Question

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

Answer 1

Ответ: во вложении Объяснение:

Answer 2

:)

Answer 3

$3); ; tg2a+ctg2a=dfrac{sin2a}{cos2a}+dfrac{cos2a}{sin2a}=dfrac{sin^2a+cos^2a}{sin2acdot cos2a}=dfrac{1}{frac{1}{2}cdot sin4a}=dfrac{2}{sin4a}\\star ; ; 2sinxcdot cosx=sin2x; ; star$

$6); ; ctg2a-sin4a=dfrac{cos2a}{sin2a}-2, sin2acdot cos2a=dfrac{cos2a-2, sin^22acdot cos2a}{sin2a}=\\=dfrac{cos2acdot (1-2sin^22a)}{sin2a}=dfrac{cos2acdot cos4a}{sin2a}=ctg2acdot cos4a\\star ; ; 1-2sin^2x=cos2x; ; star$

$7); ; 1+cos(3pi +3a)cdot cos2a-cos(frac{3pi}{2}-3a)cdot sin2a=\\=1-cos3acdot cos2a+sin3acdot sin2a=1-(cos3acdot cos2a-sin3acdot sin2a)=\\=1-cos(3a+2a)=1-cos5a=2sin^2frac{5a}{2}=2sin^22,5a\\star; ; cosxcdot cosy-sinxcdot siny=cos(x+y); ; star \\star ; ; 1-cosx=2sin^2frac{x}{2}; ; star$

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

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

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

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