1\\\\0 \leq 1-cos2x \leq 2\; \; \to \; \; \; y_{naibol.}=2\\\\2)\; \; y=cos^2a\cdot tg^2a+5cos^2a-1=sin^2a+5cos^2a-1=\\\\=(1-cos^2a)+5cos^2a-1=4cos^2a\\\\-1 \leq cosa \leq 1\\\\0 \leq cos^2a \leq 1\; \; \; \to \; \; \; 0 \leq 4cos^2a \leq 4\\\\y_{naibol.}=4$

$428.\; \; 1)\; 1+sin \frac{\pi}{6}+sin^2 \frac{\pi}{6}+sin^3\frac{\pi}{6} =1+ \frac{1}{2}+\frac{1}{4} +\frac{1}{8}=\\\\= \frac{3}{2}+\frac{3}{8}=\frac{15}{8}\\\\3)\; \; 1-tg \frac{\pi}{6}+tg^2\frac{\pi}{6}-tg^3\frac{\pi}{6} =1-\frac{1}{\sqrt3}+\frac{1}{3}-\frac{1}{3\sqrt3}=\\\\=\frac{4}{3}-\frac{4}{3\sqrt3}= \frac{4\sqrt3-4}{3\sqrt3} =\frac{4(\sqrt3-1)}{3\sqrt3}= \frac{4\sqrt3(\sqrt3-1)}{9}$

$429.\; \; 1)\; \frac{cosx}{1+sinx} +tgx= \frac{cosx}{1+sinx} +\frac{sinx}{cosx} =\\\\=\frac{cos^2x+sinx+sin^2x}{1+sinx} = \frac{1+sinx}{1+sinx} =1\\\\3)\; \; \frac{1-sin^2x}{1-cos^2x} +tgx\cdot ctgx= \frac{cos^2x}{sin^2x} +1=ctg^2x+1=\frac{1}{sin^2x}\\\\5)\; \; (ctga+tga)^2-(ctga-tga)^2=\\\\=(ctga+tga-(ctga-tga))\cdot (ctga+tga+(ctga-tga))=\\\\=2tga\cdot 2ctga=4tga\cdot ctga=4\cdot 1 =4$