Question

найдите:
1) 4sin a - 6cos a=
если tg a=1
2) -20 корней из 3 * tg (-210)=
3) 5sin a= если cos a= 2 корня из 6 / 5 и a (3pi/2; 2pi)
4) 4sin (a-2pi) + 7cos ( -pi/2 + a)=
5) 4 корня из 2 * cos pi/3 * cos 9pi/4=
6) 60 / (sin (-32pi/3) * cos (25pi/6))=

Answer 1

1. $\dfrac{1}{\cos^2 x} = \mathrm{tg}^2 \: x + 1 = 2, \quad \cos x = \pm \dfrac{1}{\sqrt{2}}$
$4 \sin x - 6 \cos x = (4 \: \mathrm{tg} \: x - 6)\cos x = \pm (4 \: \mathrm{tg} \: x - 6)\cdot \dfrac{1}{\sqrt{2}} = \pm \sqrt{2}$

2. $-20\sqrt{3}\: \mathrm{tg} \: (-210^\circ) = -20\sqrt{3} \: \mathrm{tg}\: (-180^\circ - 30^\circ) = 20\sqrt{3} \: \mathrm{tg} \:30^\circ = 20.$

3. $5 \sin \alpha = -5\sqrt{1 - \cos^2 \alpha} = -5\sqrt{1 - \left(\dfrac{2\sqrt{6}}{5}\right)^2} = -5\sqrt{\dfrac{25-24}{25}} = -1.$

4. $4 \sin (\alpha - 2\pi) + 7\cos{(\alpha - \frac{\pi}{2})} = 4\sin \alpha + 7 \sin \alpha = 11 \sin \alpha.$

5.$4\sqrt{2} \cos {\dfrac{\pi}{3}} \cos {\dfrac{9\pi}{4}} = 4\sqrt{2} \cdot \dfrac{1}{2} \cdot \cos \dfrac{\pi}{4} = 2\sqrt{2} \cdot \dfrac{1}{\sqrt{2}} = 2.$

6. $\dfrac{60}{\sin{\frac{-32\pi}{3}}} \cdot \cos \frac{25\pi}{6} = \dfrac{60}{\sin \frac{-2\pi}{3}} \cos \frac{\pi}{6} = -\dfrac{60}{\sin \frac{\pi}{3}}\cdot \cos \frac{\pi}{6} = -60.$