Question

Тригонометрия. Помогите решить, пожалуйста

Answer 1

1. 0-(-1)=1

2. 0+1=1

3.

$\frac{3}{2} + \frac{2 \sqrt{3} }{2} - \sqrt{3} = 1.5$

4.

$(2 \times \frac{ \sqrt{3} }{3} - \sqrt{3} ) \div \frac{ \sqrt{3} }{2} = - \frac{ \sqrt{3} }{3} \times \frac{2}{ \sqrt{3} } = \frac{ - 2}{3}$

5.

$\frac{2 \times 3}{4} - \frac{3}{4} + \frac{ \sqrt{3} }{3} \times \frac{ \sqrt{3} }{3} = \frac{13}{12}$

Answer 2

Ответ:

$sin\pi -cos\pi =0-(-1)=1\\\\\\sin0+cos2\pi =0+1=1\\\\\\3sin\dfrac{\pi}{6}+2cos\dfrac{\pi}{6}-tg\dfrac{\pi}{3}=3\cdot \dfrac{1}{2}+2\cdot \dfrac{\sqrt3}{2}-\sqrt3=\dfrac{3}{2}=1,5\\\\\\\Big(2tg\dfrac{\pi}{6}-tg\dfrac{\pi}{3}\Big):cos\dfrac{\pi}{6}=\Big(2\cdot \dfrac{\sqrt3}{3}-\sqrt3\Big):\dfrac{\sqrt3}{2}=\dfrac{2\sqrt3-3\sqrt3}{3}\cdot \dfrac{2}{\sqrt3}=\\\\=-\dfrac{\sqrt3}{3}\cdot \dfrac{2}{\sqrt3}=-\dfrac{2}{3}$

$2cos^2\dfrac{\pi}{6}-sin^2\dfrac{\pi}{3}+tg\dfrac{\pi}{6}\cdot ctg\dfrac{\pi}{3}=2\cdot \dfrac{3}{4}-\dfrac{3}{4}+\dfrac{\sqrt3}{3}\cdot \dfrac{\sqrt3}{3}=\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{13}{12}=1\dfrac{1}{12}$