Упр.1171 ГДЗ Никольский Потапов 9 класс (Алгебра)

Задача

а) 2sin(?/6)+3cos(?/6)-2tg(?/6)-1,5ctg(?/6)+3/sin(?/6)+1/cos(?/6);
б) 2sin(45°)+2cos(45°)-3tg(45°)-10ctg(45°)+2/sin(45°)-4/cos(45°);
в) sin(?/3)+cos(?/3)-2tg(?/3)-1,5ctg(?/3)+1/cos(?/3)+3/sin(?/3);
г) sin(?/2)-6cos(?/2)+3tg(0)+5/cos(0);
д) 2sin(180°)-3cos(180°)+4tg(180°)+2/cos(180°);
е) 3sin(3?/2)-4cos(3?/2)+5ctg(3?/2)+1/sin(3?/2);
ж) (sin(?/4)cos(?/3)-cos(?/4)sin(-?/3))/(tg^2(?/4)-sin^2(?/2)cos^4(3?/2));
з) (1/cos(-30°)+1/sin(-30°))/(sin^2(-60°)+1/sin(-30°)).

Подробный ответ

а) Подставим значения тригонометрических функций:
$$ 2\sin\frac{\pi}{6}+3\cos\frac{\pi}{6}-2\tg\frac{\pi}{6}-1{,}5\ctg\frac{\pi}{6}+\frac{3}{\sin\frac{\pi}{6}}+\frac{1}{\cos\frac{\pi}{6}} $$
$$ =2\cdot\frac12+3\cdot\frac{\sqrt3}{2}-2\cdot\frac{\sqrt3}{3}-1{,}5\cdot\sqrt3+\frac{3}{1/2}+\frac{1}{\sqrt3/2} $$
$$ =1+\frac{3\sqrt3}{2}-\frac{2\sqrt3}{3}-\frac{3\sqrt3}{2}+6+\frac{2}{\sqrt3}=7. $$
б)
$$ 2\sin45^\circ+2\cos45^\circ-3\tg45^\circ-10\ctg45^\circ+\frac{2}{\sin45^\circ}-\frac{4}{\cos45^\circ} $$
$$ =2\cdot\frac{\sqrt2}{2}+2\cdot\frac{\sqrt2}{2}-3\cdot1-10\cdot1+\frac{2}{\sqrt2/2}-\frac{4}{\sqrt2/2} $$
$$ =\sqrt2+\sqrt2-3-10+2\sqrt2-4\sqrt2=-13. $$
в)
$$ \sin\frac{\pi}{3}+\cos\frac{\pi}{3}-2\tg\frac{\pi}{3}-1{,}5\ctg\frac{\pi}{3}+\frac{1}{\cos\frac{\pi}{3}}+\frac{3}{\sin\frac{\pi}{3}} $$
$$ =\frac{\sqrt3}{2}+\frac12-2\sqrt3-1{,}5\cdot\frac{\sqrt3}{3}+\frac{1}{1/2}+\frac{3}{\sqrt3/2} $$
$$ =\frac{\sqrt3}{2}+\frac12-2\sqrt3-\frac{\sqrt3}{2}+2+2\sqrt3=2{,}5. $$
г)
$$ \sin\frac{\pi}{2}-6\cos\frac{\pi}{2}+3\tg0+\frac{5}{\cos0} $$
$$ =1-6\cdot0+3\cdot0+\frac{5}{1}=6. $$
д)
$$ 2\sin180^\circ-3\cos180^\circ+4\tg180^\circ+\frac{2}{\cos180^\circ} $$
$$ =2\cdot0-3\cdot(-1)+4\cdot0+\frac{2}{-1}=1. $$
е)
$$ 3\sin\frac{3\pi}{2}-4\cos\frac{3\pi}{2}+5\ctg\frac{3\pi}{2}+\frac{1}{\sin\frac{3\pi}{2}} $$
$$ =3\cdot(-1)-4\cdot0+5\cdot0+\frac{1}{-1}=-4. $$
ж)
$$ \frac{\sin\frac{\pi}{4}\cos\frac{\pi}{3}-\cos\frac{\pi}{4}\sin\left(-\frac{\pi}{3}\right)} {\tg^2\frac{\pi}{4}-\sin^2\frac{\pi}{2}\cos^4\frac{3\pi}{2}} $$
$$ =\frac{\frac{\sqrt2}{2}\cdot\frac12-\frac{\sqrt2}{2}\cdot\left(-\frac{\sqrt3}{2}\right)} {1^2-1^2\cdot0^4} =\frac{\frac{\sqrt2}{4}+\frac{\sqrt6}{4}}{1} =\frac{\sqrt2+\sqrt6}{4}. $$
з)
$$ \frac{\frac{1}{\cos(-30^\circ)}+\frac{1}{\sin(-30^\circ)}} {\sin^2(-60^\circ)+\frac{1}{\sin(-30^\circ)}} $$
$$ =\frac{\frac{1}{\sqrt3/2}+\frac{1}{-1/2}} {\left(-\frac{\sqrt3}{2}\right)^2+\frac{1}{-1/2}} =\frac{\frac{2}{\sqrt3}-2}{\frac34-2} $$
$$ =\frac{\frac{2-2\sqrt3}{\sqrt3}}{-\frac54} =\frac{24-8\sqrt3}{15}. $$