Question

6sin(-240°) · cos 315º - 4cos 135º · tg(-225°) · tg 120°

Answer 1

Для упрощения были использованы:

- Формулы приведения

- Свойства триг. функций

см. файл  

Answer 2

Let's calculate each part step by step:

1. sin(-240°): Since the sine function has a period of 360°, sin(-240°) is equal to sin(-240° + 360°), which is equal to sin(120°). The sine of 120° is √3/2.

2. cos(315°): The cosine function has a period of 360°, so cos(315°) is equal to cos(315° - 360°), which is equal to cos(-45°). The cosine of -45° is equal to √2/2.

3. cos(135°): The cosine of 135° is equal to -√2/2.

4. tg(-225°): The tangent function has a period of 180°, so tg(-225°) is equal to tg(-225° + 180°), which is equal to tg(-45°). The tangent of -45° is -1.

5. tg(120°): The tangent of 120° is √3.

Now, let's put it all together:

6sin(-240°) · cos 315º - 4cos 135º · tg(-225°) · tg 120°

= (6 * √3/2) * (√2/2) - (4 * (-√2/2)) * (-1) * √3

= (3√3/2) * (√2/2) - (2√2) * (√3)

= (3√6/4) - (2√6)

Now, let's simplify further:

= (3√6 - 8√6)/4

= (-5√6)/4

So, the final result is (-5√6)/4.