Задание №19 Т/Р №197 А. Ларина

Смотрите также №13; №14; №15; №16; №17; №18  Тренировочной работы №197 А. Ларина.

19. а) Найдите значение выражения $tg1^{\circ}\cdot tg2^{\circ}\cdot tg3^{\circ}\cdot …\cdot tg88^{\circ}\cdot tg89^{\circ}.$

б) Докажите, что $tg40^{\circ}+tg55^{\circ}+tg85^{\circ}=tg40^{\circ}\cdot tg55^{\circ}\cdot tg85^{\circ}$.

в) Найдите значение выражения $(1+tg1^{\circ})\cdot (1+tg2^{\circ})\cdot …\cdot (1+tg44^{\circ})$.

Решение:

a)

$tg1^{\circ}\cdot tg2^{\circ}\cdot tg3^{\circ}\cdot …\cdot tg88^{\circ}\cdot tg89^{\circ}=$

$=tg1^{\circ}\cdot tg2^{\circ}\cdot tg3^{\circ}\cdot …\cdot tg44^{\circ}\cdot tg45^{\circ}\cdot tg(90^{\circ}-44^{\circ})\cdot…\cdot tg(90^{\circ}-2^{\circ})\cdot tg(90^{\circ}-1^{\circ})=$

$=tg1^{\circ}\cdot tg2^{\circ}\cdot tg3^{\circ}\cdot …\cdot tg44^{\circ}\cdot 1\cdot ctg44^{\circ}\cdot…\cdot ctg2^{\circ}\cdot ctg1^{\circ}=$

$=(tg1^{\circ}\cdot ctg1^{\circ})\cdot (tg2^{\circ}\cdot ctg2^{\circ})\cdot  …\cdot (tg44^{\circ}\cdot ctg44^{\circ})=1$.

б) Покажем, что

$tg40^{\circ}+tg55^{\circ}+tg85^{\circ}-tg40^{\circ}\cdot tg55^{\circ}\cdot tg85^{\circ}=0.$

$tg40^{\circ}(1-tg55^{\circ}\cdot tg85^{\circ})+tg55^{\circ}+tg85^{\circ}=$

$=(1-tg55^{\circ}\cdot tg85^{\circ})(tg40^{\circ}+\frac{tg55^{\circ}+tg85^{\circ}}{1-tg55^{\circ}\cdot tg85^{\circ}})=$

$=(1-tg55^{\circ}\cdot tg85^{\circ})(tg40^{\circ}+tg140^{\circ})=$

$=(1-tg55^{\circ}\cdot tg85^{\circ})(tg40^{\circ}+tg(180^{\circ}-40^{\circ}))=$

$=(1-tg55^{\circ}\cdot tg85^{\circ})(tg40^{\circ}-tg40^{\circ})=0.$

Что и требовалось доказать.

в) 

$(1+tg1^{\circ})\cdot (1+tg2^{\circ})\cdot …\cdot (1+tg44^{\circ})=$

$\large=(\frac{cos1^{\circ}}{cos1^{\circ}}+\frac{sin1^{\circ}}{cos1^{\circ}})(\frac{cos2^{\circ}}{cos2^{\circ}}+\frac{sin2^{\circ}}{cos2^{\circ}})\cdot …\cdot (\frac{cos44^{\circ}}{cos44^{\circ}}+\frac{sin44^{\circ}}{cos44^{\circ}})=$

$\large=\frac{cos1^{\circ}+sin1^{\circ}}{cos1^{\circ}}\cdot \frac{cos2^{\circ}+sin2^{\circ}}{cos2^{\circ}}\cdot …\cdot\frac{cos44^{\circ}+sin44^{\circ}}{cos44^{\circ}}=$

$\large=\frac{\frac{\sqrt2}{2}cos1^{\circ}+\frac{\sqrt2}{2}sin1^{\circ}}{\frac{\sqrt2}{2}cos1^{\circ}}\cdot \frac{\frac{\sqrt2}{2}cos2^{\circ}+\frac{\sqrt2}{2}sin2^{\circ}}{\frac{\sqrt2}{2}cos2^{\circ}}\cdot …\cdot \frac{\frac{\sqrt2}{2}cos44^{\circ}+\frac{\sqrt2}{2}sin44^{\circ}}{\frac{\sqrt2}{2}cos44^{\circ}}=$

$\large=\frac{cos(1^{\circ}-45^{\circ})}{\frac{\sqrt2}{2}cos1^{\circ}}\cdot \frac{cos(2^{\circ}-45^{\circ})}{\frac{\sqrt2}{2}cos2^{\circ}}\cdot …\cdot \frac{cos(44^{\circ}-45^{\circ})}{\frac{\sqrt2}{2}cos44^{\circ}}=$

$\large=\frac{\sqrt2\cdot cos44^{\circ}}{cos1^{\circ}}\cdot \frac{\sqrt2\cdot cos43^{\circ}}{cos2^{\circ}}\cdot …\cdot \frac{\sqrt2\cdot cos1^{\circ}}{cos44^{\circ}}=(\sqrt2)^{44}=2^{22}.$

Ответ: а) $1$; б) $2^{22}.$