Question

Помогиите(Sin 600°+tg 480°)*Cos (-330°)

Answer 1

= (sin(540°+60°)+ tg(450+30))cos(360-30)=(-sin60-ctg30)cos30=-1,5

Answer 2

I can help you with this math problem. Here is the detailed solution:

First, we need to simplify the trigonometric functions using the periodicity and symmetry properties. We have:

$$\sin 600^\circ = \sin (360^\circ + 240^\circ) = -\sin 240^\circ = -\frac{\sqrt{3}}{2}$$

$$\tan 480^\circ = \tan (360^\circ + 120^\circ) = -\tan 120^\circ = \frac{\sqrt{3}}{3}$$

$$\cos (-330^\circ) = \cos (360^\circ - 330^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$$

Then, we can substitute these values into the original expression and simplify further. We get:

$$(\sin 600^\circ \tan 480^\circ) \cos (-330^\circ) = \left(-\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{3}\right) \cdot \frac{\sqrt{3}}{2} = -\frac{1}{4}$$

Therefore, the final answer is $$-\frac{1}{4}$$.

I hope this helps you understand the problem better. If you have any other questions, feel free to ask me.Ответь+подробно.+Помогиите(Sin+600°+tg+480°)*Cos+(-330°)