Question

Срочно нужна помощь!!!№26.16(3,6,7,8,9,10).

Answer 1

Ответ: во вложении Объяснение:

Answer 2

$3)\; \; tg2a+ctg2a=\dfrac{sin2a}{cos2a}+\dfrac{cos2a}{sin2a}=\dfrac{sin^2a+cos^2a}{sin2a\cdot cos2a}=\dfrac{1}{\frac{1}{2}\cdot sin4a}=\dfrac{2}{sin4a}\\\\\star \; \; 2sinx\cdot cosx=sin2x\; \; \star$

$6)\; \; ctg2a-sin4a=\dfrac{cos2a}{sin2a}-2\, sin2a\cdot cos2a=\dfrac{cos2a-2\, sin^22a\cdot cos2a}{sin2a}=\\\\=\dfrac{cos2a\cdot (1-2sin^22a)}{sin2a}=\dfrac{cos2a\cdot cos4a}{sin2a}=ctg2a\cdot cos4a\\\\\star \; \; 1-2sin^2x=cos2x\; \; \star$

$7)\; \; 1+cos(3\pi +3a)\cdot cos2a-cos(\frac{3\pi}{2}-3a)\cdot sin2a=\\\\=1-cos3a\cdot cos2a+sin3a\cdot sin2a=1-(cos3a\cdot cos2a-sin3a\cdot sin2a)=\\\\=1-cos(3a+2a)=1-cos5a=2sin^2\frac{5a}{2}=2sin^22,5a\\\\\star\; \; cosx\cdot cosy-sinx\cdot siny=cos(x+y)\; \; \star \\\\\star \; \; 1-cosx=2sin^2\frac{x}{2}\; \; \star$

$8)\; \; tg^4a\cdot \Big(8cos^2(\pi -a)-cos(\pi +4a)-1\Big)=\\\\=tg^4a\cdot (8cos^2a+cos4a-1)=tg^4a\cdot \Big (8cos^2a-(1-cos4a)\Big)=\\\\=tg^4a\cdot \Big(8cos^2a-2sin^22a\Big)=2\, tg^4a\cdot \Big(4cos^2a-(2sina\cdot cosa)^2\Big)=\\\\=2\, tg^4a\cdot \Big(4cos^2a-4sin^2a\cdot cos^2a\Big)=2\, tg^4a\cdot 4\, cos^2a\cdot (1-sin^2a)=\\\\=2\cdot \dfrac{sin^4a}{cos^4a}\cdot4\, cos^2a\cdot cos^2a=8\cdot sin^4a$

$9)\; \; \dfrac{1-2sin^2a}{2\, ctg(\frac{\pi}{4}-a)\cdot cos^2(\frac{\pi}{4}+a)}=\dfrac{cos2a}{2\, ctg(\frac{\pi}{2}-(\frac{\pi}{4}+a))\cdot cos^2(\frac{\pi}{4}+a)}=\\\\\star \; \; ctg(\frac{\pi}{2}-x)=tgx\; \; \star \\\\=\dfrac{cos2a}{2\, tg(\frac{\pi}{4}+a)\cdot cos^2(\frac{\pi}{4}+a)}=\dfrac{cos2a}{2\cdot \frac{sin(\pi/4+a)}{cos(\pi/4+a)}\cdot cos^2(\frac{\pi}{4}+a)}=\\\\=\dfrac{cos2a}{2\, sin(\frac{\pi}{4}+a)\cdot cos(\frac{\pi}{4}+a)}=\dfrac{cos2a}{sin(\frac{\pi}{2}+2a)}=\dfrac{cos2a}{cos2a}=1$

$10)\; \; 2\cdot tg\frac{a}{2}\cdot (tga+ctga)\cdot (1-tg^2\frac{a}{2})=\\\\=2\cdot tg\frac{a}{2}\cdot \Big(\dfrac{sina}{cosa}+\dfrac{cosa}{sina}\Big)\cdot \Big(1-\dfrac{sin^2\frac{a}{a}}{cos^2\frac{a}{2}}\Big)=\\\\\\=2\cdot \dfrac{sin\frac{a}{2}}{cos\frac{a}{2}}\cdot \dfrac{sin^2a+cos^2a}{sina\cdot cosa}\cdot \dfrac{cos^2\frac{a}{2}-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=\\\\\\\star \; \; cos^2x-sin^2x=cos2x\; \; ,\; \; 2\, sinx\cdot cosx=sin2x\; \; \star$

$=2\cdot \dfrac{sin\frac{a}{2}}{cos\frac{a}{2}}\cdot \dfrac{1}{2\, sin\frac{a}{2}\cdot cos\frac{a}{2}\cdot cosa}\cdot \dfrac{cosa}{cos^2\frac{a}{2}}=\dfrac{2\cdot sin\frac{a}{2}\cdot cosa}{2\cdot sin\frac{a}{2}\cdot cos^4\frac{a}{2}\cdot cosa}=\\\\\\=\dfrac{1}{cos^4\frac{a}{2}}$